In previous posts, we have discussed how the best way to tackle a difficult looking quantitative question on the GMAT is to clean it up – consolidating like terms, adding or subtracting inside of parentheses, or reorganizing variables to where it is easier not substitute equations. Not only is cleaning up quantitative questions helpful to cutting down on the amount of time you spend solving a question, but often, can be essential to seeing conceptually how to get to a correct answer choice. Consider the below data sufficiency question: Given n > 5, when (n! + n + 1) is divided by (n+1), what is the remainder? • (n+2) is a prime number • (n-2) is a prime number Our first step should be to see what this problem actually looks like – (n! + n + 1) ————— (n+1) Yikes, right? With just variables in the statements provided, it is likely you feel this answer is completely impossible and want to immediately jump to (E) as your answer choice. But, if you understand how to properly simplify the fraction, then you will be in much better shape to evaluate statements (1) and (2). Think of a simpler parallel example: x + x + 1 ——- x + 1 If we were to plug in, say, x = 2 it would come out to – 2 + 2+ 1 / 2 + 1 = 5/3. This is the same as recognizing that the fraction is also the same as: x x+1 — + —– x+1 x+1 Where (x+1)/(x+1) cancel out to equal 1 + x/x+1 which then equals 1 and ⅔, the same as 5/3. So, back to our impossible looking question. We can apply the same logic of the easier parallel example to say that (n! + n + 1) n! (n+1) n! ————— can be rearranged to be —- + ——— = —– + 1 (n+1) (n+1) (n+1) (n+1) Which is MUCH easier to evaluate. From here, if n! is divisible by (n+1), then the remainder is 0. If not, then we don’t know the remainder. Considering if (n+1) is a composite number, it can be split into two factors smaller than (n+1). Both will be included in n! since n! has all factors smaller than (n+1). So if (n+1) is composite, n! is divisible by (n+1). If (n+1) is prime, n! does not contain (n+1) and hence is not divisible by (n+1). Knowing this, if we evaluate (n+2) is prime, then (n+1) must be even and hence composite – the only even prime number is 2, so the remainder must be 0. Statement 1 is sufficient on its own. Moving to assess Statement 2, the same pattern of thought follows. If (n-2) is prime, the (n-1) and (n+1) must be even and, therefore, composite. Again, the remainder must be 0 and Statement 2 is sufficient on its own. Without understanding how to appropriately reduce fractions, it makes it pretty impossible for us to approach this more challenging quantitative question appropriately. By simplifying many of these questions become much, much more manageable. | 
