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Re: How many numbers that are not divisible by 6 divide evenly [#permalink]
Bunuel wrote:
enigma123 wrote:
How many numbers that are not divisible by 6 divide evenly into 264,600?
(A) 9
(B) 36
(C) 51
(D) 63
(E) 72

Any idea how to solve this please?

Finding the Number of Factors of an Integer

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.
For more on number properties check: https://gmatclub.com/forum/math-number-theory-88376.html

BACK TO THE ORIGINAL QUESTION:
How many numbers that are not divisible by 6 divide evenly into 264,600?
(A) 9
(B) 36
(C) 51
(D) 63
(E) 72

\(264,600=2^3*3^3*5^2*7^2\), thus it has total of \((3+1)(3+1)(2+1)(2+1)=144\) differernt positive factors, including 1 and the number itself.

# of factors that ARE divisible by 6 will be \(3*3*(2+1)(2+1)=81\): we are not adding 1 to the powers of 2 and 3, this time, to exclude all the cases with 2^0*... and 3^0*... (thus to exclude all the factors which are not divisible by 2 or 3), hence ensuring that at least one 2 and at least one 3 are present to get at least one 6 in the factors we are counting.

So, # of factors that ARE NOT divisible by 6 is 144-81=63.

Answer: D.

Another solution here: https://gmatclub.com/forum/new-set-of-go ... ml#p642384

Hope now it's clear.


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*?
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Re: How many numbers that are not divisible by 6 divide evenly [#permalink]
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Expert Reply
Krishchamp wrote:
Bunuel wrote:
enigma123 wrote:
How many numbers that are not divisible by 6 divide evenly into 264,600?
(A) 9
(B) 36
(C) 51
(D) 63
(E) 72

Any idea how to solve this please?

Finding the Number of Factors of an Integer

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.
For more on number properties check: https://gmatclub.com/forum/math-number-theory-88376.html

BACK TO THE ORIGINAL QUESTION:
How many numbers that are not divisible by 6 divide evenly into 264,600?
(A) 9
(B) 36
(C) 51
(D) 63
(E) 72

\(264,600=2^3*3^3*5^2*7^2\), thus it has total of \((3+1)(3+1)(2+1)(2+1)=144\) differernt positive factors, including 1 and the number itself.

# of factors that ARE divisible by 6 will be \(3*3*(2+1)(2+1)=81\): we are not adding 1 to the powers of 2 and 3, this time, to exclude all the cases with 2^0*... and 3^0*... (thus to exclude all the factors which are not divisible by 2 or 3), hence ensuring that at least one 2 and at least one 3 are present to get at least one 6 in the factors we are counting.

So, # of factors that ARE NOT divisible by 6 is 144-81=63.

Answer: D.

Another solution here: https://gmatclub.com/forum/new-set-of-go ... ml#p642384

Hope now it's clear.


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*?


Consider 2^3. When you add 1 to the power of 2, it gives all power of 2: 2^0, 2^1, 2^2 and 2^3. But 2^0 is 1 and we want the factors which are divisible by 2, so 2^0 is not ok. Thus, we count only 2^1, 2^2 and 2^3.
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Re: How many numbers that are not divisible by 6 divide evenly [#permalink]
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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Re: How many numbers that are not divisible by 6 divide evenly [#permalink]
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