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If two distinct numbers m and n are chosen at random form the set {1, 2, 3, …, 100}, then find the probability that `2^(m) + 2^(n) + 1` is divisible by 3.

A

`(131)/(300)`

B

`(31)/(300)`

C

`(21)/(200)`

D

`(62)/(300)`

Text Solution

Verified by Experts

The correct Answer is:
B
`(b)` `log_(a)b=(logb)/(loga)`
Let `b=2^(m)` and `a=2^(n)` where `m` and `n` denote the exponents on the base `2` in the given set then `log_(a)b=(m)/(n)`
Therefore, `log_(a)b` is an integer only if `n` divides `m`.
Now, total no. of ways `m` and `n` can be chosen `=25xx24=600`
For favourable cases
`{:("Let"n=1,"So m can take values 1,2,3,4,5,6,24",=24),("If"n=2,"m=4,6,8,10,12,14,18,20,22,24",=11),(n=3,"m=6,9,12,15,18,21,24",=7),(n=4,"m=8,12,16,20,24",=5),(n=5,"m=10,15,20,25",=4),(n=6,"m=12,18,24",=3),(n=7,"m=12,21",=2),(n=8,"m=16,24",=2),("n=9,10,11,12","m=1 for each",=4),(,,62):}`
Hence, required probability `=(62)/(600)=(31)/(300)`
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