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Is the integer x divisible by 36? (1) x is divisible by 12

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Is the integer x divisible by 36? (1) x is divisible by 12 [#permalink] New post 22 Feb 2011, 10:28
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Is the integer x divisible by 36?
(1) x is divisible by 12
(2) x is divisible by 9

My approach was as follows
(1)24, 48 is divisible by 12 but not by 36 so NS
(2) 27, 81 is divisible by 9 but noty 36. So NS

considering two: the integer which is divisible by 12 and 9 is multiple 3 and 36 is multiple 3. So any digit divisible 12 and 9 is also divisible by 36.

Please help with good approach.
[Reveal] Spoiler: OA

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Re: 92. Integer [#permalink] New post 22 Feb 2011, 10:41
Prime factors of 36 = 2^2*3^2

So if x is divided by a number whose factors at least comprise of all the prime factors of 36; i.e. two 2's and two 3's then x will be divisible by 36.

1.Prime factors of 12 = 2^2*3
So; only two 2's and one 3. We require 2 two's and two 3's for the number to be divisible by 36.
Not Sufficient.


2.Prime factors of 9 = 3^2
So; only two 3's. We require 2 two's and two 3's for the number to be divisible by 36.
Not Sufficient.

Combining both;
x is divisible by; 2^2*3*3^2
The factors comprise of at least two 2's and two 3's. Must be divisible by 36.
Sufficient.

Ans: "C"
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Re: 92. Integer [#permalink] New post 23 Feb 2011, 12:41
Mate, you can download my flashcards and take a look at slides 81-84. All the approaches to find GCF and LCM are there. Hope it helps.

Link: my-comprehensive-quant-flashcards-109836.html

Solving the question:
Always find the prime factors, as fluke showed. Then, remember that GCF is formed by the shared prime factors.

The 36's prime factors box has: 2, 2, 3, 3

(1) 12's prime box has 2, 2, 3. A 3 is missing, so not sufficient

(2) 9's prime box has 3, 3. Not sufficient as well (two 2's missing)

If you combine them, you will have two shared 2's and two shared 3's, which are the same as the prime factors of 36. So, both are sufficient. A number that is divisible by 9 and by 12 will also be divisible by 36.

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Re: 92. Integer [#permalink] New post 23 Feb 2011, 18:32
Find whether 36 is factor of X......

Break down 36 to prime factors....2*2*3*3

If 36 is factor of x, then the 2*2*3*3 are factors (equivalent)

This is true if we know other factors that also exhaust the prime factors.

A. 12 is a factor of x. 2*3*2 are the prime factors. This covers all the factors of 36 except for the second 3. We don't have proof that x has a second 3 as a factors so A is insufficient

B. 9 is a prime factors. 3*3 are the prime factors. This covers only 3*3 and doesn't tell us whether 2*2 are also factors. Insufficient.

Both A&B tells us that we have 3*3 and 3*2*2 as factors.

The maximum number of a certain prime factor need to come from one of the factors, as it does. Both 3's come from the 9, both 2s come from the 12. Sufficient C
Re: 92. Integer   [#permalink] 23 Feb 2011, 18:32
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Is the integer x divisible by 36? (1) x is divisible by 12

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