the problems in this booklet are organized into strands. a...

The problems in this booklet are organized into strands. A problem often appears in multiple strands. The problems are suitable for most students in Grade 7 or higher.

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Problem of the Week

Problem C

Retiring and Hiring

A small technology company is beginning to expand. The company currentlyhas 100 employees. At the end of each year for the next 4 years, 25 employeeswill retire and a new employee will be hired for each of the remainingemployees.

After this four year cycle of retiring and hiring is complete, determine theaverage number of employees that the company grows by each year.

Extension: Using your results and assuming the pattern continues, predictthe number of employees if this cycle of retiring and hiring were to continue forten years.


Problem of the Week

Problem C and Solution

Retiring and Hiring

Problem

A small technology company is beginning to expand. The company currently has 100employees. At the end of each year for the next 4 years, 25 employees will retire and a newemployee will be hired for each of the remaining employees.

After this four year cycle of retiring and hiring is complete, determine the average number ofemployees that the company grows by each year.

Solution

The following chart will look at the process of retiring and hiring over the four year period.The number remaining will be 25 less than the number at the start of the year. The numberhired in a year will be the same as the number remaining. The number of employees at the endof the year will be twice as many as the number of employees remaining.

Year # of Employees # Retiring # Remaining # Hired # of Employeesat Start of Year at End of Year

1 100 25 75 75 1502 150 25 125 125 2503 250 25 225 225 4504 450 25 425 425 850

After four years of retiring and hiring the company employs 850 people. The number ofemployees increased by 850− 100 = 750 people in four years. The average increase per yearwas 750÷ 4 = 187.5 employees.

Another way to look at this problem would be to take the average number of people hired peryear and subtract the average number of people retiring in a year. The average number hiredper year is (75 + 125 + 225 + 425)÷ 4 = 850÷ 4 = 212.5. The average retiring each year was25. Therefore the average increase per year was 212.5− 25 = 187.5, as above.

Therefore the number of employees increased by approximately 188 per year.

Solution to Extension

Observe a couple of patterns in the column labelled “# of Employees at End of Year”. First,each number ends in 50. The leading digits are 1, 2, 4, and 8. These digits are powers of 2. Inyear 2, the leading digit is 21. In year 3, the leading digit is 22. In year 4, the leading digit is23. The exponent appears to be one less than the year number. So in year 10, a goodprediction for the leading digits would be 29 = 512. After 10 years of retiring and hiring, therewould be 51 250 employees. (We can verify this prediction by continuing the table.)


Problem of the Week

Problem C

Who Brought What When?

Xavier, Yvonne and Zak arrived at school at three different times. They eachbrought one of their favourite snacks to share with the other two (one broughtpretzels; one brought cookies; one brought licorice), and their favourite sportsapparatus (one brought a baseball; one brought a soccer ball; one brought afootball).

We know a few other facts:

1. The first to arrive did not bring cookies.

2. Xavier arrived second and brought a football.

3. Yvonne arrived before Zak.

4. The person who brought cookies also brought a baseball.

5. The person who brought pretzels did not bring a soccer ball.

Determine the order they arrived in, what they each brought for a snack, andwhich sports apparatus they each brought.


Problem of the Week



Problem

Xavier, Yvonne and Zak arrived at school at three different times. They each brought one oftheir favourite snacks to share with the other two (one brought pretzels; one brought cookies;one brought licorice), and their favourite sports apparatus (one brought a baseball; onebrought a soccer ball; one brought a football).







Determine the order they arrived in, what they each brought for a snack, and which sportsapparatus they each brought.

Solution

When solving logic problems, setting up a chart to fill in is generally a good way to start.

Order of Arrival Snack Sports ApparatusXavierYvonne

Zak

Some of the given information is often more helpful than other information. For example, inthe second statement we learn that Xavier arrived second and brought a football. Now thethird statement gets us the fact that Yvonne arrived first and Zak arrived third. (This is truesince Yvonne arrived before Zak and she could not arrive second leaving only the first andthird spots left.) We can add this information to the chart.

Order of Arrival Snack Sports ApparatusXavier 2nd footballYvonne 1st

Zak 3rd

We can combine the first statement and the fourth statement. Yvonne did not bring cookies.The person who brought cookies also brought a baseball. This cannot be Xavier since hebrought a football. Therefore, Zak brought cookies and a baseball. Since there is only onepiece of sports apparatus unaccounted for, Yvonne must have brought the soccer ball. We willadd this new information to our chart.


Order of Arrival Snack Sports ApparatusXavier 2nd footballYvonne 1st soccer ball

Zak 3rd cookies baseball

We can now use the fifth statement to conclude that Xavier brought pretzels since the personbringing pretzels did not bring a soccer ball and Xavier is the only one without a snackaccounted for other than Yvonne (who brought the soccer ball).

Order of Arrival Snack Sports ApparatusXavier 2nd pretzels footballYvonne 1st soccer ball


The only snack unaccounted for is the licorice and Yvonne is the only one whose snack isunknown. Therefore, Yvonne brought licorice and our chart can be completed.

Order of Arrival Snack Sports ApparatusXavier 2nd pretzels footballYvonne 1st licorice soccer ball


The information is summarized in the chart but will be stated below for completeness.

• Yvonne arrived first bringing licorice and a soccer ball.

• Xavier arrived second bringing pretzels and a football.

• Zak arrived third bringing cookies and a baseball.


Problem of the Week

Problem C

Quick Change

A die has the numbers 1, 2, 3, 4, 6, and 8 on its six faces. When this die isrolled, if an odd number appears on the top face, all of the odd numbers on thedie magically double. For example, if the number appearing on the top facewas a 1, the 1 and 3 on the original die would double and the other fournumbers would remain the same. (This is illustrated below.)

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However, if an even number appears on the top face as a result of the first roll,all of the even numbers on the die are halved. For example, if the numberappearing on the top face was an 8, the 2, 4, 6, and 8 on the original die wouldchange to half of their initial value and the other two numbers would remainthe same.

Suppose the die with 1, 2, 3, 4, 6, and 8 on its six faces is rolled once andchanges as described above. The changed die is rolled again. No change occurswhen the die is rolled this time. What is the probability that the numberappearing on the top face after this roll is a 2?


Problem of the Week

Problem C and Solutions

Quick Change

Problem

A die has the numbers 1, 2, 3, 4, 6, and 8 on its six faces. When this die is rolled, if an oddnumber appears on the top face, all of the odd numbers on the die magically double. Forexample, if the number appearing on the top face was a 1, the 1 and 3 on the original diewould double and the other four numbers would remain the same. However, if an even numberappears on the top face as a result of the first roll, all of the even numbers on the die arehalved. For example, if the number appearing on the top face was an 8, the 2, 4, 6, and 8 on theoriginal die would change to half of their initial value and the other two numbers would remainthe same. Suppose the die with 1, 2, 3, 4, 6, and 8 on its six faces is rolled once and changes asdescribed above. The changed die is rolled again. No change occurs when the die is rolled thistime. What is the probability that the number appearing on the top face after this roll is a 2?

Solution 1

In this solution, we will determine the possibilities for the first and second roll to count thetotal number of possible outcomes. We will then count the number of outcomes in which thesecond roll is a 2 and determine the probability.

If the first roll is odd, the numbers on the die change from {1, 2, 3, 4, 6, 8} to {2, 2, 6, 4, 6, 8}as a result of doubling the odd numbers. If we write the possible first and second rolls as anordered pair, then the following 12 combinations are possible:

First roll 1: (1, 2), (1, 2), (1, 6), (1, 4), (1, 6), (1, 8)First roll 3: (3, 2), (3, 2), (3, 6), (3, 4), (3, 6), (3, 8)

If the first roll is even, the numbers on the die change from {1, 2, 3, 4, 6, 8} to{1, 1, 3, 2, 3, 4} as a result of halving the even numbers. If we write the possible first andsecond rolls as an ordered pair, then the following 24 combinations are possible:

First roll 2: (2, 1), (2, 1), (2, 3), (2, 2), (2, 3), (2, 4)First roll 4: (4, 1), (4, 1), (4, 3), (4, 2), (4, 3), (4, 4)First roll 6: (6, 1), (6, 1), (6, 3), (6, 2), (6, 3), (6, 4)First roll 8: (8, 1), (8, 1), (8, 3), (8, 2), (8, 3), (8, 4)

There are 36 total possible outcomes. Of these outcomes, 8 have a second roll of 2. The

probability of rolling a 2 on the second roll is8

36=

2

9.


Problem

A die has the numbers 1, 2, 3, 4, 6, and 8 on its six faces. When this die is rolled, if an oddnumber appears on the top face, all of the odd numbers on the die magically double. Forexample, if the number appearing on the top face was a 1, the 1 and 3 on the original diewould double and the other four numbers would remain the same. However, if an even numberappears on the top face as a result of the first roll, all of the even numbers on the die arehalved. For example, if the number appearing on the top face was an 8, the 2, 4, 6, and 8 on theoriginal die would change to half of their initial value and the other two numbers would remainthe same. Suppose the die with 1, 2, 3, 4, 6, and 8 on its six faces is rolled once and changes asdescribed above. The changed die is rolled again. No change occurs when the die is rolled thistime. What is the probability that the number appearing on the top face after this roll is a 2?

Solution 2

In this solution, we solve the problem in a more theoretical manner using two known resultsfrom probability theory.

1. Two events are independent if the outcome of one event does not effect the outcome ofthe other. If two events are independent, then the probability of both events happeningis the product of their two probabilities.

2. Two events are disjoint if they cannot happen at the same time. The probability of twodisjoint events is the sum of their individual probabilities.

There is a2

6=

1

3chance of getting an odd number on the first roll. The die then changes from

{1, 2, 3, 4, 6, 8} to {2, 2, 6, 4, 6, 8} as a result of doubling the odd numbers. For each of the

possible odd rolls, there is now a2

6=

1

3chance of rolling a 2 on the second roll. Since the two

events are independent, we would multiply the two probabilities to obtain1

3× 1

3=

1

9. This is

the probability of rolling a 2 on the second roll when the first roll is odd.

There is a4

6=

2

3chance of getting an even number on the first roll. The die then changes from

{1, 2, 3, 4, 6, 8} to {1, 1, 3, 2, 3, 4} as a result of halving the even numbers. For each of the

possible even rolls, there is now a1

6chance of rolling a 2 on the second roll. Since the two

events are independent, we would multiply the two probabilities to obtain2

3× 1

6=

2

18=

1

9.

This is the probability of rolling a 2 on the second roll when the first roll is even.

Since the two cases are disjoint we add the probabilities to obtain1

9+

1

9=

2

9. The probability

of rolling a 2 on the second roll is2

9.


Problem of the Week

Problem C

Circle-Go-Round

A circle with centre O is drawn around 4OBD so that B and D lie on thecircumference of the circle. BO is extended to A on the circle. Chord ACintersects OD and BD at F and E, respectively.

If ∠BAC = 19◦ and ∠OFA = 99◦, determine the measure of ∠BEC.

