cicerone wrote:

Folks, here is the next question..

From the set of positive factors of 840 one factor is chosen at random. What is the probability that it is divisible by 15?

A. 1/4

B. 7/32

C. 1/2

D. 1/56

E. None of these

my approach :

total no of factors =(2^3)*3*5*7=refer powers (3+1)x(1+1)x(1+1)x(1+1)=32 factors

we want to select factors divisible by 15 ie 3 x 5 should be part of all divisors..we can ensure that by neglecting 3^0 & 5^0 in each factor i.e.

= (2^0+2^1+2^2+2^3) * (3^1) * (5^1) * (7^0+7^1) = thus no of factors factors 4x1x1x2 = 8 factors [15 , 15x7, 15x2, 15x4, 15x8, 15x7x2, 15x7x4, 15x7x8]

prob = 8/32 = 1/4 OA A _________________

Bhushan S.

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