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
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
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:
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
Was this answer helpful ?
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