(∠BEC is marked x◦ on the diagram.)

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Problem of the Week


Circle-Go-Round

Problem

A circle with centre O is drawn around 4OBD sothat B and D lie on the circumference of the circle.BO is extended to A on the circle. Chord AC in-tersects OD and BD at F and E, respectively. If∠BAC = 19◦ and ∠OFA = 99◦, determine the mea-sure of ∠BEC. (∠BEC is marked x◦ on the diagram.)

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Solution 1

In a triangle, the angles add to 180◦. So in 4OFA,

∠FOA = 180◦ − 99◦ − 19◦ = 62◦ = a.

BOA is a diameter and is therefore a straight line. Two angles on a straightline add to 180◦, so

∠BOD = 180◦ − a◦ = 180◦ − 62◦ = 118◦ = b.

O is the centre of the circle with B and D on the circumference. Therefore,OD and OB are radii of the circle and OD = OB. It follows that 4ODB isisosceles and ∠ODB = ∠OBD = c.

Then in 4ODB, c◦ + c◦ + b◦ = 180◦

2c+ 118 = 180

2c = 62

c = 31

Opposite angles are equal so it follows that ∠CEB = ∠DEF = x◦ and∠DFE = ∠AFO = 99◦ = d.

In 4FED, x◦ + c◦ + d◦ = 180◦

x+ 31 + 99 = 180

x+ 130 = 180

x = 50

∴ ∠BEC = 50◦.


Problem

A circle with centre O is drawn around 4OBD so thatB and D lie on the circumference of the circle. BO isextended to A on the circle. Chord AC intersects ODand BD at F and E, respectively. If ∠BAC = 19◦ and∠OFA = 99◦, determine the measure of ∠BEC. (∠BECis marked x◦ on the diagram.)

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Solution 2

In a triangle, the angle formed at a vertex between the extension of a side andan adjacent side is called an exterior angle. In the following diagram, ∠XZWis exterior to 4XY Z. The exterior angle theorem states: “the exterior angle ofa triangle equals the sum of the two opposite interior angles.” In the diagram,r◦ = p◦ + q◦. We will use this result and two of the pieces of information wefound in Solution 1.

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∠FOA = 180◦ − 99◦ − 19◦ = 62◦ = a.


∠FOA is exterior to 4ODB.

∴ ∠FOA = ∠ODB + ∠OBD

a◦ = c◦ + c◦

62 = 2c

31 = c

∠CEB is exterior to 4EBA.

∴ ∠CEB = ∠EAB + ∠EBA

x◦ = 19◦ + c◦

x = 19 + 31

x = 50

∴ ∠BEC = 50◦.


Problem of the Week

Problem C

Square On

ABCD is a square with area 64 m2. E,F,G, and H are points on sidesAB,BC,CD, and DA, respectively, such that AE = BF = CG = DH = 2 m.E,F,G, and H are connected to form square EFGH.

Determine the area of EFGH.

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Problem of the Week


Square On

Problem

ABCD is a square with area 64 m2. E,F,G, and H are points on sidesAB,BC,CD, and DA, respectively, such that AE = BF = CG = DH = 2 m.E,F,G, and H are connected to form square EFGH. Determine the area ofEFGH.

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Solution

The area of square ABCD is 64 m2. Therefore the side lengths are 8 m since 8× 8 = 64 andthe area is calculated by multiplying the length and the width.

Each of the smaller parts of the sides of square ABCD are 2 m so the longer parts of the sidesare 8− 2 = 6 m.

Approach #1 to finding the area of square EFGH

In right 4HAE, AE = 2 and AH = 6. We can use one side as the base and the other as theheight in the calculation of the area of the triangle since they are perpendicular to each other.Therefore the area of 4HAE = AE×AH

2= 2×6

2= 6 m2. Since each of the triangles has the same

base length and height, their areas are equal and the total area of the four triangles is4× 6 = 24 m2.

The area of square EFGH can be determined by subtracting the area of the four trianglesfrom the area of square ABCD. Therefore the area of square EFGH = 64− 24 = 40 m2.


Some students may be familiar with the Pythagorean Theorem. This theorem states that in aright triangle, the square of the length of the hypotenuse (the longest side) is equal to the sumof the squares of the other two sides. The longest side is located opposite the right angle.

In right 4HAE, AE = 2, AH = 6 and HE is the hypotenuse. Therefore,

HE2 = AE2 + AH2

= 22 + 62

= 4 + 36

= 40

Taking the square root, HE =√

40 m

But EFGH is a square so all of its side lengths equal√

40. The area is calculated bymultiplying the length and the width. The area of EFGH =

√40×

√40 = 40 m2.

Therefore the area of square EFGH is 40 m2.


Problem of the Week

Problem C

Triangle Triumph

In the diagram, 4ABC is a right triangle with ∠ABC = 90◦, BD = 6 m,AB = 8 m, and the area of 4ADC is 50% more than the area of 4ABD.

Determine the perimeter of 4ADC.

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The Pythagorean Theorem states, “In a right triangle, the square of thelength of hypotenuse (the side opposite the right angle) equals the sum of thesquares of the lengths of the other two sides.”

In the following right triangle, p2 = r2 + q2.

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Problem of the Week


Triangle Triumph

Problem

In the diagram, 4ABC is a right triangle with∠ABC = 90◦, BD = 6 m, AB = 8 m, and thearea of 4ADC is 50% more than the area of4ABD. Determine the perimeter of 4ADC.

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Solution

Let a be the length of side DC, b be the length of side AC, and c be the lengthof side AD. Draw a line through A parallel to BC. The distance between thisline and BC is 8 m and is the height of 4ABD and 4ADC.

To find the area of a triangle, multiply the length of the base by the height anddivide by 2. The area of 4ABD = AB ×BD ÷ 2 = 8× 6÷ 2 = 24 m2.

The area of 4ADC is 50% more than the area of 4ABD. Therefore, the

area of 4ADC = area of 4ABD +1

2area of 4ABD = 24 + 12 = 36 m2.

But the area of 4ADC = (AB)(DC)÷ 2 = 8(DC)÷ 2 = 4(DC). Therefore,4(DC) = 36 and DC = 9 m. Then BC = BD + DC = 6 + 9 = 15 m.

Since 4ABD has a right angle, AD2 = AB2 + BD2 = 82 + 62 = 100. ThenAD =

√100 = 10 since AD > 0.

Also, 4ABC has a right angle, soAC2 = AB2 + BC2 = 82 + 152 = 64 + 225 = 289. Then AC =

√289 = 17 since

AC > 0.

∴ The perimeter of 4ADC = a + b + c

= DC + AC + AD

= 9 + 17 + 10

= 36 m

The perimeter of 4ADC is 36 m.


Problem of the Week

Problem C

The Great Tricycle Race

Two paths are built from A to F as shown.

! " # $ % &The distance from A to F in a straight line is 100 m. Points B, C, D, and Elie along AF such that AB = BC = CD = DE = EF .

The upper path, shown with a dashed line, is a semi-circle with diameter AF .The lower path, shown with a solid line, consists of five semi-circles withdiameters AB, BC, CD, DE, and EF .

Starting at the same time, Bev and Mike ride their tricycles along these pathsfrom A to F . Bev rides along the upper path from A to F while Mike ridesalong the lower path from A to F . If they ride at the same speed, who will getto F first?


Problem of the Week



Problem

Two paths are built from A to F as shown. The distance from A to F in a straight line is 100m. Points B, C, D, and E lie along AF such that AB = BC = CD = DE = EF . The upperpath, shown with a dashed line, is a semi-circle with diameter AF . The lower path, shownwith a solid line, consists of five semi-circles with diameters AB, BC, CD, DE, and EF .Starting at the same time, Bev and Mike ride their tricycles along these paths from A to F .Bev rides along the upper path from A to F while Mike rides along the lower path from A toF . If they ride at the same speed, who will get to F first?

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Solution

The circumference of a circle is found by multiplying its diameter by π. To find thecircumference of a semi-circle, divide its circumference by 2.

The length of the upper path is equal to half the circumference of a circle with diameter 100 m.The length of the upper path equals π × 100 ÷ 2 = 50π m. (This is approximately 157.1 m.)

Each of the semi-circles along the lower path have the same diameter. The diameter of each ofthese semi-circles is 100 ÷ 5 = 20 m. The length of the lower path is equal to half thecircumference of five circles, each with diameter 20 m. The distance along the lower pathequals 5 × (π × 20 ÷ 2) = 5 × (10π) = 50π m.

Since both Bev and Mike ride at the same speed and both travel the same distance, they willarrive at point F at the same time. Neither wins the race since both arrive at the same time.The answer to the problem may surprise you. Most people, at first glance, would think thatthe upper path is longer.

If you were to extend the problem so that Bev travels the same route but Mike travels along alower path made up of 100 semi-circles of equal diameter from A to F , they would still bothtravel exactly the same distance, 50π m. Check it out!


Problem of the Week

Problem C

One Step at a Time

A circle with centre O has a point A on the circumference. Radius OA isrotated 20◦ clockwise about the centre, resulting in the image OB. Point A isthen connected to point B. Radius OB is rotated 20◦ clockwise about thecentre, resulting in the image OC. Point B is then connected to point C.

The process of clockwise rotations continues until some radius rotates backonto OA. Every point on the circumference is connected to the pointsimmediately adjacent to it as a result of the process. A polygon is created.

O

A

B

C

D

EF

2020

2020

20

A

B

C

D

E

F

nth point

=⇒

Construction Resulting Polygon

a) Determine the number of sides of the polygon.

b) Determine the sum of the angles in the polygon. That is, determine thesum of the angles at each of the vertices of the polygon.


Problem of the Week


One Step at a Time

Problem

A circle with centre O has a point A on the circumference. Radius OA is rotated 20◦ clockwiseabout the centre, resulting in the image OB. Point A is then connected to point B. RadiusOB is rotated 20◦ clockwise about the centre, resulting in the image OC. Point B is thenconnected to point C. The process of clockwise rotations continues until some radius rotatesback onto OA. Every point on the circumference is connected to the points immediatelyadjacent to it as a result of the process. A polygon is created.


b) Determine the sum of the angles in the polygon. That is, determine the sum of theangles at each of the vertices of the polygon.

Solution

Each time the process is repeated, another congruent triangle is created. Each of thesetriangles has a 20◦ angle at O, the centre of the circle. But a complete rotation at the centre is360◦. Since each angle in the triangles at the centre of the circle is 20◦ and the total measureat the centre is 360◦, then there are 360÷ 20 = 18 triangles formed. This means that there are18 distinct points on the circumference of the circle and the polygon has 18 sides. An 18-sidedpolygon is called an octadecagon, from octa meaning 8 and deca meaning 10.

The other two angles in the each of the congruent triangles are equal. (Two sides of thetriangle are radii of the circle. The triangles are therefore isosceles.) The angles in a trianglesum to 180◦ so after the 20◦ angle is removed, there is 160◦ remaining for the other two angles.It follows that each of the other two angles in each triangle measures 160◦ ÷ 2 = 80◦. Thefollowing diagram illustrates this information for the two adjacent triangles AOB and BOC.

