How many numbers that are not divisible by 6 divide evenly

Numbers that divide 264600 are the factors of 264600

So we have to find out the numbers that are factors of 264600 but are not multiples of 6 (2*3)

a Number won't be a multiple of 6 if it last either 2 or 3 or both as it's prime factors

264600 = 2646*100 = 1323*2*2^2*5^2 = 9*147*(2^3)*(5^2) = (3^2)*49*3*(2^3)*(5^2) = (3^3)*(7^2)*(2^3)*(5^2)


Factors of (7^2)*(2^3)*(5^2) without a multiple of 3 = (2+1)*(3+1)*(2+1)=36
Factors of (7^2)*(2^3)*(5^2) without a multiple of 2 = (3+1)*(2+1)*(2+1)=36
Factors of (7^2)*(2^3)*(5^2) without primes 2 and 3 both = (2+1)*(2+1)=9

So total required numbers = 36+36-9 = 63

ANswer: Option D

Can you please elaborate on why are you not adding 1 to the powers of 2 and 3, this time, to exclude all the cases with 2^0*... and 3^0*?

At heart, this is a counting problem in which you need to carefully separate the scenarios so you don’t over-count the factors and understand how a positive factor is “built” which will divide evenly into that number.


(1st) Prime Factorize 264, 600

= (2)^3 * (3)^3 * (5)^2 * (7)^2


Each Factor that will divide evenly into the number will be made up of some combination of 1 or all of these prime factors up to each prime factors exponent.

When all the powers are 0, the factor will = 1

(2nd)Split up the factors we are looking for into different scenarios

In order for one of the factors to be divisible by 6, the number must be divisible by at least ——(2) * (3)

So, only factors that do not have one of each prime factor (or neither) will NOT be divisible by 6

Scenario 1: all the factors that are possible that are divisible by 2 alone and NOT 3

Prime factor 2: we have 3 available options—- (2)^1 or (2)^2 or (2)^3


Prime Factor 3: it must be (3)^0 or 1 - 1 available option


Prime Factor 5: 3 available options

Prime factor 7: 3 available options


Total possible factors divisible by 2 but NOT divisible by 6 = 3 * 1 * 3 * 3 =

27


Scenario 2: factor is divisible by 3, but NOT by 6

Same logic as above, but this time the prime factor of 2 has only 1 available options from which to choose

1 * 3 * 3 * 3 = 27


Scenario 3: a factor that so neither divisible by 2 nor divisible by 3

Prime factor of 2: must be to 0 power

Prime factor of 3: must be to 0 power

Prime factor of 5: have 3 available options

Prime factor of 7: have 3 available options

1 * 1 * 3 * 3 = 9


Total factors of 264,600 that are NOT divisible by 6 is =

27 + 27 + 9 = 63

B

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