Great question here - but I have a few things to point out toward the correct answer:
1) Beware of the 'too-easy' choice A here...26 (choice E) is a long, long way from 5, so that should tip you off that maybe you're missing something.
2) What the initial explanation misses is that we don't simply need "naturally occurring" 6s - 6, 12, 18, etc. We can "manufacture" 6s out of any combination of 2 and 3.
Consider this - is 3*4 divisible by 6? It is - 3*4 is 12, which divides by 6. It just doesn't have a "natural" 6 in it, but if we break it down to prime factors it's 3*2*2, and in order to be divisible by 6 we only need a 2 and a 3.
So...for problems like these, break the divisor down to prime factors and find as many pairs of that factorization as you can.
Here, we need 2*3. Strategically, there will be many more 2s than 3s - every SECOND number gives us a 2, but every THIRD number gives us a 3. 3s are going to be our "constraint" then, so all we really need to do is find the number of 3s in 25!. Looking at that, we have as muttiples of 3:
3, 6, 9, 12, 15, 18, 21, 24
But in keeping in step with the "prime factors" consideration - any 3 will do - we need to break these out to maximize our 3s. 9 = 3*3, for example, so it gives us two 3s, not just one. From our list we have:
3
3*2
3*3
3*4
3*5
3*3*2
3*7
3*8
Because each of 9 and 18 gives us an extra 3, we have a total of ten 3s, so the correct answer is 10.
A few big takeaways here:
1) With divisibility-related questions, it's usually helpful to break numbers down to prime factors.
2) Use your judgment with this one, but if a question almost seems too easy and you can't justify any other answer choices as being remotely possible (here if you picked 5 the others seem way, way far away with no basis for picking them) you probably forgot to consider something and you may want to return to the question more thoroughly.
Last edited by
Brian@VeritasPrep on Tue Sep 14, 2010 11:04 am, edited 1 time in total.
Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep
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