O

A

B

C

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80

8080

80

Each angle in the polygon is formed by an 80◦ angle from one triangle and the adjacent 80◦

angle from the next triangle. For example, ∠ABC = ∠ABO + ∠OBC = 80◦ + 80◦ = 160◦.There are 18 vertices in the octadecagon and the angle at each vertex is 160◦. Therefore thesum of the angles in the octadecagon is 18× 160◦ = 2880◦.

Diagrams are provided on the next page to further support the solution.


Angle at the Centre of a Circle

360

Completing the Construction

20

2020

20202020

2020

20

2020

20 20 2020

2020

80

8080

80

80

80

80

8080

80

8080 80

8080 80 80 80 80 80

80808080

8080

8080

8080

8080

80808080

A

B

C

D

E

F

N

M

L

K

J

IH G

P Q RS

The Octadecagon - 18 sided Polygon

A

B

C

D

E

F

N

M

L

K

J

IH G

P Q RS

160

160

160

160160 160

160

160

160

160

160160 160 160

160

160

160

160

Notice the vertices of the octadecagon are labelled A to S, but the letter O is missing since itwas used in the original construction as the centre of the circle.


Problem of the Week

Problem C

Overlapping Areas

The area of 4ACD is twice the area of square BCDE. AC and AD meet BEat K and L respectively such that KL = 6 cm.

If the side length of the square is 8 cm, determine the area of 4AKL.

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Problem of the Week


Overlapping Areas

Problem

The area of 4ACD is twice the area of square BCDE. AC andAD meet BE at K and L respectively such that KL = 6 cm.If the side length of the square is 8 cm, determine the area of4AKL.

Solution 1

In the first solution we will find the area of square BCDE, thearea of 4ACD and the area of trapezoid KCDL.

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To find the area of a square, multiply the length times the width. To find thearea of a trapezoid, multiply the sum of the lengths of the two parallel sides bythe height and divide the product by 2.

Area of square BCDE = 8× 8

= 64 cm2

Area 4ACD = 2× Area of Square BCDE

= 2× 64

= 128 cm2

In trapezoid KCDL, the two parallel sides are KL and CD, and the height isthe width of square BCDE, namely BC.

Area of trapezoid KCDL = (KL + CD)×BC ÷ 2

= (6 + 8)× 8÷ 2

= 14× 8÷ 2

= 56 cm2

Area 4AKL = Area 4ACD − Area of trapezoid KCDL

= 128− 56

= 72 cm2

Therefore, the area of 4AKL is 72 cm2.


Problem

The area of 4ACD is twice the area of square BCDE.AC and AD meet BE at K and L respectively such thatKL = 6 cm. If the side length of the square is 8 cm,determine the area of 4AKL.

Solution 2

Construct the altitude of 4ACD intersecting BE at Pand CD at Q. In this solution we will find the height of4AKL and then use the formula for the area of a triangleto find the required area.

To find the area of a square, multiply the length times thewidth. To find the area of a triangle, multiply the lengthof the base times the height and divide the product by 2.

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(

)Area of square BCDE = 8× 8

= 64 cm2


= 2× 64

= 128 cm2

But Area 4ACD = CD × AQ÷ 2

128 = 8× AQ÷ 2

128 = 4× AQ

∴ AQ = 32 cm

We know that AQ = AP + PQ, AQ = 32 cm and PQ = 8 cm, the side lengthof the square. It follows that AP = AQ− PQ = 32− 8 = 24 cm.

∴ Area 4AKL = KL× AP ÷ 2

= 6× 24÷ 2

= 72 cm2



Problem of the Week

Problem C

Floor Plan

The rectangular floor plan of the first level of a house is shown in the followingdiagram.

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Both the laundry room and the dining room are square with areas of 4 m2 and25 m2, respectively. The living room is rectangular with an area of 30 m2.

Determine the area of the kitchen.


Problem of the Week


Floor Plan

Problem

The rectangular floor plan of the first level of a house is shown in the followingdiagram. Both the laundry room and the dining room are square with areas of4 m2 and 25 m2, respectively. The living room is rectangular with an area of30 m2. Determine the area of the kitchen.

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Solution

Let the width of a room be the distance represented top to bottom on thediagram. Let the length of a room be the distance represented horizontally onthe diagram.

The dining room is a square and has an area of 25 m2. Its length and widthmust both be 5 m since Area = 5 × 5 = 25 m2. The width of the dining roomand living room are the same. So the width of the living room is 5 m. But thearea of the living room is 30 m2 so the length of the living room is 6 m sinceArea = 5 × 6 = 30 m2.

The laundry room is a square and has an area of 4 m2. Its length and widthmust both be 2 m since Area = 2 × 2 = 4 m2. The width of the laundry roomand kitchen are the same. So the width of the kitchen is 2 m.

(Length of Laundry Room (Length of Living Room

+ Length of Kitchen) = + Length of Dining Room)

2 + Length of Kitchen = 6 + 5

Length of Kitchen = 9 m

Since the width of the kitchen is 2 m and the length of the kitchen is 9 m, thearea of the kitchen is 2 × 9 = 18 m2.


Problem of the Week

Problem C

Square On

ABCD is a square with area 64 m2. E,F,G, and H are points on sidesAB,BC,CD, and DA, respectively, such that AE = BF = CG = DH = 2 m.E,F,G, and H are connected to form square EFGH.

Determine the area of EFGH.

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Problem of the Week


Square On

Problem

ABCD is a square with area 64 m2. E,F,G, and H are points on sidesAB,BC,CD, and DA, respectively, such that AE = BF = CG = DH = 2 m.E,F,G, and H are connected to form square EFGH. Determine the area ofEFGH.

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Solution

The area of square ABCD is 64 m2. Therefore the side lengths are 8 m since 8× 8 = 64 andthe area is calculated by multiplying the length and the width.

Each of the smaller parts of the sides of square ABCD are 2 m so the longer parts of the sidesare 8− 2 = 6 m.


In right 4HAE, AE = 2 and AH = 6. We can use one side as the base and the other as theheight in the calculation of the area of the triangle since they are perpendicular to each other.Therefore the area of 4HAE = AE×AH

2= 2×6

2= 6 m2. Since each of the triangles has the same

base length and height, their areas are equal and the total area of the four triangles is4× 6 = 24 m2.

The area of square EFGH can be determined by subtracting the area of the four trianglesfrom the area of square ABCD. Therefore the area of square EFGH = 64− 24 = 40 m2.


Some students may be familiar with the Pythagorean Theorem. This theorem states that in aright triangle, the square of the length of the hypotenuse (the longest side) is equal to the sumof the squares of the other two sides. The longest side is located opposite the right angle.

In right 4HAE, AE = 2, AH = 6 and HE is the hypotenuse. Therefore,

HE2 = AE2 + AH2

= 22 + 62

= 4 + 36

= 40

Taking the square root, HE =√

40 m

But EFGH is a square so all of its side lengths equal√

40. The area is calculated bymultiplying the length and the width. The area of EFGH =

√40×

√40 = 40 m2.

Therefore the area of square EFGH is 40 m2.


Problem of the Week

Problem C

Triangle Triumph



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Problem of the Week


Triangle Triumph

Problem


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Solution





2area of 4ABD = 24 + 12 = 36 m2.



√100 = 10 since AD > 0.


√289 = 17 since

AC > 0.


= DC + AC + AD

= 9 + 17 + 10

= 36 m



Problem of the Week

Problem C


Two paths are built from A to F as shown.

! " # $ % &The distance from A to F in a straight line is 100 m. Points B, C, D, and Elie along AF such that AB = BC = CD = DE = EF .

The upper path, shown with a dashed line, is a semi-circle with diameter AF .The lower path, shown with a solid line, consists of five semi-circles withdiameters AB, BC, CD, DE, and EF .

Starting at the same time, Bev and Mike ride their tricycles along these pathsfrom A to F . Bev rides along the upper path from A to F while Mike ridesalong the lower path from A to F . If they ride at the same speed, who will getto F first?


Problem of the Week



Problem

Two paths are built from A to F as shown. The distance from A to F in a straight line is 100m. Points B, C, D, and E lie along AF such that AB = BC = CD = DE = EF . The upperpath, shown with a dashed line, is a semi-circle with diameter AF . The lower path, shownwith a solid line, consists of five semi-circles with diameters AB, BC, CD, DE, and EF .Starting at the same time, Bev and Mike ride their tricycles along these paths from A to F .Bev rides along the upper path from A to F while Mike rides along the lower path from A toF . If they ride at the same speed, who will get to F first?

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Solution

The circumference of a circle is found by multiplying its diameter by π. To find thecircumference of a semi-circle, divide its circumference by 2.

The length of the upper path is equal to half the circumference of a circle with diameter 100 m.The length of the upper path equals π × 100 ÷ 2 = 50π m. (This is approximately 157.1 m.)

Each of the semi-circles along the lower path have the same diameter. The diameter of each ofthese semi-circles is 100 ÷ 5 = 20 m. The length of the lower path is equal to half thecircumference of five circles, each with diameter 20 m. The distance along the lower pathequals 5 × (π × 20 ÷ 2) = 5 × (10π) = 50π m.

Since both Bev and Mike ride at the same speed and both travel the same distance, they willarrive at point F at the same time. Neither wins the race since both arrive at the same time.The answer to the problem may surprise you. Most people, at first glance, would think thatthe upper path is longer.

If you were to extend the problem so that Bev travels the same route but Mike travels along alower path made up of 100 semi-circles of equal diameter from A to F , they would still bothtravel exactly the same distance, 50π m. Check it out!


Problem of the Week

Problem C

Slightly Irregular

In the following slightly irregular shape,

• AB = 50 cm, CD = 15 cm, EF = 30 cm;

• the area of the shaded triangle, 4DEF , is 210 cm2; and

• the area of the entire figure, ABCDE, is 1000 cm2.

Determine the length of AE.

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Problem of the Week


Slightly Irregular

Problem

In the following slightly irregular shape, AB = 50 cm, CD = 15 cm, EF = 30 cm, the area ofthe shaded triangle, 4DEF , is 210 cm2; and the area of the entire figure, ABCDE, is1000 cm2. Determine the length of AE.

Solution

The first task is to mark the given information on the diagram.This has been completed on the diagram to the right.

To find the area of a triangle, multiply the base length by theheight and divide by 2. In 4DEF , the base, EF , has length 30cm. The height of 4DEF is the perpendicular distance fromEF (extended) to vertex D, namely GD. The area is given. So

Area 4DEF =30×GD

2210 = 15×GD

14 = GD

We know that EH = AB = 50, GH = DC = 15, andEH = EF +FG+GH. It follows that 50 = 30 +FG+ 15 andFG = 5 cm.

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Now we can relate the total area to the areas contained inside.

