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### Probability of a Sum

```Date: 08/16/2002 at 23:13:03
From: Mudit
Subject: Probability

If the integers m and n are chosen at random between 1 and 100, what
is the probability that a number of the form 7^m + 7^n is divisible
by 5?
```

```
Date: 08/17/2002 at 02:51:20
From: Doctor Greenie
Subject: Re: Probability

Hello, Mudit -

If you look at the congruences 7^n (mod 5) for n = 1 to 100, you will
find that

7^n == 2 (mod 5) for integers n of the form 4k+1;
7^n == 4 (mod 5) for integers n of the form 4k+2;
7^n == 3 (mod 5) for integers n of the form 4k+3;
7^n == 1 (mod 5) for integers n of the form 4k

Of the numbers from 1 to 100, exactly 1/4 of them are of the form
4k+1, exactly 1/4 of them are of the form 4k+2, exactly 1/4 of them
are of the form 4k+3, and exactly 1/4 of them are of the form 4k.
Once you choose one of the numbers between 1 and 100 for m, the
choice for n, to make the sum 7^m + 7^n divisible by 5, is limited to
the numbers in exactly one of these groups:

If you choose m so that 7^m == 1 (mod 5), then to have 7^m + 7^n
divisible by 5, you need to choose n so that 7^n == 4 (mod 5).

If you choose m so that 7^m == 2 (mod 5), then to have 7^m + 7^n
divisible by 5, you need to choose n so that 7^n == 3 (mod 5).

If you choose m so that 7^m == 3 (mod 5), then to have 7^m + 7^n
divisible by 5, you need to choose n so that 7^n == 2 (mod 5).

If you choose m so that 7^m == 4 (mod 5), then to have 7^m + 7^n
divisible by 5, you need to choose n so that 7^n == 1 (mod 5).

Therefore, no matter what number you choose for the first of your two
choices for m or n, the probability is 1/4 that the number you choose
for the second one will make your sum divisible by 5.

The probability that two numbers m and n chosen at random from the
numbers 1 to 100 will make the sum 7^m + 7^n divisible by 5 is 1/4.

I hope this helps.  Please write back if you have any further

- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Number Theory
College Probability

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