workout wrote:
GMATbuster's Weekly Quant Quiz#11 Ques #7
For Questions from earlier quizzes: Click HereFor any integer P greater than 1, P! denotes the product of all the integers from 1 to P, inclusive. If one integer is selected at random between 5! And 5! + 10, inclusive, what is the probability that the chosen number will have only two factors?
A)0
B)1/11
C)1/10
D)2/11
E)2/10
We are looking at numbers between 5! and 5!+10 or between 120 and 130, inclusive
So, 130-120+1 = 11 numbers. We can eliminate C and E
The only numbers with 2 factors are prime numbers.
We can eliminate 120, 121, 122, 123, 124, 125, 126, 128, 129, 130 (divisible by 2,3,5 and 11).
So we're left with 127, which it turns out is a prime (tested 7 to make sure), so B.
For reference 101–200 has 21 prime numbers (101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 183, 193, 197, 199)
It could've easily been D if the range had been slightly different (e.g. 130-140 inclusive)