Area ABCDE = Area ABHE + Area CDGH + Area 4DFG + Area 4DEF

1000 = AB × AE + DG×DC +FG×GD

2+ 210

1000 = 50× AE + 14× 15 +5× 14

2+ 210

1000 = 50× AE + 210 + 35 + 210

1000 = 50× AE + 455

1000− 455 = 50× AE

545 = 50× AE545

50= AE

∴ AE = 10.9 cm.


Problem of the Week

Problem C

A Little Tiling

Tyler, the tiler, has an unlimited supply of square tiles. He has 1 cm by 1 cmtiles, 2 cm by 2 cm tiles, 3 cm by 3 cm tiles, and so on. Every tile has integerside lengths.

A rectangular table top with an 84 cm by 112 cm surface is to be completelycovered by identical square tiles, none of which can be cut. Tyler knows thathe can cover the table top with 1 cm × 1 cm tiles, 9 408 in total, since84× 112 = 9 408 cm2. However, Tyler wants to use the minimum number ofidentical tiles to complete the job in order to reduce his overall material cost.

Determine the minimum number of tiles required to completely cover the tabletop.


Problem of the Week


A Little Tiling

Problem

Tyler, the tiler, has an unlimited supply of square tiles. He has 1 cm by 1 cmtiles, 2 cm by 2 cm tiles, 3 cm by 3 cm tiles, and so on. Every tile has integerside lengths. A rectangular table top with an 84 cm by 112 cm surface is to becompletely covered by identical square tiles, none of which can be cut. Tylerknows that he can cover the table top with 1 cm × 1 cm tiles, 9 408 in total,since 84 × 112 = 9 408 cm2. However, Tyler wants to use the minimum numberof identical tiles to complete the job in order to reduce his overall materialcost. Determine the minimum number of tiles required to completely cover thetable top.

Solution

To use the smallest number of tiles we must use the largest tile possible. Thesquare tile must have sides less than or equal to 84 cm. If it was greater than84 cm, the tile would have to be cut to fit the width of the table.

Since the tiles are square and must completely cover the top surface, the sidelength of the tile must be a number that is a factor of both 84 and 112. In fact,since we need the largest side length, we are looking for the greatest commonfactor of 84 and 112.

The factors of 84 are

1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, and 84.


1, 2, 4, 7, 8, 14, 16, 28, 56, and 112.

The largest number common to both lists is 28. Therefore the greatestcommon factor of 84 and 112 is 28. The required tiles are 28 cm × 28 cm.Since 84 ÷ 28 = 3, the surface is 3 tiles wide. Since 112 ÷ 28 = 4, the surface is4 tiles long. The minimum number of tiles required is 3 × 4 = 12 tiles.

The number of 28 cm × 28 cm tiles required to cover the top of the table is 12.This is the minimum number of tiles required.


Problem of the Week

Problem C

One Step at a Time

A circle with centre O has a point A on the circumference. Radius OA isrotated 20◦ clockwise about the centre, resulting in the image OB. Point A isthen connected to point B. Radius OB is rotated 20◦ clockwise about thecentre, resulting in the image OC. Point B is then connected to point C.

The process of clockwise rotations continues until some radius rotates backonto OA. Every point on the circumference is connected to the pointsimmediately adjacent to it as a result of the process. A polygon is created.

O

A

B

C

D

EF

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B

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D

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Construction Resulting Polygon


b) Determine the sum of the angles in the polygon. That is, determine thesum of the angles at each of the vertices of the polygon.


Problem of the Week


One Step at a Time

Problem

A circle with centre O has a point A on the circumference. Radius OA is rotated 20◦ clockwiseabout the centre, resulting in the image OB. Point A is then connected to point B. RadiusOB is rotated 20◦ clockwise about the centre, resulting in the image OC. Point B is thenconnected to point C. The process of clockwise rotations continues until some radius rotatesback onto OA. Every point on the circumference is connected to the points immediatelyadjacent to it as a result of the process. A polygon is created.


b) Determine the sum of the angles in the polygon. That is, determine the sum of theangles at each of the vertices of the polygon.

Solution

Each time the process is repeated, another congruent triangle is created. Each of thesetriangles has a 20◦ angle at O, the centre of the circle. But a complete rotation at the centre is360◦. Since each angle in the triangles at the centre of the circle is 20◦ and the total measureat the centre is 360◦, then there are 360÷ 20 = 18 triangles formed. This means that there are18 distinct points on the circumference of the circle and the polygon has 18 sides. An 18-sidedpolygon is called an octadecagon, from octa meaning 8 and deca meaning 10.

The other two angles in the each of the congruent triangles are equal. (Two sides of thetriangle are radii of the circle. The triangles are therefore isosceles.) The angles in a trianglesum to 180◦ so after the 20◦ angle is removed, there is 160◦ remaining for the other two angles.It follows that each of the other two angles in each triangle measures 160◦ ÷ 2 = 80◦. Thefollowing diagram illustrates this information for the two adjacent triangles AOB and BOC.

O

A

B

C

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Each angle in the polygon is formed by an 80◦ angle from one triangle and the adjacent 80◦

angle from the next triangle. For example, ∠ABC = ∠ABO + ∠OBC = 80◦ + 80◦ = 160◦.There are 18 vertices in the octadecagon and the angle at each vertex is 160◦. Therefore thesum of the angles in the octadecagon is 18× 160◦ = 2880◦.

Diagrams are provided on the next page to further support the solution.


Angle at the Centre of a Circle

360

Completing the Construction

20

2020

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2020

20

2020

20 20 2020

2020

80

8080

80

80

80

80

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80

8080 80

8080 80 80 80 80 80

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8080

8080

8080

8080

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A

B

C

D

E

F

N

M

L

K

J

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P Q RS

The Octadecagon - 18 sided Polygon

A

B

C

D

E

F

N

M

L

K

J

IH G

P Q RS

160

160

160

160160 160

160

160

160

160

160160 160 160

160

160

160

160

Notice the vertices of the octadecagon are labelled A to S, but the letter O is missing since itwas used in the original construction as the centre of the circle.


Problem of the Week

Problem C

This is Sum Problem

The number 90 can be expressed as the sum of 3 consecutive whole numbers.That is, 90 = 29 + 30 + 31. The number 90 can also be written as the sum of 4consecutive whole numbers. That is, 90 = 21 + 22 + 23 + 24.

Express the number 220 as the sum of 5 consecutive whole numbers and thenas the sum of 8 consecutive whole numbers.

This problem is extended this week in Problem D and Problem E of Problemof the Week.


Problem of the Week


This is Sum Problem

Problem

The number 90 can be expressed as the sum of 3 consecutive whole numbers. That is,90 = 29 + 30 + 31. The number 90 can also be written as the sum of 4 consecutive wholenumbers. That is, 90 = 21 + 22 + 23 + 24. Express the number 220 as the sum of 5 consecutivewhole numbers and then as the sum of 8 consecutive whole numbers.

Solution

First, we want to express 220 as the sum of 5 consecutive whole numbers.

When 90 is expressed as the sum of 3 consecutive whole numbers, the average is 90÷ 3 = 30.Since 30 is a whole number and there is an odd number of consecutive whole numbers in thesum, 29, 30 and 31 will produce the correct sum.

If we apply the same idea to 220, the average would be 220÷ 5 = 44. Since 44 is a wholenumber and there is an odd number of consecutive whole numbers in the sum, 42, 43, 44, 45,and 46, should produce the correct sum. Checking, 42 + 43 + 44 + 45 + 46, we obtain 220 asrequired.

Next, we want to express 220 as the sum of 8 consecutive whole numbers.

When 90 is expressed as the sum of 4 consecutive whole numbers, the average is 90÷ 4 = 22.5.We need consecutive whole numbers such that two are below and two are above 22.5. Thisgives the numbers 21, 22, 23, and 24, as in the example. This would only work for an evennumber of consecutive whole numbers if the average is half way between two consecutive wholenumbers.

Applying the same idea to 220, the average would be 220÷ 8 = 27.5. This number is half waybetween 27 and 28. We would need four consecutive whole numbers below the average and fourconsecutive whole numbers above the average giving us the eight numbers 24, 25, 26, 27, 28,29, 30, and 31. Checking, 24 + 25 + 26 + 27 + 28 + 29 + 30 + 31, we obtain 220 as required.

As a concluding note, it is not possible to express 220 as the sum of four consecutive wholenumbers. 220÷ 4 = 55. The number 55 is the average of the four numbers but 55 is a wholenumber. We could try 54 + 55 + 56 + 57 = 222 6= 220 or 53 + 54 + 55 + 56 = 218 6= 220.

It is possible to express 220 as the sum of n consecutive whole numbers when n is odd providedthat the average 220÷ n is a whole number.

It is possible to express 220 as the sum of n consecutive whole numbers when n is evenprovided that the average 220÷ n is half way between two consecutive whole numbers.


Problem of the Week

Problem C

Playing With Blocks

Ten blocks are arranged as illustrated in the following diagram. Each lettershown on the front of a block represents a number.

The sum of the numbers on any three consecutive blocks is 19.

Determine the value of S.


Problem of the Week


Playing With Blocks

Problem

Ten blocks are arranged as illustrated in the following diagram. Each lettershown on the front of a block represents a number. The sum of the numbers onany three consecutive blocks is 19. Determine the value of S.

Solution

Since the sum of the numbers on any three consecutive blocks is the same,

4 + P +Q = P +Q+R

4+ 6P+ 6Q = 6P+ 6Q+R since P +Q is common to both sides

∴ R = 4

Again, since the sum of the numbers on any three consecutive blocks is thesame,

T + U + V = U + V + 8

T+ 6U+ 6V = 6U+ 6V + 8 since U + V is common to both sides

∴ T = 8

Since the sum of any three consecutive numbers is 19:

R + S + T = 19

4 + S + 8 = 19 substituting R = 4 and T = 8

S + 12 = 19

∴ S = 7

The value of S is 7.


Problem of the Week

Problem C

Show Me The Money

Over the past several years, Ima Saver collected nickels (5 cent coins) anddimes (10 cent coins) and put them in his pink piggy bank. Only nickels anddimes went in his bank.

Finally, one day, the bank was full so Ima counted his money and discoveredthat he had exactly $10 in the bank. He also observed that he had 11 morenickels than dimes in his bank.

How many coins were in Ima Saver’s pink piggy bank?


Problem of the Week


Show Me The Money

Problem

Over the past several years, Ima Saver collected nickels (5 cent coins) and dimes (10 centcoins) and put them in his pink piggy bank. Only nickels and dimes went in his bank. Finally,one day, the bank was full so Ima counted his money and discovered that he had exactly $10 inthe bank. He also observed that he had 11 more nickels than dimes in his bank. How manycoins were in Ima Saver’s pink piggy bank?

Solution 1

In this solution we will solve the problem without using equations.

