666 + 666 + 666Author(s): Monte J. ZergerSource: The Mathematics Teacher, Vol. 91, No. 2 (FEBRUARY 1998), pp. 151-152Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/27970458 .

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(Continued from page 120)

that a calculator can produce inaccurate or misleading graphs of trigonometric functions of the form y = sin (ax). As a result of our class discussion, one of my students, Oliver Kosut, who was a freshman, investigated the

problem of predicting the number of waves in the window from Xmin = 0 to Xmax = 2/rthat the calculator (in this case, a TI-83) would produce. His work follows.

Oliver Kosut Zoya Voskoboynikov ps96157@itsa. ucsf.edu Lick-Wilmerding High School

San Francisco, CA 94112

666 + 666 + 666 What a perfect way to wind down the millennium! After all, 6 is the first perfect number, and 666 has been unfairly portrayed far too

long.

Calculator magic 1. Enter 666 into your calculator,

and turn the calculator upside down. The least common mul tiple of the two numbers dis

played in this manner is 1998. Since 1998 = LCM (666, 999), this year is the first year in the common era that is divisible by both 666 and its "calculator opposite." The common era

began with 1 A.D. because no 0 A.D. occurred. The numbers 6 and 9 may be opposites, but they seem to agree on some

thing?1998. It can easily be

represented using either nine 6s or six 9s, that is, 666 + 666 + 666 or 999+ 999.

2. Next, enter 1998 into your cal culator, and again rotate it 180

degrees. Compute the difference between the two numbers.

8661-1998 = 6663.

Inserting a multiplication sign between the 666 and the 3 will return you to 1998.

3. For any positive integer , <j>(n) represents Euler's ^-function, the number of numbers less than and relatively prime to it. For example 0(9) = 6, since the six numbers 1,2,4,5, 7, and 8 are relatively prime to 9.

0(1998) can be computed to be 648. Compare these equations:

1998 = 666+666 + 666

( 998) =

(6?6?6) + (6?6?6)

+ (6?6*6)

4. Consider the rectangle formed

by the 1,2,4,5,7, and 8 keys on your calculator:

Form any six-digit number by starting with any odd digit and reading the number cyclically, either clockwise or counter clockwise, around the rectangle. These numbers are all exactly divisible by 1998. For example 125,874= 1998-63.

Niven numbers A Niven number is any number that is divisible by the sum of its

digits. Since the digits in 1998 total 27 and since 1998 = 27 ? 74, 1998 is a Niven number. When was the last Niven year? When was the last Niven year divisible by 27? What will the next Niven year be? What is the next Niven year divisible by 27?

The reversal of 1998 is also divisible by 27.

Powers

1. An amazing property of 1998 was apparently discovered by I. A. Sakmar. First, cube it as in the first equation. Next, con vert this equation into the sub sequent equation merely by deleting the exponent and converting the commas to plus signs:

19983 = 7,976,023,992

1998 = 7 + 976+023 + 992

Curiously, 18,198,1998, 19 998,... all share this prop erty if the digits in their cubes are appropriately grouped.

183 = 5,832

18 = 5+8+3+2

1983 =7,762,392

198 = 7 + 76+23 + 92

199983= 7,997,600,239,992

19998 = 7+9976+0023+9992

2, Michael Keith has discovered another surprising property of 1998:

il

Vol 91, No. 2 ? February 1998 151

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1998 = (1 + 9 + 9 + 8) + (l3 + 93 + 93 + 83)

This number is the largest one

equal to the sum of its digits and the cubes of its digits, and only five other such numbers exist: 12,30,870,960, and? oh, yes?666 itself.

3. By moving from cubes to fourth powers, the first four natural numbers can be com bined in a beautiful way to yield the abbreviated form of this amazing year:

l4 + 24 + 34 = 98 Pi and e

Dividing 1998 by 636 gives the first five digits of , whereas dividing it by 735 produces the first 4 digits of e:

^?3.1415... 636

^?2.718... 735

Number fun First note that 1998 is the (999 + 999)th year of the common era, the 999th year of the second millennium, and the 9th year of this decade. Then assign the value 9n to the nth letter of the alphabet, as follows:

A = 9,? = 18,C = 27,...,Z = 234

Next sum the values of the letters in NINETEEN NINETY EIGHT, and you will obtain exactly 1998.

