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2010-2011 SMML PLAYOFF ROUND 1 1. What is the smallest two-digit number having an odd number of distinct factors? 2. How many positive integers are factors of both 360 and 2700? 3. The smallest number divisible by each of the digits 1, 2, 3, … , 9 can be written in the form: a w b x c y d z where a, b, c and d are prime. Find a + b + c + d + w + x + y + z .

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Page 1: 2010-2011 SMML PLAYOFF ROUND 1doversherborn.comcastbiz.net/highschool/academics/math/...2010-2011 SMML PLAYOFF ROUND 3 1. The base of an isosceles triangle is one side of a square

2010-2011 SMML PLAYOFF ROUND 1

1. What is the smallest two-digit number having an odd number of distinct factors? 2. How many positive integers are factors of both 360 and 2700? 3. The smallest number divisible by each of the digits 1, 2, 3, … , 9 can be written in the form:

awbxcydz where a, b, c and d are prime. Find

a + b + c + d + w + x + y + z .

Page 2: 2010-2011 SMML PLAYOFF ROUND 1doversherborn.comcastbiz.net/highschool/academics/math/...2010-2011 SMML PLAYOFF ROUND 3 1. The base of an isosceles triangle is one side of a square

2010-2011 SMML PLAYOFF ROUND 2

1. Pete found a used car listed with a dealer for $2400. He was able to talk the dealer into selling him the car at a 15% discount. Pete later sold the vehicle to Luke for 15% more than he paid. What was Pete’s profit after selling the car to Luke? 2. Fred earns $4 more in 3 hours than Ethel earns in 2 hours. At the end of an 8-hour day, Ethel earns $14 more than Fred. How much does Fred earn in a 40-hour week? 3. Alan can build one “Kiddie Play” playhouse in 12 hours. Bob can build one “Kiddie Play” playhouse in 6 hours. Carl can build one “Kiddie Play” playhouse in 8 hours. How long would it take them to build 6 “Kiddie Play” playhouses if they were to all work together?

Page 3: 2010-2011 SMML PLAYOFF ROUND 1doversherborn.comcastbiz.net/highschool/academics/math/...2010-2011 SMML PLAYOFF ROUND 3 1. The base of an isosceles triangle is one side of a square

2010-2011 SMML PLAYOFF ROUND 3

1. The base of an isosceles triangle is one side of a square as shown. The triangle and square have the same area. Find the ratio of a side of the square to the height of the triangle.

2. Three spheres of radius 6 are stacked in a cylinder with a cross-section as shown. What is the volume, in cubic units, of the space inside the cylinder, but outside the spheres? Leave your answer in terms of π. 3. An isosceles triangle rotated about the perpendicular bisector of its base forms a cone. An isosceles trapezoid rotated about the perpendicular bisector of its bases forms a frustrum. Find the volume, in cubic units, of a frustrum formed as described from an isosceles trapezoid having height of 3 and bases of 12 and 16. Leave your answer in terms of π.

Page 4: 2010-2011 SMML PLAYOFF ROUND 1doversherborn.comcastbiz.net/highschool/academics/math/...2010-2011 SMML PLAYOFF ROUND 3 1. The base of an isosceles triangle is one side of a square

2010-2011 SMML PLAYOFF ROUND 4

1. The fifth term of a geometric sequence is 81 times the first term. What is the product of all possible values of the common ratio in this sequence? 2. Find the sum of all three-digit numbers divisible by 6. 3. For what value of y will the coefficient of x5 in the binomial expansion of

3x 2 +yx

7

equal

−105?

Page 5: 2010-2011 SMML PLAYOFF ROUND 1doversherborn.comcastbiz.net/highschool/academics/math/...2010-2011 SMML PLAYOFF ROUND 3 1. The base of an isosceles triangle is one side of a square

2010-2011 SMML PLAYOFF ROUND 5

1. Let

P x( ) = x 2 − 5x + 6 and

Q x( ) = x 2 + 6x + 8 . Find the sum of the roots of

R x( ) if

R x( ) = P x( )Q x( ) . 2. Find the sum of the coefficients in the polynomial expansion of:

P x( ) = 2x − 3( )5 3x +1( )3 3. The three cube roots of 8 form an equilateral triangle when plotted in the complex plane. Find the area of this triangle.

