Anindya Madhudor wrote:How many numbers that are not divisible by 6 divide evenly into 264,600?

a. 16
b. 36
c. 51
d. 63
e. 72

OA: D

264,600 = 2³ * 3³ * 5² * 7².
At its core, this is a COMBINATORICS question.
A factor of 264,600 can COMBINE up to three 2's, three 3's, two 5's, and two 7's.
Any FACTOR COMBINATION that does not include both 2 and 3 will not be a multiple of 6.

Case 1: Factor combinations that include NO 2's OR 3's
Number of options for 5 = 3. (No 5's, one 5, or two 5's.)
Number of options for 7 = 3. (No 7's, one 7, or two 7's.)
To combine these options, we multiply:
3*3 = 9.

Case 2: Factors combinations that include AT LEAST ONE 2
Number of options for 2 = 3. (One 2, two 2's, or three 2's.)
Number of options for 5 = 3. (No 5's, one 5, or two 5's.)
Number of options for 7 = 3. (No 7's, one 7, or two 7's.)
To combine these options, we multiply:
3*3*3 = 27.

Case 3: Factor combinations that include AT LEAST ONE 3
Number of options for 3 = 3. (One 3, two 3's, or three 3's.)
Number of options for 5 = 3. (No 5's, one 5, or two 5's.)
Number of options for 7 = 3. (No 7's, one 7, or two 7's.)
To combine these options, we multiply:
3*3*3 = 27.

Total options = 9+27+27 = 63.

The correct answer is D.

264,600 = 2³ * 3³ * 5² * 7²

Total number of factors are (3+1)*(3+1)*(2+1)*(2+1) = 144
Formula for above reasoning is, if (x = a^p * b^q) where a and b are distinct prime numbers then number of factors of x are (p+1) * (q+1)

Number of factors with at least one 2 and 3(divisible by 6) are :-
3(can be 2^1,2^2,2^3) *3 (can be 3^1,3^2,3^3) *3 (can be 5^0,5^1,5^2) *3 (can be 7^0,7^1,7^2) = 81

Remaining(not divisible by 6)are = 144-81 = 63

Answer D