vcbabu wrote:

find all the numbers less than 100 with exactly 8 divisors.

This is an interesting problem. Let's approach this logically. Here is a big hint:

because there need to be exactly 8 divisors, we need one of two possible factorizations:

A^1*B^1*C^1 < 100 where A, B, and C are prime. (2 x 2 x 2 combinations of exponents -- don't forget the zeroth power!)

or A^3 * B^1 < 100 where A, and B are prime. (4 x 2 combinations of exponents).

Note that there is no other way to factor this where you will get exactly 8 (well, technically there is A^7, but 2^7, which is the smallest number, = 128 so that won't work).

With a little bit of common sense and cut and try, we can narrow this down quickly since there are only so many possible combinations that yield a result < 100. I'm going to leave the rest of the solution for you to solve.

_________________

Best,

AkamaiBrah

Former Senior Instructor, Manhattan GMAT and VeritasPrep

Vice President, Midtown NYC Investment Bank, Structured Finance IT

MFE, Haas School of Business, UC Berkeley, Class of 2005

MBA, Anderson School of Management, UCLA, Class of 1993