Page 6: 2010-2011 SMML PLAYOFF ROUND 1doversherborn.comcastbiz.net/highschool/academics/math/...2010-2011 SMML PLAYOFF ROUND 3 1. The base of an isosceles triangle is one side of a square

2010-2011 SMML PLAYOFF

TEAM ROUND - No Calculator 1. The mean, median and mode of a set of 6 numbers is 5. If two of the members of this set are 3’s, what is the largest number in the set?

2. Find all values of x such that

4x 2 + 6x − 24 < x 3 + x 2 − 3x + 3. 3. The digits 1 through 9 are written on individual slips of paper and placed in a hat. 3 slips of paper are chosen from the hat at random and in no particular order. What is the probability that the product of the three digits chosen will be a perfect square? State your answer as a fraction in simplest form.

Page 7: 2010-2011 SMML PLAYOFF ROUND 1doversherborn.comcastbiz.net/highschool/academics/math/...2010-2011 SMML PLAYOFF ROUND 3 1. The base of an isosceles triangle is one side of a square

2010-2011 SMML PLAYOFF ROUND 1

1. What is the smallest two-digit number having an odd number of distinct factors? When looking at a factor table for any number, all factors are paired. Thus, the only way for there to be an odd number of factors is for one factor to be paired with itself. This implies the number must be a perfect square. The smallest two-digit perfect square is 16. Answer: 16 2. How many positive integers are factors of both 360 and 2700?

360 = 23 ⋅ 32 ⋅ 52700 = 22 ⋅ 33 ⋅ 52

Common factors have: 0, 1, or 2 factors of 2 = 3 possibilities 0, 1, or 2 factors of 3 = 3 possibilities 0 or 1 factor of 5 = 2 possibilities

3⋅ 3⋅ 2 =18 Answer: 18 common factors 3. The smallest number divisible by each of the digits 1, 2, 3, … , 9 can be written in the form:

awbxcydz where a, b, c and d are prime. Find

a + b + c + d + w + x + y + z . To be divisible by each of the digits 1 through 9, a number must have at least 3 prime factors of 2, 2 prime factors of 3, one prime factor of 5 and one prime factor of 7. Thus, the smallest number divisible by each of these digits would be:

23 ⋅ 32 ⋅ 51⋅ 71. Thus we are looking for 2 + 3 + 3 + 2 + 5 + 1 + 7 + 1 = 24. Answer: 24

Page 8: 2010-2011 SMML PLAYOFF ROUND 1doversherborn.comcastbiz.net/highschool/academics/math/...2010-2011 SMML PLAYOFF ROUND 3 1. The base of an isosceles triangle is one side of a square

2010-2011 SMML PLAYOFF ROUND 2

1. Pete found a used car listed with a dealer for $2400. He was able to talk the dealer into selling him the car at a 15% discount. Pete later sold the vehicle to Luke for 15% more than he paid. What was Pete’s profit after selling the car to Luke? Pete bought the car at 85% of the list price of $2400.

.85⋅ 2400 = 2040 Therefore, Pete’s purchase price was $2,040. Pete then sold the car for 15% more than he paid.

1.15⋅ 2040 = 2346 Therefore, Luke purchased the car for $2,346.

2346 − 2040 = 306 Answer: $306 2. Fred earns $4 more in 3 hours than Ethel earns in 2 hours. At the end of an 8-hour day, Ethel earns $14 more than Fred. How much does Fred earn in a 40-hour week? Let x = Fred’s hourly wage and y = Ethel’s hourly wage. Then:

3x = 2y + 4 ⇒ 3x − 2y = 4 and

8x +14 = 8y ⇒ 8x − 8y = −14 ⇒ 4x − 4y = −7

Multiply the first equation by 2 and subtract the second equation to yield:

2x =15 ⇒ x = 7.5 Thus, Fred’s 40 hour earnings is

40⋅ 7.5 = 300 . Answer: $300 3. Alan can build one “Kiddie Play” playhouse in 12 hours. Bob can build one “Kiddie Play” playhouse in 6 hours. Carl can build one “Kiddie Play” playhouse in 8 hours. How long would it take them to build 6 “Kiddie Play” playhouses if they were to all work together?

