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 Post subject: GMAT Number TheoryPosted: Sat Mar 03, 2012 9:07 am

Joined: Sun May 30, 2010 3:15 am
Posts: 424
The least common multiple of positive integer m and 3-digit integer n is 690. If n is not divisible by 3 and m is not divisible by 2, what is the value of n?
A. 115
B. 230
C. 460
D. 575
E. 690

(B) We factorize 690 = 2 × 3 × 115 = 2 × 3 × 5 × 23.
Dividing 690 starting by the smallest factor we get possible values for 3-digit integer n:
690, 690 / 2 = 345, 690 / 3 = 230, 690 / 5 = 138, 690 / 6 = 115.
690 / 10 = 69 which is a 2-digit number, therefore n must be one of the following:
690, 345, 230, 138, 115.

n is not divisible by 3. So we eliminate 690, 345, 138.
m is not divisible by 2, but the least common multiple of m and n is. So n must be divisible by 2. We eliminate 115. That leaves us only one option for n, 230. The answer is (B).
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Hi,
I understand why 690, 345, 138 are eliminated. But I do not understand why 115 was eliminated. Since this number is not divisible by 2 or 3. Where as 230 is divisible by 2 but not by 3.

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 Post subject: Re: GMAT Number TheoryPosted: Sat Mar 03, 2012 9:22 am

Joined: Sun May 30, 2010 2:23 am
Posts: 498
Quote:
But I do not understand why 115 was eliminated. Since this number is not divisible by 2 or 3.
So we have 230 and 115 left.
Now, forget about divisibility by 3. We've already used this fact to rule out the other 3-digit divisors of 690.

The question statement tells us "m is not divisible by 2". If n (the number we are looking for) was also not divisible by 2, then the LCM of m and n (690) would not be divisible by 2 as well.
But 690 IS divisible by 2. So n must be divisible by 2.

We rule 115 out because it is NOT divisible by 2.

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 Post subject: Re: GMAT Number TheoryPosted: Sat Oct 27, 2012 7:41 am

Joined: Sun May 30, 2010 3:15 am
Posts: 424
Your attempt to divide by ten in the explanation does not make any sense.

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 Post subject: Re: GMAT Number TheoryPosted: Sat Oct 27, 2012 7:53 am

Joined: Sun May 30, 2010 2:23 am
Posts: 498
questioner wrote:
Your attempt to divide by ten in the explanation does not make any sense.
We have prime-factorized 690 into 690 = 2 × 3 × 5 × 23.

These are the prime factors. Any other factor will be a combination (a product) of the given prime factors. So the divisors of 690, starting from the smallest, are:
1, 2, 3, 5, 6 (which is 2 × 3), 10 (which is 2 × 5), 15 (which is 3 × 5), etc.

We divide 690, starting by the smallest factor:
690 / 1 = 690
690 / 2 = 345
690 / 3 = 230
690 / 5 = 138
690 / 6 = 115
690 / 10 = 69
690 / 15 = 46

First of all note, that on the right we get the divisors (factors) of 690. Every factor of 690, n, has a corresponding factor of 690, (690/n). If we continue to increase the factor we divide by we will decrease the factor on the right side of the equality. So we do not need to consider any more factors of 690 and we already have a full list of 3-digit factors of 690:
690, 345, 230, 138, 115.

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