Ima had 11 more nickels than dimes. These 11 nickels are worth 11× 5 = 55¢ or $0.55. Theremaining $10.00− $0.55 = $9.45 would be made up using an equal number of nickels anddimes. Each nickel-dime combination is worth 15¢ or $0.15. By dividing $9.45 by $0.15 wedetermine the number of 15 cent combinations that are required to make the total. Since$9.45÷ $0.15 = 63 we need 63 nickel-dime pairs. That is, we need 63 nickels and 63 dimes tomake $9.45. But there are 11 more nickels. Therefore, there is a total of 63 + 63 + 11 = 137coins in his bank.

Solution 2

In this solution we will solve the problem using an equation.

Let n represent the number of nickels and (n− 11) represent the number of dimes. Since eachnickel is worth 5¢, the value of n nickels is (5n)¢. Since each dime is worth 10¢, the value ofn− 11 dimes is 10(n− 11)¢. The bank contains a total value of $10 or 1 000¢. Therefore,

Value of Nickels (in ¢) + Value of Dimes (in ¢) = Total Value (in ¢)5n+ 10(n− 11) = 1 000

5n+ 10n− 110 = 1 000

15n = 1000 + 110

15n = 1110

n = 74

n− 11 = 63

There are 74 nickels and 63 dimes for a total of 74 + 63 = 137 coins in his bank.


Problem of the Week

Problem C

That’s Odd

1 = 12

1 + 3 = 4 = 22

1 + 3 + 5 = 9 = 32

1 + 3 + 5 + 7 = 16 = 42

Did you know that the sum of the first n positive odd integers is n2? Thediagram above illustrates the first four possible sums. The sum of the firstfive positive odd integers would be 52 or 25. We can easily check to see that1 + 3 + 5 + 7 + 9 = 25.

When adding the first a positive odd integers to the first b positive oddintegers, the sum is 180. If p is the largest odd number in the first set ofnumbers and q is the largest odd number in the second set of numbers, thendetermine the sum p + q.


Problem of the Week


That’s Odd

Problem

Did you know that the sum of the first n positive odd integers is n2? The sum of the first fivepositive odd integers would be 52 or 25. We can easily check to see that 1 + 3 + 5 + 7 + 9 = 25.When adding the first a positive odd integers to the first b positive odd integers, the sum is180. If p is the largest odd number in the first set of numbers and q is the largest odd numberin the second set of numbers, then determine the sum p + q.

Solution

Since there are a positive odd integers and the largest is p, then 1 + 3 + 5 + · · · + p = a2.Since there are b positive odd integers and the largest is q, then 1 + 3 + 5 + · · · + q = b2.We also know that when these two sets of odd numbers are added together, the sum is 180 so

(1 + 3 + 5 + · · · + p) + (1 + 3 + 5 + · · · + q) = a2 + b2 = 180.

One way to proceed is to pick values for a, determine a2 and then determine if the remainingnumber required to sum to 180 is a perfect square. The results are summarized in the tablebelow.

a a2 b2 = 180 − a2 b (b > 0) Solution?1 1 180-1=179 13.4 no2 4 180-4=176 13.3 no3 9 180-9=171 13.1 no4 16 180-16=164 12.8 no5 25 180-25=155 12.4 no6 36 180-36=144 12 yes7 49 180-49=131 11.4 no8 64 180-64=116 10.8 no9 81 180-81=99 9.9 no10 100 180-100=80 8.9 no11 121 180-121=59 7.7 no12 144 180-144=36 6 yes13 169 180-169=11 3.3 no

If a = 14, then a2 = 196. This produces a value greater than 180 and cannot be a possiblesolution.

There appear to be two possible solutions. When a = 6 and b = 12, thena2 + b2 = 36 + 144 = 180. This means that adding the first 6 odd positive integers to the first12 odd positive integers results in a sum of 180. So p is the sixth odd positive integer, namely11, and q is the twelfth odd positive integer, namely 23. The sum, p + q, is 11 + 23 or 34. Thesecond solution, a = 12 and b = 6, produces p = 23 and q = 11. The sum, p + q, is still 34.


Problem of the Week

Problem C

Clock Talk

My clock is a perfectly good clock. It keeps exact time. But it only has anhour hand. Today, in the afternoon, I looked at my clock and discovered thatthe hour hand was 7

8 of the distance between the “4” and the “5”.

Determine the exact time (hours, minutes and seconds).


Problem of the Week


Clock Talk

Problem

My clock is a perfectly good clock. It keeps exact time. But it only has anhour hand. Today, in the afternoon, I looked at my clock and discovered thatthe hour hand was 7

8 of the distance between the “4” and the “5”. Determinethe exact time (hours, minutes and seconds).

Solution

To solve this problem we note that in one hour, the hour hand travels 112 of a

complete revolution while the minute hand travels a complete revolution or60 minutes.

Since the hour hand is 78 of the distance between the “4” and the “5”, the

minute hand will travel 78 of a complete revolution or 7

8 of 60 minuteswhich is 7

8 × 60 or 5212 minutes.

Since we want the time in hours, minutes and seconds, we need to convert 12

minute to seconds.

The number of seconds may be obvious but the calculation,12 minute× 60 seconds

1 minute = 30 seconds, is provided for completeness.

Therefore the precise time is 30 seconds after 4:52 p.m. This can be written4:52:30 p.m. or 16:52:30 using the twenty-four hour clock.


Problem of the Week

Problem C

Ride On, Ride On!

A motorcycle and a delivery truck left a roadside diner at the same time. Aftertravelling in the same direction for one and one-quarter hours, the motorcyclehad travelled 25 km farther than the delivery truck. If the average speed of themotorcycle was 60 km/h, find the average speed of the delivery truck.


Problem of the Week


Ride On, Ride On!

Problem

A motorcycle and a delivery truck left a roadside diner at the same time. Aftertravelling in the same direction for one and one-quarter hours, the motorcyclehad travelled 25 km farther than the delivery truck. If the average speed of themotorcycle was 60 km/h, find the average speed of the delivery truck.

Solution

We can calculate distance by multiplying the average speed by the time.

In one and one-quarter hours at 60 km/h, the motorcycle would travel60 × 11

4 = 60 × 54 = 75 km.

In the same time, the delivery truck travels 25 km less. The delivery truck hastravelled 75 − 25 = 50 km. Since the distance travelled equals the averagespeed multiplied by the time, then the average speed will equal the distancetravelled divided by the time travelled. Thus, the average speed of the deliverytruck equals 50 ÷ 11

4 = 50 ÷ 54 = 50 × 4

5 = 40 km/h.

Therefore the average speed of the delivery truck is 40 km/h.

The calculations in this problem could be done using decimals by convertingone and one-quarter hours to 1.25 hours.


Problem of the Week

Problem C

Balloon Breaker

A wall is covered with balloons. Each balloon has the number 5, 3 or 2 printedon it. You are given 7 darts to throw at the wall. If your dart breaks a balloon,then you earn the number of points printed on the balloon. If your dart doesnot break a balloon, then you are awarded 0 points for that shot. You win aprize if your total score is exactly 16 points on seven shots. If your total is over16 or under 16, then you lose.

Determine the number of different point combinations that can be used to winthe game.


Problem of the Week


Balloon Breaker

Problem

A wall is covered with balloons. Each balloon has the number 5, 3 or 2 printed on it. You aregiven 7 darts to throw at the wall. If your dart breaks a balloon, then you earn the number ofpoints printed on the balloon. If your dart does not break a balloon, then you are awarded 0points for that shot. You win a prize if your total score is exactly 16 points on seven shots. Ifyour total is over 16 or under 16, then you lose. Determine the number of different pointcombinations that can be used to win the game.

Solution 1

Let us consider cases.

1. You break three balloons with a 5 printed on them. You have a total of 3 × 5 = 15points. There is no possible way to get 16 points since the other balloon values are 2 or3. There is no way to win by breaking three (or more) balloons with a 5 printed on them.

2,. You break two balloons with a 5 printed on them. You have a total of 2 × 5 = 10 points.You need to get 16 − 10 = 6 points by breaking balloons with 2 or 3 printed on them.There are two ways to do this. Break three balloons with a 2 printed on them and misson two shots or break two balloons with 3 printed on them and miss on three shots.There are 2 ways to win if you break two balloons with a 5 printed on them.

3. You break one balloon with a 5 printed on it. You have a total of 5 points. You need toget 16 − 5 = 11 points by breaking balloons with 2 or 3 printed on them. You cannot get11 points breaking only balloons with a 2 printed on them and you cannot get 11 pointsbreaking only balloons with a 3 printed on them. However, you can get 11 points bybreaking one 3 and four 2’s or by breaking three 3’s and one 2. There are 2 ways to winif you break one balloon with a 5 printed on it.

4. You break no balloons with a 5 printed on it. You need to make 16 points by breakingonly balloons with a 2 or 3 printed on them. You cannot get 16 points breaking onlyballoons with a 3 printed on them. You cannot get 16 points breaking only balloons witha 2 printed on them because you only have seven darts giving a maximum of 14 points.It is possible to get 16 points using combinations of 3 point and 2 point balloons. If youbreak two 3 point balloons and five 2 point balloons, then you win in seven shots. If youbreak four 3 point balloons and two 2 point balloons, then you have 16 points in six shotsand would have to miss on one of your shots. There are 2 ways to win if you do notbreak any 5 point balloons.

There are 0 + 2 + 2 + 2 = 6 combinations that allow you to win by getting 16 points in sevenshots.


Problem


Solution 2

In this solution we will complete a chart to determine the valid possibilities.

Let the number of 5 point balloons be a, the number of 3 point balloons be b, and the numberof 2 point balloons be c. We want 5a + 3b + 2c = 16 and a + b + c ≤ 7.

Number Number Number Total Number WINof 5 Point of 3 Point of 2 Point Points of Missed Shots orBalloons Balloons Balloons Scored Needed LOSE

a b c 5a + 3b + 2c 16 − a− b− c3 0 0 15 LOSE3 0 1 17 LOSE3 1 0 18 LOSE2 2 0 16 3 WIN2 1 1 15 LOSE2 1 2 17 LOSE2 0 3 16 2 WIN1 4 0 17 LOSE1 3 0 14 LOSE1 3 1 16 2 WIN1 2 2 15 LOSE1 2 3 17 LOSE1 1 4 16 1 WIN0 6 0 18 LOSE0 5 0 15 LOSE0 5 1 17 LOSE0 4 2 16 1 WIN0 3 3 15 LOSE0 3 4 17 LOSE0 2 5 16 0 WIN0 1 6 15 LOSE0 0 7 14 LOSE

There are only 6 combinations that allow you to win by getting 16 points in seven shots. (Infollowing a method like the above method, one must be careful to systematically examine allpossible cases.)


Problem of the Week

Problem C

Photo Fun

Five of Santa’s elves: Alpha, Beta, Delta, Epsilon and Gamma, are lined up inalphabetical order from left to right. They can each choose one of five festivehats to wear for a photo. The hats are identical except for colour. Three of thehats are red and two of the hats are green.

How many different photos can be taken?


Problem of the Week


Photo FunProblem

Five of Santa’s elves: Alpha, Beta, Delta, Epsilon and Gamma, are lined up inalphabetical order from left to right. They can each choose one of five festivehats to wear for a photo. The hats are identical except for colour. Three of thehats are red and two of the hats are green. How many different photos can betaken?

