If the integers m and n are chosen at random | vedclass

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(A) The last digit of $7^k$ follows a cycle of length $4$: $7^1=7, 7^2=9, 7^3=3, 7^4=1, 7^5=7, \dots$ For $7^m + 7^n$ to be divisible by $5$,the last

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$\frac{1}{4}$

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(A) The last digit of $7^k$ follows a cycle of length $4$: $7^1=7, 7^2=9, 7^3=3, 7^4=1, 7^5=7, \dots$ For $7^m + 7^n$ to be divisible by $5$,the last digit of the sum must be $0$ or $5$. Since the last digits of powers of $7$ are ${1, 3, 7, 9}$,the possible sums of last digits are: $1+1=2, 1+3=4, 1+7=8, 1+9=10$ (divisible by $5$) $3+1=4, 3+3=6, 3+7=10$ (divisible by $5$),$3+9=12$ $7+1=8, 7+3=10$ (divisible by $5$),$7+7=14, 7+9=16$ $9+1=10$ (divisible by $5$),$9+3=12, 9+7=16, 9+9=18$ In each cycle of $4$ for $m$ and $n$,there are $4 	imes 4 = 16$ pairs. The pairs $(m, n) \pmod 4$ that result in a sum divisible by $5$ are $(1, 3), (3, 1), (2, 4), (4, 2)$. There are $4$ such pairs out of $16$ total possibilities in each $4 	imes 4$ block. Since $100$ is a multiple of $4$,the probability is $\frac{4}{16} = \frac{1}{4}$.

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

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