In one hour, Alan can complete

112

of a playhouse, Bob can complete

16

of a playhouse,

and Carl can complete

18

of a playhouse. Together, they can complete

112

+16

+18

=224

+424

+324

=924

=38

of a playhouse in one hour.

1 hour38

playhouse⋅ 6 playhouses =

8 hour3 playhouses

⋅ 6 playhouses =16 hours

Answer: 16 hours

Page 9: 2010-2011 SMML PLAYOFF ROUND 1doversherborn.comcastbiz.net/highschool/academics/math/...2010-2011 SMML PLAYOFF ROUND 3 1. The base of an isosceles triangle is one side of a square

2010-2011 SMML PLAYOFF ROUND 3

1. The base of an isosceles triangle is one side of a square as shown. The triangle and square have the same area. Find the ratio of a side of the square to the height of the triangle. Let x be the length of one side of the square. Then, the area of the square = the area of the triangle = x2.

Since the area of the triangle, is

12bh and b = x, we have

areatriangle =

12bh =

12xh = x 2 ⇒

12

=xh

Answer:

12

or 1:2

2. Three spheres of radius 6 are stacked in a cylinder with a cross-section as shown. What is the volume, in cubic units, of the space inside the cylinder, but outside the spheres? Leave your answer in terms of π.

Volumesphere =

43π⋅ 63 = 288π Total volume of spheres =

3⋅ 288π = 864π

The height of the cylinder is 6 times the radius of the sphere = 36. Volumecylinder =

π⋅ 62(36) =1296π . The difference is

1296π − 864π = 432π Answer:

432π 3. An isosceles triangle rotated about the perpendicular bisector of its base forms a cone. An isosceles trapezoid rotated about the perpendicular bisector of its bases forms a frustrum. Find the volume, in cubic units, of a frustrum formed as described from an isosceles trapezoid having height of 3 and bases of 12 and 16. Leave your answer in terms of π. The trapezoid described is the lower portion of the triangle shown. If the entire triangle is rotated around the axis, the resulting volume will be:

Volumecone =

13π⋅ 82 ⋅ 12 = 256π

Likewise, if the upper triangle is rotated around the axis, the resulting volume will be:

Volumecone =

13π⋅ 62 ⋅ 9 =108π

The difference in the two volumes is: Volumefrustrum =

256π −108π =148π

Answer:

148π

8

3 6

9

Page 10: 2010-2011 SMML PLAYOFF ROUND 1doversherborn.comcastbiz.net/highschool/academics/math/...2010-2011 SMML PLAYOFF ROUND 3 1. The base of an isosceles triangle is one side of a square

2010-2011 SMML PLAYOFF ROUND 4

1. The fifth term of a geometric sequence is 81 times the first term. What is the product of all possible values of the common ratio in this sequence?

t5 = t1r4 ⇒ 81t1 = t1r

4 ⇒ r4 = 81⇒ r = ±3 also (3)(-3)(3i)(-3i) => -81

Answer: - 9 or -81 2. Find the sum of all three-digit numbers divisible by 6. The 3-digit numbers divisible by 6 form an arithmetic sequence with first term 102 and final term 996. 996 = 102 + 6(n – 1)

⇒ n = 150 The sum of all three-digit numbers divisible by 6 is therefore:

Sum =150 102 + 996( )

2= 82,350

Answer: 82,350 3. For what value of y will the coefficient of x5 in the binomial expansion of

3x 2 +yx

7

equal

−105?

The x5 term in the binomial expansion will be of the form:

74

3x 2( )4 y

x

7−4

⇒ 35 81x 8( ) y 3

x 3

= 2835x 5y 3 = −105x 5 ⇒ y 3 = −

127

⇒ y = −13

Answer:

y = −13

Page 11: 2010-2011 SMML PLAYOFF ROUND 1doversherborn.comcastbiz.net/highschool/academics/math/...2010-2011 SMML PLAYOFF ROUND 3 1. The base of an isosceles triangle is one side of a square

2010-2011 SMML PLAYOFF ROUND 5

1. Let

P x( ) = x 2 − 5x + 6 and

Q x( ) = x 2 + 6x + 8 . Find the sum of the roots of

R x( ) if

R x( ) = P x( )Q x( ) .