Solution

Since the elves are already organized in alphabetical order, we are looking forthe number of different ways that we can distribute the hats among the elves.We will consider cases:

1. If the first elf gets a green hat, there are four ways to give out the secondgreen hat. Once the green hats are distributed, the remaining three elvesmust each get a red hat. Therefore, there are 4 ways to distribute the hatsso that the first elf receives a green hat.

2. If the first elf gets a red hat and the second elf gets a green hat, there arethree ways to give out the second green hat. Once the green hats aredistributed, the remaining two elves must each get a red hat. Therefore,there are 3 ways to distribute the hats so that the first elf receives a redhat and the second elf receives a green hat.

3. If the first two elves each get a red hat and the third elf gets a green hat,there are two ways to give out the second green hat. Once the green hatsare distributed, the remaining elf must get a red hat. Therefore, there are2 ways to distribute the hats so that the first two elves each receive a redhat and the third elf receives a green hat.

4. If the first three elves each get a red hat and the fourth elf gets a greenhat, the fifth elf must get the second green hat. Therefore, there is only1 way to distribute the hats so that the first three elves each receive a redhat and the fourth elf receives a green hat.

There are no other cases to consider. The total number of ways to distributethe hats is the sum of the number of ways from each of the cases. Therefore,there are 4 + 3 + 2 + 1 = 10 ways to distribute the hats. There are 10 differentphotos that can be taken.


Problem of the Week

Problem C

A Powerful Problem For The New Year

53 is a power with base 5 and exponent 3.

53 means 5× 5× 5 and equals 125 when expressed as an integer.

When 52013 is expressed as an integer, what are the last three digits?


Problem of the Week



Problem


53 means 5 × 5 × 5 and equals 125 when expressed as an integer.


Solution

Let’s start by examining the last three digits of various powers of 5.

51 = 005

52 = 025

53 = 125

54 = 625

55 = 3 125

56 = 15 625

57 = 78 125

58 = 390 625

Notice that there is a pattern for the last three digits after the first two powersof 5. For every odd integral exponent greater than 2, the last three digits are“125”. For every even integral exponent greater than 2, the last three digitsare “625”. The pattern continues so 59 will end “125” since the exponent 9 isodd and 510 will end “625” since the exponent 10 is even. This is easily verifiedsince 59 = 1 953 125 and 510 = 9 765 625.

For 52013, the exponent 2013 is greater than 2 and is an odd number.

∴ the last three digits of 52013 are 125.


Problem of the Week

Problem C

Triangle Triumph



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Problem of the Week


Triangle Triumph

Problem


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Solution





2area of 4ABD = 24 + 12 = 36 m2.



√100 = 10 since AD > 0.


√289 = 17 since

AC > 0.


= DC + AC + AD

= 9 + 17 + 10

= 36 m



Problem of the Week

Problem C

Positively Perfect Prime Products

A prime number is any number that has exactly two positive integer factors, 1and the number itself. A composite number has more than two positive integerfactors. The number 1 is neither prime nor composite.

Four distinct prime numbers have a product of d10: a three-digit number withhundreds digit d. Determine all possible values of the sum of these four primenumbers.


Problem of the Week


Positively Perfect Prime Products

Problem

A prime number is any number that has exactly two positive integer factors, 1 and the numberitself. A composite number has more than two positive integer factors. The number 1 is neitherprime nor composite. Four distinct prime numbers have a product of d10: a three-digit numberwith hundreds digit d. Determine all possible values of the sum of these four prime numbers.

Solution

Since the product d10 ends in 0, it must be divisible by 10, which is the product of the twoprimes 2 and 5.

When d10 is divided by 10, the quotient is d10÷ 10 = d1. This two-digit number must becomposite and must be the product of two distinct prime numbers, neither of which is 2 or 5.We can rule out any two-digit prime numbers for d1 since these numbers would only have oneprime factor. Therefore, we can rule out 11, 31, 41, 61, and 71: the five prime numbersending with a 1. Then, d cannot be 1, 3, 4, 6, or 7. Since d10 is a three-digit number, d 6= 0because d10 = 010 = 10 is not a three-digit number. The remaining possibilities for d are2, 5, 8, and 9.

If d = 2, then the two-digit number would be 21, which has prime factors 7 and 3. The fourprime factors of d10 = 210 are 2, 3, 5, and 7, producing a sum of 2 + 3 + 5 + 7 = 17.

If d = 5, then the two-digit number would be 51, which has prime factors 17 and 3. The fourprime factors of d10 = 510 are 2, 3, 5, and 17, producing a sum of 2 + 3 + 5 + 17 = 27.

If d = 8, then the two-digit number would be 81, which is the product 9× 9. There are alreadyfour factors, two of which are composite so d10 = 810 cannot be expressed as the product offour distinct prime numbers. Therefore, 8 can be ruled out as a possible value for d. (Note that810 = 2× 3× 3× 3× 3× 5, which is the product of six prime numbers not all of which aredistinct.

If d = 9, then the two-digit number would be 91, which has prime factors 7 and 13. The fourprime factors of d10 = 910 are 2, 5, 7, and 13, producing a sum of 2 + 5 + 7 + 13 = 27.However, we already have the sum 27.

Since there are no other possible cases to consider, the only possible sums of the four distinctprime factors that multiply to d10 are 17 and 27.


Problem of the Week

Problem C


Xavier, Yvonne and Zak arrived at school at three different times. They eachbrought one of their favourite snacks to share with the other two (one broughtpretzels; one brought cookies; one brought licorice), and their favourite sportsapparatus (one brought a baseball; one brought a soccer ball; one brought afootball).







Determine the order they arrived in, what they each brought for a snack, andwhich sports apparatus they each brought.


Problem of the Week



Problem

Xavier, Yvonne and Zak arrived at school at three different times. They each brought one oftheir favourite snacks to share with the other two (one brought pretzels; one brought cookies;one brought licorice), and their favourite sports apparatus (one brought a baseball; onebrought a soccer ball; one brought a football).







Determine the order they arrived in, what they each brought for a snack, and which sportsapparatus they each brought.

Solution

When solving logic problems, setting up a chart to fill in is generally a good way to start.

Order of Arrival Snack Sports ApparatusXavierYvonne

Zak

Some of the given information is often more helpful than other information. For example, inthe second statement we learn that Xavier arrived second and brought a football. Now thethird statement gets us the fact that Yvonne arrived first and Zak arrived third. (This is truesince Yvonne arrived before Zak and she could not arrive second leaving only the first andthird spots left.) We can add this information to the chart.

Order of Arrival Snack Sports ApparatusXavier 2nd footballYvonne 1st

Zak 3rd

We can combine the first statement and the fourth statement. Yvonne did not bring cookies.The person who brought cookies also brought a baseball. This cannot be Xavier since hebrought a football. Therefore, Zak brought cookies and a baseball. Since there is only onepiece of sports apparatus unaccounted for, Yvonne must have brought the soccer ball. We willadd this new information to our chart.


Order of Arrival Snack Sports ApparatusXavier 2nd footballYvonne 1st soccer ball


We can now use the fifth statement to conclude that Xavier brought pretzels since the personbringing pretzels did not bring a soccer ball and Xavier is the only one without a snackaccounted for other than Yvonne (who brought the soccer ball).

Order of Arrival Snack Sports ApparatusXavier 2nd pretzels footballYvonne 1st soccer ball


The only snack unaccounted for is the licorice and Yvonne is the only one whose snack isunknown. Therefore, Yvonne brought licorice and our chart can be completed.

Order of Arrival Snack Sports ApparatusXavier 2nd pretzels footballYvonne 1st licorice soccer ball


The information is summarized in the chart but will be stated below for completeness.

• Yvonne arrived first bringing licorice and a soccer ball.

• Xavier arrived second bringing pretzels and a football.

• Zak arrived third bringing cookies and a baseball.


Problem of the Week

Problem C

Good to Plan and Prepare but ...

In preparing to write an examination, Stu Deus made the followingobservations:

• The exam had 20 questions.

• He estimated that he would spend 6 minutes per question.

• The exam would take him 2 hours to complete.

However, during the actual examination, Stu discovered some difficultquestions which each required 15 minutes to complete. He also discovered somequestions which were much easier than he expected and only took him 2minutes per question to complete. Seven of the questions, however, stillrequired 6 minutes each to complete. Surprisingly, Stu was still able tocomplete the exam in 2 hours.

How many of the 20 examination questions did Stu find difficult?


Problem of the Week



Problem

In preparing to write an examination, Stu Deus made the following observations: the exam had20 questions, he would spend 6 minutes per question, and the exam would take him 2 hours tocomplete. However, during the actual examination, Stu discovered some difficult questionswhich each required 15 minutes to complete. He also discovered some questions which weremuch easier than he expected and only took him 2 minutes per question to complete. Seven ofthe questions, however, still required 6 minutes each to complete. Surprisingly, Stu was stillable to complete the exam in 2 hours. How many of the 20 examination questions did Stu finddifficult?

Solution 1: Systematic Trial

Since 7 of the questions required 6 minutes each to complete, it took Stu 7 × 6 = 42 minutes tocomplete these questions. The total exam took 2 hours or 120 minutes. He had 120 − 42 = 78minutes to complete 20 − 7 = 13 questions.

Let d represent the number of difficult questions and e represent the number of easierquestions. We know that d + e = 13.

Since each difficult question took 15 minutes, it took 15d minutes to complete all of the difficultquestions. Since each easier question took 2 minutes, it took 2e minutes to complete all of theeasier questions. Since Stu’s total remaining time was 78 minutes, 15d + 2e = 78 minutes.

At this point we can pick values for d and e that add to 13 and substitute into the equation15d + 2e = 78 to find the combination that works. (We can observe that d < 6 since15 × 6 = 90 > 78. If this were the case, then e would have to be a negative number.)If d = 3 then e = 13 − 3 = 10. The time to complete these would be15 × 3 + 2 × 10 = 45 + 20 = 65 minutes and he would complete the exam in less than 2 hours.

If d = 4 then e = 13 − 4 = 9. The time to complete these would be15 × 4 + 2 × 9 = 60 + 18 = 78 minutes and he would complete the exam in exactly 2 hours.

Therefore, Stu found 4 of the questions to be more difficult and time-consuming than heexpected.

Solution 2 involves algebra and equation solving.


Problem


Solution 2: Using Algebra and Equations


Let d represent the number of difficult questions and (13 − d) represent the number of easierquestions.

Since each difficult question took 15 minutes, it took 15d minutes to complete all of thedifficult questions. Since each easier question took 2 minutes, it took 2(13 − d) minutes tocomplete all of the easier questions.

Since Stu’s total remaining time was 78 minutes,

15d + 2(13 − d) = 78

15d + 26 − 2d = 78

13d + 26 = 78

Subtracting 26 from both sides: 13d = 52

Dividing both sides by 13: d = 4



Problem of the Week

Problem C

CUT!

