Sanderson M. Smith

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[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)


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.


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)


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."


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





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