Sanderson M. Smith

Home | About Sanderson Smith | Writings and Reflections | Algebra 2 | AP Statistics | Statistics/Finance | Forum

[This is a note I sent to my students relating to a problem on the California Mathematics League competition sheet in November, 2001.]

Many of you have asked about problem #6 on the CML competition. It is a very interesting problem. For those of you who may be interested, I'll provide a bit of an analysis.

The problem -->

What are both primes p > 0 for which 1/p has a purely periodic decimal expansion with a period 5 digits long?

[NOTE: 1/37 = 0.027027027027 .....starts to repeat immediately, so it is purely periodic. Its period is 3 digits long.]

Of course one must know what a prime number is... and I'm sure all of you know that a prime number is an integer greater than 1 (by definition, the number 1 is not a prime) that is evenly divisible only by itself and 1.

The first few primes are 2,3,5,7,11,13,17,19,23,29, ...

I found the two requested prime numbers, but I didn't do it in the most sophisticated fashion. I used technology (my wonderful TI-83 graphics calculator) and a trial-and-error approach. (Some great historical mathematical discoveries came as a result of trial and error.) The statement of the problem indicates there are two primes satisfying the stated condition, so I just set out to find them. Here's how I used my calculator. I started by entering storing the number 7 in x, and then calculating 1/x and looking at the decimal representation. (Note the use of the : so that I could get two commands for the price of one.)

7->x: 1/x

This produced .1428571429....

The period here is 6... doesn't satisfy stated requirements.

Using ENTRY, I got my commands back, changed the 7 to 11, pushed ENTER, and found that

1/11 = .0909090909....

The period here is 2.

By continuing this process and substituting primes, I found that

1/41 = .0243902439 .... (period = 5)

and

1/271 = .0036900369.... (period = 5).

OK, this definitely wasn't the most sophisticated approach to the problem, but given the information in the PREMISE of the problem, the solution was obtainable by trial and error.

ANY EXERCISE (TEST) SHOULD BE A LEARNING EXPERIENCE....

A bit later, I did some fiddling with the problem and came up with this (sophisticated) approach. [ALGEBRA IIers... if you will read carefully, I know you can follow the reasoning. MATH IS A LANGUAGE, and I am communicating in the language.]

Let P = a prime satisfying the stated conditions.

Then 1/P = 0.abcdeabcdeabcde.....

Note: The letters a,b,c,d,e represent digits. The expression abcde is not a product.

OK, 1/P = 0.abcdeabcdeabcde...

==> 10000(1/P) = abcde.abcdeabcdeabcde.....

Hence, 100000(1/P) - (1/P) = 99999(1/P) =(abcde) ==> 1/P = (abcde)/99999.

Now factoring 99999 yields 99999 = (3)(3)(41)(271).

Hence 1/P =(abcde)/[(3)(3)(41)(271)].

This means that the only possibilities for P (remember, it is a prime) are 3, 41, and 271, BECAUSE....

the only way abcde can be "reduced" to 1 will be if

(i) it is divisible by (3)(41)(271), yielding 1/3;

(ii) it is divisible by (3)(3)(271), yielding 1/41;

(iii) it is divisible by (3)(3)(41), yielding 1/271.

Remember now... these are only possibilities. They have to be checked. Since 1/3 = 0.333333... (period = 1), it doesn't satisfy required condition. As indicated above

1/41 = .0243902439 .... (period = 5)

and

1/271 = .0036900369.... (period = 5)

do satisfy the stated conditions. Hence, the prime numbers requested are 41 and 271.

"The primary question is not 'WHAT do we know, but HOW do we know it.' "

-ARISTOTLE (around 350 B.C.)

"For the things of this world cannot be made known without a knowledge of mathematics."

-ROGER BACON

[HISTORICAL NOTE: Euclid (300 B.C.) produced a very simple proof that the number of prime numbers is infinite.]

Home | About Sanderson Smith | Writings and Reflections | Algebra 2 | AP Statistics | Statistics/Finance | Forum

Previous Page | Print This Page

Copyright © 2003-2009 Sanderson Smith