Kuri Uz is given some red licorice that is wrapped in a coil. Upon flatteningthe licorice, Kuri discovers that the total length is 60 cm. She then cuts thelicorice into two pieces such that the ratio of the lengths of the two pieces is7 : 3. Each piece is then bent to form a square.

What is the total area of the two squares?


Problem of the Week


CUT!Problem

Kuri Uz is given some red licorice that is wrapped in a coil. Upon flattening thelicorice, Kuri discovers that the total length is 60 cm. She then cuts the licoriceinto two pieces such that the ratio of the lengths of the two pieces is 7 : 3. Eachpiece is then bent to form a square. What is the total area of the two squares?

Solution

Since the licorice is cut in the ratio 7 : 3, let the longer piece be 7x cm and theshorter piece be 3x cm. Then 7x + 3x = 60 or 10x = 60 and x = 60 ÷ 10 = 6.Therefore, the longer piece is 7x = 7 × 6 = 42 cm and the shorter piece is3x = 3 × 6 = 18 cm.

Each of the two pieces is then bent to form a square. The perimeter of eachsquare is the length of the licorice used to form it. The side length of thelonger square is 42 ÷ 4 = 10.5 cm and the side length of the shorter square is18 ÷ 4 = 4.5 cm.

To find the area of each square, we multiply length by width. In effect, to findthe area of the square, we square the side length. The area of the larger squareis 10.5 × 10.5 = 10.52 = 110.25 cm2 and the area of the smaller square is4.5 × 4.5 = 4.52 = 20.25 cm2.

Therefore, if Kuri Uz cuts the licorice into pieces of length 42 cm and 18 cm,she can form two squares with area 110.25 cm2 and 20.25 cm2, respectively.The total area of the two squares is 110.25 + 20.25 = 130.5 cm2.

As a slight extension, the ratio of the area of the larger square to the area ofthe smaller square is

110.25 : 20.25 = 11025 : 2025 = 441 : 81 = 49 : 9 = 72 : 32.

The ratio of the perimeters is 7 : 3 and the ratio of the areas is 72 : 32. Ingeneral, if the ratio of the perimeters of two squares is a : b, is it true that theratio of the areas of the two squares is a2 : b2?


Problem of the Week

Problem C

Trickling Down

The number in an unshaded square in the grid below is obtained by adding thenumbers connected to it from the row above. For example, the 5 in the secondrow is obtained by adding the numbers, 2 and 3, connected to it in the rowabove. The numbers trickle down until the final square containing 2x.

Determine the value of x.

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Problem of the Week


Trickling Down

Problem

The number in an unshaded square in the grid is ob-tained by adding the numbers connected to it from therow above. For example, the 5 in the second row isobtained by adding the numbers, 2 and 3, connectedto it in the row above. The numbers trickle down un-til the final square containing 2x. Determine the value of x.

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Solution 1

In the first solution a trial and error type solution will be presented. We willpick a value for x and then complete the table.

Let x = 0. We would expect 2x = 2(0) = 0. However,after completing the table, 2x = 12 6= 0. Therefore,x 6= 0.

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& '#!

Let x = 5. We would expect 2x = 2(5) = 10. Aftercompleting the table, 2x = 27 6= 10. Therefore, x 6= 5.For our next trial we should choose a number lower thanx = 0.

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Let x = −5. We would expect 2x = 2(−5) = −10.However, after completing the table, 2x = −3 6= −10.Therefore, x 6= −5.

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Let x = −12. We would expect 2x = 2(−12) = −24. Af-ter completing the table, 2x = −24, the expected value.Therefore, x = −12.

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This solution is valid but not efficient. To reach the solution could take many,many trials. In the second solution, we will look at an algebraic solution.


Solution 2

For easy reference label all of the empty, unshaded squares in the diagram asshown.

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We can complete the second row using the addition rule for the table,a = 3 + x and b = x + 1.

Moving to the third row, c = 5 + a = 5 + 3 + x = 8 + x andd = a + b = 3 + x + x + 1 = 4 + 2x.

Finally, in the fourth row, 2x = c + d = 8 + x + 4 + 2x.

We can now solve the equation.

2x = 8 + x + 4 + 2x

2x− 2x = 8 + x + 4 + 2x− 2x Subtracting 2x from both sides

0 = 12 + x Simplifying

0− 12 = 12 + x− 12 Subtracting 12 from both sides

−12 = x

Therefore, x = −12.

An algebraic solution to this problem is much more efficient. Some studentsmay not quite have the necessary background to complete this solution ontheir own at this point.


Problem of the Week

Problem C

Can’t Get There From Here

A list consists of the integers

1, 2, 3, 5, 8, 13, 21, and 34.

A second list of integers is created that containsall possible integers that can be formed by addingtwo different integers from the first list. Thesmallest integer in this list is 1 + 2 = 3 and thelargest integer in this list is 21 + 34 = 55.

A third list of integers contains all of the integersfrom 3 to 55 that are not contained in the secondlist.

Determine the number of integers in the third list.

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For your information:

The integers in the first list are numbers found in the Fibonacci Sequence. Therectangle shown above is often referred to as The F ibonacci Rectangle. Thenumbers inside the squares correspond to the side length of the square thatcontains the number. The two smallest squares (the white one and the blackone) in the diagram each have side length 1. The rectangle spirals out fromthese two smallest squares. You may wish to investigate this sequence further.


Problem of the Week



Problem

A list consists of the integers 1, 2, 3, 5, 8, 13, 21, and 34. A second list of integers is createdthat contains all possible integers that can be formed by adding two different integers from thefirst list. The smallest integer in this list is 1 + 2 = 3 and the largest integer in this list is21 + 34 = 55. A third list of integers contains all of the integers from 3 to 55 that are notcontained in the second list. Determine the number of integers in the third list.

Solution

Notice that in the first list, each integer after the first two integers is the sum of the previoustwo integers.

It is easiest to determine the number of numbers in the second list. We are able tosystematically do this.

If 34 is added to each of the seven smaller integers in the first list, seven unique integers in therange 35 to 55 can be formed, namely 35, 36, 37, 39, 42, 47, and 55.

If 21 is added to each of the six smaller integers in the first list, six unique integers in the range22 to 34 can be formed, namely 22, 23, 24, 26, 29, and 34.

If 13 is added to each of the five smaller integers in the first list, five unique integers in therange 14 to 21 can be formed, namely 14, 15, 16, 18, and 21.

If 8 is added to each of the four smaller integers in the first list, four unique integers in therange 9 to 13 can be formed, namely 9, 10, 11, and 13.

If 5 is added to each of the three smaller integers in the first list, three unique integers in therange 6 to 8 can be formed, namely 6, 7, and 8.

If 3 is added to each of the two smaller integers in the first list, two unique integers in therange 4 to 5 can be formed, namely 4 and 5.

If the first two integers in the first list are added, the number 3 is obtained.

The second list contains 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28 integers. The number of integers from 3to 55 is the same as the number of integers from 1 to 53. There are 53 integers from 3 to 55.Therefore, the third list contains 53 − 28 = 25 integers.


Problem of the Week

Problem C

A Little Tiling

Tyler, the tiler, has an unlimited supply of square tiles. He has 1 cm by 1 cmtiles, 2 cm by 2 cm tiles, 3 cm by 3 cm tiles, and so on. Every tile has integerside lengths.

A rectangular table top with an 84 cm by 112 cm surface is to be completelycovered by identical square tiles, none of which can be cut. Tyler knows thathe can cover the table top with 1 cm × 1 cm tiles, 9 408 in total, since84× 112 = 9 408 cm2. However, Tyler wants to use the minimum number ofidentical tiles to complete the job in order to reduce his overall material cost.

Determine the minimum number of tiles required to completely cover the tabletop.


Problem of the Week


A Little Tiling

Problem

Tyler, the tiler, has an unlimited supply of square tiles. He has 1 cm by 1 cmtiles, 2 cm by 2 cm tiles, 3 cm by 3 cm tiles, and so on. Every tile has integerside lengths. A rectangular table top with an 84 cm by 112 cm surface is to becompletely covered by identical square tiles, none of which can be cut. Tylerknows that he can cover the table top with 1 cm × 1 cm tiles, 9 408 in total,since 84 × 112 = 9 408 cm2. However, Tyler wants to use the minimum numberof identical tiles to complete the job in order to reduce his overall materialcost. Determine the minimum number of tiles required to completely cover thetable top.

Solution

To use the smallest number of tiles we must use the largest tile possible. Thesquare tile must have sides less than or equal to 84 cm. If it was greater than84 cm, the tile would have to be cut to fit the width of the table.

Since the tiles are square and must completely cover the top surface, the sidelength of the tile must be a number that is a factor of both 84 and 112. In fact,since we need the largest side length, we are looking for the greatest commonfactor of 84 and 112.


1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, and 84.


1, 2, 4, 7, 8, 14, 16, 28, 56, and 112.

The largest number common to both lists is 28. Therefore the greatestcommon factor of 84 and 112 is 28. The required tiles are 28 cm × 28 cm.Since 84 ÷ 28 = 3, the surface is 3 tiles wide. Since 112 ÷ 28 = 4, the surface is4 tiles long. The minimum number of tiles required is 3 × 4 = 12 tiles.

The number of 28 cm × 28 cm tiles required to cover the top of the table is 12.This is the minimum number of tiles required.


Problem of the Week

Problem C

E - I - E - I - O

“Old MacDonald had a farm, E-I-E-I-O”, says the old children’s song. But OldMacDonald did have a farm! And on that farm he had some horses, cows, pigsand 69 water troughs for the animals to drink from. Only horses drank fromthe horse troughs, exactly two horses for each trough. Only cows drank fromcow troughs, exactly three cows per trough. And only pigs drank from pigtroughs, exactly eight pigs per trough. Old MacDonald’s farm has the samenumber of cows, horses and pigs.

How many animals does Old MacDonald have on his farm?


Problem of the Week


E - I - E - I - O

Problem

“Old MacDonald had a farm, E-I-E-I-O”, says the old children’s song. But OldMacDonald did have a farm! And on that farm he had some horses, cows, pigsand 69 water troughs for the animals to drink from. Only horses drank fromthe horse troughs, exactly two horses for each trough. Only cows drank fromcow troughs, exactly three cows per trough. And only pigs drank from pigtroughs, exactly eight pigs per trough. Old MacDonald’s farm has the samenumber of cows, horses and pigs. How many animals does Old MacDonaldhave on his farm?

Solution 1

Let n represent the number of each type of animal.

Since there are two horses for every horse trough, then n must be divisible by2. Since there are three cows for every cow trough, then n must be divisible by3. Since there are eight pigs for every pig trough, then n must be divisible by8. Therefore, n must be divisible by 2, 3 and 8. The smallest number divisibleby 2, 3, and 8 is 24. (This number is called the lowest common multiple orLCM , for short.)

