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

Options

A. `(1)/(4)`
B. `(1)/(2)`
C. `(1)/(8)`
D. `(1)/(3)`

Answer

A

Explanation

Let `l=7^(n)+7^(m)`, then we observe that `7^(1),7^(2),7^(3)` and `7^(4)` ends in 7,9,3 and 1, respectively. Thus, `7^(i)` ends in 7,9,3 or 1 according as i is of the form 4k - 3, 4k - 2, 4k - 1 for 4k, respectively.
If S is the sample space, then `n(S)=(100)^(2)`.
`7^(m)+7^(n)` is divisible by 5, if
(i) m is of the form 4k - 3 and n is of the form 4k - 1 for
(ii) m is of the form 4k - 2 and n is of the form 4k or
(iii) m is of the form 4k - 1 and n is of the form 4k - 3 or
(iv) m is of the form 4k and n is of the form 4k - 2.
Thus, number of favourable ordered pairs
`(m,n)=4xx25xx25`
`therefore` Required probability `=(4xx25xx25)/((100)^(2))=(1)/(4)`