VITEEE 2014 Paper

Let I = 7n+7m, then we observe that 7i,
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 respectively.
If S is the sample space, then n(S) = (100)2
7m+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 + 1 or
(iv) m is of the form 4k and n is of the form 4k + 1 or So, number of favourable ordered pairs (m,n) = 4 x 25 x 25
∴ Required probability =

4×25×25

(100)2

=

1

4