R x( ) = x 2 − 5x + 6( ) x 2 + 6x + 8( ) = x − 3( ) x − 2( ) x + 4( ) x + 2( ) Therefore, the roots are 3, 2, -4 and -2. Their sum is -1. Answer: -1 2. Find the sum of the coefficients in the polynomial expansion of:

P x( ) = 2x − 3( )5 3x +1( )3 The sum of the coefficients is simply the result when x = 1. Therefore, the sum of the coefficients in the expansion of the given polynomial is:

2⋅ 1− 3( )5 3⋅ 1+1( )3 = −1( )5 4( )3 = −64 Answer: -64 3. The three cube roots of 8 form an equilateral triangle when plotted in the complex plane. Find the area of this triangle. Each of the cube roots of 8 lies on a circle of radius 2 centered at the origin in the complex plane. Each of the three triangles shown within the equilateral triangle is an isosceles triangle with legs of length 2 and included angle of

120 . Therefore, the area of

each isosceles triangle is

122( ) 2( )sin120 = 3⇒ the area of the equilateral triangle is

3 3 . Answer:

3 3

Page 12: 2010-2011 SMML PLAYOFF ROUND 1doversherborn.comcastbiz.net/highschool/academics/math/...2010-2011 SMML PLAYOFF ROUND 3 1. The base of an isosceles triangle is one side of a square

2010-2011 SMML PLAYOFF

TEAM ROUND - No Calculator 1. The mean, median and mode of a set of 6 numbers is 5. If two of the members of this set are 3’s, what is the largest number in the set? If the mode is 5, and there are two 3’s, there must be at least three 5’s in the set. If the mean of the 6 numbers is 5, the sum of the numbers must be 30. Since

2⋅ 3+ 3⋅ 5 = 21, the remaining member of the set must be 9. Answer: 9 2. Find all values of x such that

4x 2 + 6x − 24 < x 3 + x 2 − 3x + 3. This inequality is equivalent to

−x 3 + 3x 2 + 9x − 27 < 0⇒ −x 2 x − 3( ) + 9 x − 3( ) < 0⇒ − x + 3( ) x − 3( )2 < 0 When graphed, this yields a cubic with roots at x = 3 and x = -3 and which approaches infinity in QII and negative infinity in QIV. It is tangent to the x-axis at x = 3. Therefore, the values of x which satisfy this inequality are x > -3 and

x ≠ 3. This is equivalent to -3 < x < 3 or x > 3. Answer: -3 < x < 3 or x > 3 will also accept (-3<x<3) U (x>3) or (-3,3) U (3, ∞) 3. The digits 1 through 9 are written on individual slips of paper and placed in a hat. 3 slips of paper are chosen from the hat at random and in no particular order. What is the probability that the product of the three digits chosen will be a perfect square? State your answer as a fraction in simplest form.

There are

93

= 84 total ways to choose 3 slips from the 9 in the hat. In order to

determine which products will yield perfect squares, we consider the prime factorization of each.

1, 2, 3, 22, 5, 2⋅ 3, 7, 23, 32

There are only 6 ways to make a perfect square by multiplying any 3 of these numbers:

1⋅ 2⋅ 23,

1⋅ 22 ⋅ 32,

2⋅ 3⋅ 2⋅ 3( ),

2⋅ 22 ⋅ 23,

2⋅ 23 ⋅ 32,

3⋅ 2⋅ 3( )⋅ 23 Thus, the probability that the product of the 3 numbers will be a perfect square

is

684

=114

.

Answer:

114

Page 13: 2010-2011 SMML PLAYOFF ROUND 1doversherborn.comcastbiz.net/highschool/academics/math/...2010-2011 SMML PLAYOFF ROUND 3 1. The base of an isosceles triangle is one side of a square

Answers for all rounds Round 1

1. 16 2. 18 common factors 3. 24

Round 2

1. $306 2. $300 3. 16 hours

Round 3

1. ½ or 1:2 2. 432π 3. 148π

Round 4

1. -9 or -81 2. 82,350 3. -1/3

Round 5

1. -1 2. -64 3.

3 3 Team Round

1. 9 2. -3<x<3 or x>3

will also accept (-3<x<3) U (x>3) or (-3,3) U (3, ∞) 3. 1/14