If the integers m and n are chosen at random between 1 and 100, then the probability that the number of the form 7 + 7 is divisible by 5 equals

Dear Shiva,

7m + 7n = 7(m+ n). If this is divisible by 5, then m+n must be divisible by 5, since 7 is not. So the question is asking: if m and n are chosen randomly from between 1 and 100, what is the probability that m+n is divisible by 5.

Now, I assume the question means "If m and n are chosen randomly from between 1 and 100 inclusive"; the question should make clear whether 1 and 100 are permissible values of m and n. If the range is inclusive, then the answer will be 1/5. There are a few ways to see this; we could consider units digits, since we''re dealing with division by 5. No matter what I choose for m, there will be two possible units digits for n that will make m+n a multiple of 5. So for example, if m=23, then the units digit of n must either be 2 or 7 in order that m+n ends in 5 or 0. Since there are ten possible units digits for n in total, and two that make m+n a multiple of five, there is a 2/10 = 1/5 chance I choose a value of n that makes m+n divisible by 5.

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