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
Options:
A .  
14
B .  
17
C .  
18
D .  
149
Answer: Option A : A Since m and n are selected between 1 and 100, hence sample space = 100×100. Also 71=7,72=49,73=343,74=2401,75=16807 etc. Hence 1, 3, 7 and 9 will be the last digits in the powers of 7. Hence for favorable cases nm→↓1,11,21,3....1,1002,12,22,3...2,100...........................................100,1100,2100,3100,100 For m = 1; n = 3, 7, 11….. 97 ∴ Favorable cases = 25 Similarly for every m, favorable n are 25. ∴ Total favourable cases = 100×25 Hence required probability = 100×25100×100=14
Submit Your Solution hear: