Each of seven consecutive integers, all greater than positive integer

MANHATTAN GMAT OFFICIAL SOLUTION:

First, just get the info straight. You’re looking for n, which is a positive integer. There are also 7 other integers. Those 7 integers are both larger than n and consecutive, but n may not be consecutive with the rest.

STATEMENT (1) NOT SUFFICIENT: Try some actual numbers in order to understand how all of this works.

At this point, it’s useful to think about what patterns exist for remainders. First of all, any remainders have to be less than the divisor. If the divisor is 3, then the largest possible remainder is 2.

Second of all, for a given n, the pattern of remainders will be consistent, but that pattern could start on a different number. For example, above, the n = 3 pattern started with 0. If list L were 7, 8, 9, 10, 11, 12, 13, though, then the remainders would be 1, 2, 0, 1, 2, 0, 1. The pattern is the same; the sequence just starts with 1 instead of 0.

Okay. So for n = 4, the pattern would be 0, 1, 2, 3, 0, 1, 2. The number 3 was added to the set. Further, the pattern could start on any number, depending upon the exact numbers in List L. As n increases by 1, one new remainder gets added to the pattern.

Start thinking about what would need to be true in order to get two odd values and five even values. Have any of the patterns above already returned this result? Yes! When n = 3 and the pattern starts on the number 0, then there are exactly two instances of the odd value 1. How come that worked for n = 3 but not for n = 4?

Interesting! In n = 3, the pattern is even, odd, even, even, odd, even, even. There are more evens than odds, so it’s possible to have only 2 odds. The pattern for n = 4, though, is even, odd, even, odd, even, odd, even. This time, there’s no way to get only 2 odds while having 5 evens. Okay, so for statement 1 to be true, the pattern has to be “asymmetric,” with more evens than odds.

Where else is that going to happen? Not for any n = even cases. (Not sure about that? Try out n = 6 and see.) The variable n, then has to be odd.

You already saw that n = 3 works. What about n = 5? The remainder pattern would be 0, 1, 2, 3, 4, 0, 1, 2, 3, 4. If you start the pattern at 4 (so that you have two evens in a row), then you’d have 4, 0, 1, 2, 3, 4, 0. Bingo! Exactly two odds.

You’re done with this statement (for now, anyway) because there are at least two possible values for n. Statement 1 is not sufficient.

STATEMENT (2) NOT SUFFICIENT: Many values of n are possible with this statement. For instance, if n is any integer greater than 6, then seven consecutive integers may be found to produce the remainders 0, 1, 2, 3, 4, 5, and 6, in which no integer (regardless of odd/even) appears more than once.

STATEMENTS (1) AND (2) SUFFICIENT: We know that n = 3 and n = 5 satisfy the first statement. Are other odd numbers possible? No, it turns out. If n = 7, then the remainder pattern is 0, 1, 2, 3, 4, 5, 6. That list contains three odd numbers, not just two. For any larger n, the remainder pattern will have to include at least three odd numbers because the remainder pattern does not repeat within any set of 7 numbers. You’re not going to be able to produce the odd, even, even pattern that allowed n = 3 and n = 5 work. (Not sure about that? Try n = 9 and n = 11 to see how it works.)

So only 3 and 5 are possible values for n, according to statement 1. If n = 3, though, the only possible odd remainder is 1, so the two odd numbers in list L must both be equal to 1. Therefore, n = 3 cannot satisfy the second statement, leaving only n = 5. When n = 5, the two odd integers 1 and 3 can each appear exactly once. The list has two distinct odd values and 5 even values, so n does indeed equal 5. The two statements together are sufficient.

The correct answer is C.