it can happen in following way.
(7,3) or (9,1)
From 1 to 100, there are 25 numbers which when put as 7^m give 7 at last, similarly 25 which give 3, 25 which give 1 and 25 which give 1 as the unit digit when raised onto 7.
Number which give 7 = 25
Number which give 3= 25
Numbers which give 1=25
Numbers which give 9 = 25

Hence now, we can either have (7,3) or (9,1)
Consider, for 7 and 3
This means one number from 7 category and other from the one which gives 3 . This can be done in 25C1 X 25C1 ways.
Similarly for (9,1) Number of ways are 25C1X25C1

Ways to select m and n , considering they are replaced is 100C1 X 100 C1

Hence the probability is (25C1 X 25C1 + 25C1 X 25C1 ) / (100C1 X 100C1) = 1/8

sanju09 wrote:If the integers m and n are chosen at random from 1 to 100, then what is the probability that a number of the form 7^m + 7^n is divisible by 5?
(A) ½
(B) ¼
(C) 1/6
(D) 1/8
(E) 1/16

For 7^m + 7^n to be a multiple of 5, its units digit must be 0 or 5.

When an integer is raised to consecutive powers, the resulting units digits repeat in a CYCLE:
7¹ --> units digit of 7.
7² --> units digit of 9. (Since the product of the preceding units digit and 7 = 7*7 = 49.)
7³ --> units digit of 3. (Since the product of the preceding units digit and 7 = 9*7 = 63.)
7� --> units digit of 1. (Since the product of the preceding units digit and 7 = 3*7 = 21.)
From here, the units digits will repeat in the same pattern: 7, 9, 3, 1.

The units digits repeat in a CYCLE OF 4.
For exponents 1 through 100, this cycle of 4 will repeat 25 TIMES.
The result:
For exponents 1 through 100, each of the 4 units digits -- 7, 9, 3 and 1 -- will appear the SAME NUMBER OF TIMES.

Question rephrased:
If the units digits of 7^m and 7^n have an EQUAL CHANCE of being 7, 9, 3, or 1, what is the probability that 7^m + 7^n is divisible by 5?

Total options for 7^m + 7^n:
The number of options for the units digit of 7^m = 4. (7, 9, 3, or 1.)
The number of options for the units digit of 7^n = 4. (7, 9, 3 or 1.)
To combine these options, we multiply:
Total options = 4*4 = 16.

Good options for 7^m + 7^n:
For 7^m + 7^n to be divisible by 5, the sum of the units digit of 7^m and the units digit of 7^n must be 0 or 5.
The following combinations are good:
7+3
9+1
3+7
1+9
Good options = 4.

Thus:
(good options)/(total options) = 4/16 = 1/4.

The correct answer is B.