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If a, b, and c are consecutive even integers and a < b < c, all of the following must be divisible by 4 EXCEPT

(A) a + c (B) b + c (C) ac (D) (bc)/2 (E) (abc)/4

Is there a way to solve it quickly without picking numbers? Algebraicaly, requires more than 2 min.

You can rule out C, D and E quite quickly if you know one thing: if you have a list of consecutive even numbers, every second number in that list will be a multiple of 4. So if a, b and c are consecutive even numbers, in order, then either b is divisible by 4, or a and c are both divisible by 4. Either way, ac will be divisible by at least 4, bc will be divisible by at least 8, and abc will divisible by at least 16, from which C, D and E must all be divisible by 4.

To decide between A and B, you can pick numbers, or you can write the three consecutive even integers as 2k, 2k + 2 and 2k + 4. Then a + c = (2k) + (2k + 4) = 4k + 4 = 4(k + 1), so a+c must be divisible by 4. That leaves B as the only possible answer.

I think picking numbers is perfectly fine here, since if you use 2, 4 and 6, say, it doesn't take too long. In questions with setups different from the one here, however, you may want to use a more conceptual approach.
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Re: a, b, and c are consecutive even integers... [#permalink]

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13 Dec 2011, 15:47

IanStewart wrote:

In questions with setups different from the one here, however, you may want to use a more conceptual approach.

Thank you! Do you have some rules or principles that could help us in deciding whether we should picki numbers or solve it algebraically?

IMO, I think that always we have to try to solve it algebraicaly because it is safer. We have to pick numbers when we are sure that one case or few cases represent every possible situation.
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Re: a, b, and c are consecutive even integers... [#permalink]

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13 Dec 2011, 21:26

1

This post was BOOKMARKED

I did get the correct answer within 80 seconds and it was not by luck either. I did not pick numbers but just used the concept stated by Ian.

The 3 numbers can be written as a, (a + 2) & (a + 4). If 'a' is divisible by 4, then even 'c' or 'a + 4' is divisible by 4. However, is 'b' is divisible by 4, then both 'a' and 'a + 4' are still divisible by 2.

A - (a + c) = a + (a + 4) = 2a + 4 = 2(a + 2) = 2b. 2b will always be divisible by 4 even if 'b' is not divisible by 4. Reason: 'b' already has a prime factorization of at least a '2'. Hence '2b' has two 2s. C - ac = a(a+4). If, as stated above, one of them is divisible by 4, then the product is divisible. If both of them are not divisible by 4, then the product is still divisible by 4 because of the presence of two 2s again in the prime factorization. D - bc/2 = (a + 2)(a + 4)/2. Either b or c is divisible by 2. Hence, if we assume that b is divisible by 2 and not divisible by 4, then it leaves us just one possibility. Is c divisible by 4? It has to be because c is the next consecutive even integer. E - abc/4 = a(a + 2)(a + 4)/4. One of these integers is divisible by 4 already. If we again assume 'b' to be that integer divisible by 4, then we are left with the question - Is a(a + 4) divisible by 4? This is the same as option C.

B - b + c = (a + 2) + (a + 4) = 2a + 6 = 2(a + 3). (a + 3) will never be divisible by 2 because it is an odd integer. Hence, 2(a + 3), although divisible by 2, will not be divisible by 4 because it has just one 2 in its prime factorization.

As a whole, whether you choose numbers (2, 4 & 6 being the easiest) or solve conceptually, the answer is still easily obtainable within 2 minutes.

Great reply Ian. How much do these books cost. How will I get these books. What does "advanced" mean - 50, 51 level

I'll make a more detailed announcement about my materials soon, but I have problem sets designed for test takers in about the 45-51 range, and topic-specific books which are aimed at above average test takers. If you'd like more information, feel free to send me an email at the address in my signature.
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If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

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