If there are 24 of each kind of animal, there would be 24÷ 2 = 12 troughs forhorses, 24÷ 3 = 8 troughs for cows and 24÷ 8 = 3 troughs for pigs. Thiswould require a total of 12 + 8 + 3 = 23 troughs. Since there are 69 troughsand 69÷ 23 = 3, we require 3 times more of each type of animal. That is, therewould be 24× 3 = 72 of each type of animal. The total number of animals is72 + 72 + 72 or 216.

We can check the correctness of this solution. Since there are two horses forevery horse trough, there are 72÷ 2 = 36 horse troughs. Since there are threecows for every cow trough, there are 72÷ 3 = 24 cow troughs. Since there areeight pigs for every pig trough, there are 72÷ 8 = 9 pig troughs. The totalnumber of troughs is 36 + 24 + 9 = 69, as expected.


Solution 2

In this solution, algebra and equation solving will be used to solve the problem.


Since there are n horses and there are two horses for every horse trough, then there would be n2

troughs for horses. Since there are n cows and there are three cows for every cow trough, thenthere would be n

3troughs for cows. Since there are n pigs and there are eight pigs for every pig

trough, then there would be n8troughs for pigs. Since there are 69 troughs in total,

n

2+

n

3+

n

8= 69

12n

24+

8n

24+

3n

24= 69 common denominator 24

23n

24= 69 simplify the fractions

23n = 24× 69 multiply both sides by 24

23n = 1656 simplify

n =1656

23divide both sides by 23

n = 72

There are 72 of each type of animal, a total of 216 animals.

Solution 3

In this solution, ratios will be used to solve the problem.

Let n represent the number of each type of animal. The ratio of the number of troughsrequired for the pigs to the number of troughs required for the cows is

n

8:n

3=

3n

24:8n

24= 3n : 8n = 3 : 8.

Similarly, the ratio of the number of troughs required for the cows to the number of troughsrequired for the horses is 2 : 3 = 8 : 12. So the ratio of the number of troughs required for thepigs to the number required for the cows to the number required for the horses is 3 : 8 : 12.

Let the number of troughs required for the pigs be 3k, for the cows be 8k and for the horses be12k, for some positive integer value of k.

Since the total number of troughs required is 69, then

3k + 8k + 12k = 69

23k = 69

k = 3

The number of troughs required for the pigs is 3k = 9. There are 8 pigs at each trough. Thereare a total of 9× 8 = 72 pigs. Since there are the same number of each animal, there are also 72cows and 72 horses. There are a total of 72 + 72 + 72 = 216 animals on Old MacDonald’s farm.


Problem of the Week

Problem C

The Bead Goes On

Jill has been making two types of bracelets. The smaller bracelets contain fourbirthstone-coloured beads each and the larger bracelets contain sevenbirthstone-coloured beads each. So far Jill has used a total of 99 beads.

How many of each type of bracelet has Jill made? (There may be more thanone possible answer.)


Problem of the Week


The Bead Goes On

Problem

Jill has been making two types of bracelets. The smaller bracelets contain fourbirthstone-coloured beads each and the larger bracelets contain seven birthstone-colouredbeads each. So far Jill has used a total of 99 beads. How many of each type of bracelet has Jillmade? (There may be more than one possible answer.)

Solution

Let S represent the number of small bracelets and L represent the number of large bracelets.Since the small bracelets use 4 beads each, S bracelets would use 4× S or 4S beads in total.Since the large bracelets use 7 beads each, L bracelets would use 7× L or 7L beads in total.Since Jill has used a total of 99 beads, 4S + 7L = 99, with L and S integers greater than orequal to 0.

We can also determine a maximum value for L. Each large bracelet uses 7 beads. We knowthat 99÷ 7=̇14.1 so L must be a positive integer less than or equal to 14. We could at thispoint check all of the possible integer values for L from 0 to 14. However, we can narrow downthe possibilities even more.

In the equation, 4S + 7L = 99, 4S will always be an even number since 4 times any number isalways even. We then have an even number plus 7L equals the odd number 99. This meansthat 7L must be an odd number. (An even number plus an even number would be an evennumber, not an odd number.) For 7L to be an odd number, L must be odd. (If L is even, 7Lwould be even.) This observation reduces the possible values for L to the odd positive integersbetween 0 and 14, namely 1, 3, 5, 7, 9, 11, 13. Now we can test the possible values of L to seewhich ones, if any, produce a valid possibility for S.

Number Number Number Numberof Large of Beads of Beads of Small Valid or InvalidBracelets Used Remaining Bracelets Possibility?

L 7L 99− 7L (99− 7L)÷ 41 7× 1 = 7 99-7=92 92÷ 4 = 23 valid, it is an integer3 7× 3 = 21 99-21=78 78÷ 4 = 19.5 invalid, not an integer5 7× 5 = 35 99-35=64 64÷ 4 = 16 valid, it is an integer7 7× 7 = 49 99-49=50 50÷ 4 = 12.5 invalid, not an integer9 7× 9 = 63 99-63=36 36÷ 4 = 9 valid, it is an integer11 7× 11 = 77 99-77=22 22÷ 4 = 5.5 invalid, not an integer13 7× 13 = 91 99-91=8 8÷ 4 = 2 valid, it is an integer

There are four valid possibilities. Jill has either made 1 large bracelet and 23 small bracelets,or 5 large bracelets and 16 small bracelets, or 9 large bracelets and 9 small bracelets, or 13large bracelets and 2 small bracelets.


Problem of the Week

Problem C

Road Trip

A cyclist rode 560 km in seven days. Each day she travelled 15 km more thanthe day before.

Determine how far she rode on the seventh day.


Problem of the Week


Road Trip

Problem

A cyclist rode 560 km in seven days. Each day she travelled 15 km more thanthe day before. Determine how far she rode on the seventh day.

Solution 1

We will begin by representing the information on a diagram.

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The total distance is made up of seven trips with the same length as Day 1 plus15+2(15)+3(15)+4(15)+5(15)+6(15) = 15+30+45+60+75+90 = 315 km.So seven days riding the same distance as Day 1 would total560 − 315 = 245 km. On Day 1, the cyclist rode 245 ÷ 7 = 35 km.

On Day 7, the cyclist rode the Day 1 distance plus 6 × 15 = 90 km. The totaldistance travelled on the seventh day was 35 + 90 = 125 km.

This first solution deliberately avoids algebra and equations. Many solverswould be able to reason the solution in a similar way.

The second and third solutions will present more algebraic approaches.


Solution 2

Let x be the distance travelled on the first day. Then the cyclist ridesx + 15, x + 30, x + 45, x + 60, x + 75, and x + 90 km on days two throughseven, respectively. Then,

x + (x + 15) + (x + 30) + (x + 45) + (x + 60) + (x + 75) + (x + 90) = 560

7x + 15 + 30 + 45 + 60 + 75 + 90 = 560

7x + 315 = 560

7x + 315 − 315 = 560 − 315

7x = 2457x

7=

245

7x = 35

The cyclist rode 35 km on the first day. On the seventh day, the cyclist rodex + 90 = 35 + 90 = 125 km.

Solution 3

Let m be the distance travelled on the middle day, which is day four. On dayfive the cyclist would ride (m + 15) km; on day six the cyclist would ridem + 15 + 15 = (m + 30) km; and on day seven the cyclist would ridem + 30 + 15 = (m + 45) km. We can also work backwards by reducing thedistance the cyclist rode by 15 km. On day three the cyclist would ride(m− 15) km; on day two the cyclist would ride m− 15 − 15 = (m− 30) km;and on day one the cyclist would ride m− 30 − 15 = (m− 45) km. Then,

m + (m + 15) + (m + 30) + (m + 45) + (m− 15) + (m− 30) + (m− 45) = 560

7m + 15 + 30 + 45 − 15 − 30 − 45 = 560

7m + 15 − 15 + 30 − 30 + 45 − 45 = 560

7m = 5607m

7=

560

7m = 80

The cyclist rode 80 km on the fourth day. On the seventh day, the cyclist rodem + 45 = 80 + 45 = 125 km.

This third solution always works well when there is an odd number of terms ina sequence that increases (or decreases) by a constant amount.


Problem of the Week

Problem C

This One is For the Birds

One hundred pigeons are to be housed in identical-sized cages under thefollowing conditions:

• each cage must contain at least one pigeon;

• no two cages can contain the same number of pigeons; and

• no cages can go inside any other cage.

Determine the maximum number of cages required to house the pigeons.


Problem of the Week



Problem

One hundred pigeons are to be housed in identical-sized cages under the following conditions:





Solution

In order to maximize the number of cages, each cage must contain the smallest number of birdspossible. However, no two cages can contain the same number of pigeons. The simplest way todetermine this number is to put one pigeon in the first cage and then let the number of pigeonsin each cage after that be one more than the number of pigeons in the cage before it, until all100 pigeons are housed.

Put 1 pigeon in the first cage, 2 pigeons in the second cage, 3 pigeons in the third cage, and soon. After filling 12 cages in this manner, we have 1 + 2 + 3 + · · · + 11 + 12 = 78 pigeonshoused. After putting 13 pigeons in the thirteenth cage, 78 + 13 = 91 pigeons are housed.There are 9 pigeons left to house. But we already have a cage containing 9 pigeons. Theremaining 9 pigeons must be distributed among the existing cages while maintaining thecondition that no two cages contain the same number of pigeons.

The most obvious way to do this is to put the 9 pigeons in the last cage which already contains13 pigeons. This would mean that the final cage would contain 13 + 9 = 22 birds! A bettersolution might be to increase the number of birds in each of the final nine cages by one birdeach. This solution is summarized below:

Cage # 1 2 3 4 5 6 7 8 9 10 11 12 13# of birds 1 2 3 4 6 7 8 9 10 11 12 13 14

The maximum number of cages required is 13. If you had 14 cages, with the first cage holding1 pigeon and each cage after that holding one more pigeon than the cage before, you couldhouse 105 pigeons, five more than the number of pigeons that we have.


Problem of the Week

Problem C

Overlapping Areas

The area of 4ACD is twice the area of square BCDE. AC and AD meet BEat K and L respectively such that KL = 6 cm.

If the side length of the square is 8 cm, determine the area of 4AKL.

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Problem of the Week


Overlapping Areas

Problem

The area of 4ACD is twice the area of square BCDE. AC andAD meet BE at K and L respectively such that KL = 6 cm.If the side length of the square is 8 cm, determine the area of4AKL.

Solution 1

In the first solution we will find the area of square BCDE, thearea of 4ACD and the area of trapezoid KCDL.

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To find the area of a square, multiply the length times the width. To find thearea of a trapezoid, multiply the sum of the lengths of the two parallel sides bythe height and divide the product by 2.

Area of square BCDE = 8× 8

= 64 cm2


= 2× 64

= 128 cm2

In trapezoid KCDL, the two parallel sides are KL and CD, and the height isthe width of square BCDE, namely BC.

Area of trapezoid KCDL = (KL + CD)×BC ÷ 2

= (6 + 8)× 8÷ 2

= 14× 8÷ 2

= 56 cm2

Area 4AKL = Area 4ACD − Area of trapezoid KCDL

= 128− 56

= 72 cm2
