lf the integers m and n are chosen at random from I to 100, then the probability that a number of the form 7ⁿ + 7^m is divisible by 5 equals

 Let I =7ⁿ + 7^m, then we observe that7¹ ,7²,7³ and 7⁴
ends in 7, 9,3 and 1, respectively. Thus, 7ⁱ ends in 7, 9, 3 or 1 according as i is of the form 4k+ 1,4k+ 2, 4k -1 or 4k,respectively.
lf S is the sample space, then n(S) = (100)².
7^m+ 7ⁿ is divisible by 5, if
(i) m is of the form 4k + 1 and n is of the form 4k - 1 or
(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 + 1or
(iv) m is of the form 4kand n is of the form 4k + 2.
Thus, number of favorable ordered pairs (m, n) = 4 x 25 x 25
.'. Required probability $=\frac{4\times 25\times 25}{(100{)}^2}=\frac{1}{4}$