first of all, you must understand this fact: every nth integer is divisible by n.
in other words, every other integer is divisible by 2; every third integer is divisible by 3; etc.
make sure you understand this basic principle first; if you don't, then essentially none of the following will make any sense to you.
for problems like this one, in which you're considering the divisibility of unknown numbers, it can help to do the following:
(1) set up a number line, with ..., n - 1, n, n + 1, n + 2, ... placed along it
(2) write out the different CASES for which ones are divisible by 3 (or divisible by whatever number you're talking about; in this problem it just happens to be 3)
here's what i mean.
in this case, there are 3 different cases, because every third integer is divisible by 3. (in general, for divisibility by n, there will be n different ways in which to draw the number line.)
here are the 3 cases. in each case, the multiples of 3 are in orange, and the non-multiples of 3 are in the default color (black on my view, although i'm not sure whether that's the default for everyone else).
case 1
... n-6 n-5 n-4 n-3 n-2 n-1 n n+1 n+2 n+3 n+4 n+5 n+6 ...
case 2
... n-6 n-5 n-4 n-3 n-2 n-1 n n+1 n+2 n+3 n+4 n+5 n+6 ...
case 3
... n-6 n-5 n-4 n-3 n-2 n-1 n n+1 n+2 n+3 n+4 n+5 n+6 ...
once you've done this, just look at the 3 cases for each of the answer choices. if any of the factors is in orange, then that particular product is a multiple of 3 (because, if there's a 3 anywhere in the factorization, then the whole product is a multiple of 3 - as with any other factor).
for choice (a), all three cases include one orange factor apiece: n is orange in case 1, n+1 is orange in case 3, and n-4 is orange in case 2. therefore, (a) is the correct answer.
for choice (b), case 3 contains no orange factors, so (b) doesn't have to be a multiple of 3.
for choice (c), case 2 contains no orange factors, so (b) doesn't have to be a multiple of 3.
for choice (d), case 2 contains no orange factors, so (b) doesn't have to be a multiple of 3.
for choice (e), case 3 contains no orange factors, so (b) doesn't have to be a multiple of 3.
note that the correct answer is a, not d.
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note that you could also just plug in a ton of numbers and see what happens with the factorizations: just plug in 7, 8, 9, ... (per the directions) for the choices and see what the products are. don't multiply the products if you do that; just see whether the numbers in the products are divisible by 3.
Ron has been teaching various standardized tests for 20 years.
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