Question

If the integers m and n are chosen at random between 1 and 100, then the probability that a number of the form 7m+7n is divisible by 5, equals (a) 1 4 (b) 1 7 (c) 1 8 (d) 1 49 A. 1 4 B. 1 7 C. 1 8 D. 1 49 Select the correct answer from above options

Answer 1

Correct Answer - A 71=7,72=49,73=343,74=2401… Therefore for 7r,r∈N the number ends at unit place 7,9,3,1,7… ∴7m+7n will be divisible by 5 if it end at 5 or 0. But it cannot end at 5 Also for end at 5. For this m and n should be as follows For any given value of m, there will be be 25 values of n Hence , the probability of the required event is (100xx 25)/(100xx100)=1/4 Note:Power of prime numbers have cycclic numbers in their unit place.