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 Post subject: GRE Number Theory (Select One)Posted: 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: t.4, qt.11: number theory, factorizationPosted: 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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