Probability of a SumDate: 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 questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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