Bibliography: Keith, Michael, "The Number 666." Journal of Recreational Mathematics 15 (1982-1983): 85-86. Sakmar, I. A. "Chanson San Paroles."

Mathematics Magazine 69 (February 1996): 62.

Monte J. Zerger mjzerger@adams.edu Adams State College Marnosa, CO 81102

Horizontal asymptotes: What they are not A common misconception about a horizontal asymptote of a func tion is that the function "gets closer and closer to the asymptote

but never gets there." Expressed in the appropriate "mathematese* we have the following: Faux definition. Let fix) be a

function defined for all e (-e?, oo). Then, the line y = k

is a horizontal asymptote off if, as ; increases, the graph of f gets closer and closer to y = k but f(x) t k for all e (-<?, oo).

We can easily represent several flaws in this "definition." We do so in the following two examples.

fit) ih

Fig. 1 A graph of fit) = e~t/2 cos4i

Example 1. Suppose that a

weight is attached to a vertically mounted spring and that the weight is set in motion by dis placing it vertically from its equilibrium position and then releasing it. If we let fit) be the displacement of the weight from its equilibrium position at time t, then, assuming the presence of friction and with the appropriate choice of constants and units, we find that fit) = e~m cos 4L (See fig. 1.) Notice that /"has a hori zontal asymptote of y = 0. Also, f takes on the value 0 for all values of t such that

, 2/1 + 1 t =

8

for some integer n. Certainly this example illustrates that fern "get there"! This example also illustrates the falseness of the "gets closer and closed assumption.

Example 1, which illustrates the so-called damped harmonie motion, is especially appealing because of its physical origin. Students can easily visualize masses oscillating on springs. A less physically interesting, But equally valid, example is the following.

Example2. Considerate) = e~x and the line y = -1. (See fig. 2.) Since g is a decreasing positive

function, it does in fact get closer and closer to the line y = -1 as increases. Also, g never attains the value -1. However, y = -1 is not a horizontal asymptote oig.

g(x)

3^

-1 OL 1 2 3

_I y = -i -1 :

Fig. 2 A graph oig(x)

= e~x

Example 2 illustrates that an accurate definition of horizontal asymptote must somehow replace the "gets closer and closer" con

cept with a "gets arbitrarily close" concept.

From these examples, we see that a soft approach to the defini tion of horizontal asymptote can be problematic. Only with the rigor of limits can this difficult concept be dealt with precisely.

Robert Gardner gardwrr@etsuarts.etsu-tn.edu East Tennessee State

University Johnson City, TN 37614

Creative math I In his article "From Fibonacci Numbers to Geometric Sequences and the Binet Formula by Way of the Golden Ratio" (May 1997), Angelo S. DiDomenico describes a fascinating connection among the Fibonacci numbers, the golden ratio, and geometric sequences. He explains their role in the development of the Binet formula, which is given as

where Fn is the nth Fibonacci number,

(1) F = 1 a -?

a 2

and

?= 1-V5

I was amazed at the beauty and elegance of this remarkable development. We discussed the following derivation of this for mula in our high school's creative math class.

We began with the following surprising observation:

a=1a+0

a2 =la+l

a3=2a+l

a4=3a+2

a5=5a+3

(2)

an=Fna+F^x

Similarly we discovered the following:

j8x=lj8+0

?2 =1)3+1

?3 =20+1 (3) \?*=S?+2

^=50+3

?r=FJ+F_1

We then proved the generaliza tions in (2) and (3) by mathemati cal induction. If we next subtract the respective sides of the gener alization in (2) and (3), we get

(4) an-?n = Fn{a-?). Since a - ? = V5, substituting this value into (4) and solving for Fn yields

(5) a F

By adding the respective sides of the generalizations in (2) and (3), we obtain

(6) a* + j8n = Fn(a + 0) + 2Fn_1.

Substituting a + ? = 1 into (6) and rewriting gives

(7) ?n + j8n=^ + ̂-i+^-i.

Knowing that Fn + Fn_i=Fn + 1 and Fn+1+Fn_i= Ogives

(8) a* + j3n = L?, where Ln is the nth Lucas number.

Paul </. Maiorano Framingham High School Framingham, MA 01701

152 THE MATHEMATICS TEACHER

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