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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 (a)

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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 (a) `1/4` (b) `1/7` (c) `1/8` (d) `1/49` Select the correct answer from above options

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`7^m+7^n` divisible by 5 `7^n[7^(m-n)+1]` `7^(m-n)+1` should be divisible by 5 Power of 7 has last digits revolving in 7,9,3,1. `7^(m-n)` should have last digit as 9. `7^(m-n)=7^(4p+2)` `m-n=4p+2` `4p+2<=100`<br> `p<=24.5`<br> favorable=25 total cases=100 `P=25/100=1/4`.

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