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Options

A. `(1)/(4)`
B. `(1)/(7)`
C. `(1)/(8)`
D. `(1)/(49)`

Answer

A

Explanation

We have,
Number of ways of selecting m and n `=100xx100`
We know that `7^(k), k in N ` has 1,3,9,7, at the units place for
`k=4 lamda, 4 lamda-1, 4lamda-2, 4lamda-3`.
Therefore, `7^(m)=7^(n)` can have, 0,2,4,6 and 8 at units place. But, `7^(m)+7^(n)` will be divisible by 5 if it has 0 at units place.
The digit at units place in `7^(m)+7^(n)` will be 0 if m and n have the following forms :

Clearly, for each value of m there are 25 values of n for which the digit at units place in `7^(m)+7^(n)` is 0.
Number of ways of selecting m and n so that `7^(m)+7^(n)` is divisible by 5 is
`25xx25+25xx25+25xx25+25xx25=2500`.
Hence, required probability `=(2500)/(100xx100)=(1)/(4)`