Number Properties DS question

[sajibfin06]

If M and N are integers such that M = N^2 - 1, is M divisible by 12?
(1) N-1 is the square of an even number
(2) N+1 is a two digit number and the sum of its digits is a multiple of 3

[MartyMurray]

For M to be divisible by 12, we need the factors of M to include at least one 3 and two 2's.

If M = n²- 1 then M = (n + 1)(n - 1)

Statement 1 tells us that n - 1 is even. Since n - 1 is the square of an even number, it also must include in its factors two 2's.

So we have the two 2's. Now we need to determine whether we have the 3.

Going through squares of even numbers I notice that if n - 1 = 4, then n + 1 = 6. So we would have a 3.

n - 1 = 16 n + 1 = 18 We would have a 3.

n - 1 = 36 n + 1 = 38 The 3 is in n - 1 this time.

It also works for 8, 10 and 12. So I am going to hazard a guess that it would work for any square of an even number.

I am sure I could work out a clear theory as to why this works but I am not going to bother. I am calling Statement 1 sufficient.

Statement 2 tells us that N + 1 is a two digit multiple of 3. N + 1 could be an odd multiple of 3 in which case n - 1 would also be odd and M would not be a multiple of 3.

If n + 1 even multiple of 3 then n - 1 is even also giving us two 2's and a 3. So M would be a multiple of 12.

Insufficient.

Choose A.

[Matt@VeritasPrep]

Here's a fairly technical approach.

S1:: (n - 1) = (even)² = (2k)² = 4k², where k is some integer whose value we don't care about. (2k just means "a multiple of 2".)

So n = 4k² + 1, and n² - 1 = (4k² + 1)² - 1, or 16k� + 8k². So our question becomes "Is 16k� + 8k² divisible by 12?"

For this to be divisible by 12, it has to be divisible by 4 and by 3. Let's divide by 4, giving us 4k� + 2k². If this is divisible by 4, we're set.

We can factor out 2k², which gives us 2k² * (2k² + 1). So we have two consecutive integers. Now consider k itself, for which we have three cases:

Case 1: k is divisible by 3. Then 2k² is divisible by 3.
Case 2: k is 1 greater than a multiple of 3. Then 2k² + 1 is divisible by 3.
Case 2: k is 2 greater than a multiple of 3. Then 2k² + 1 is divisible by 3.

So in every case, n² - 1 gives a number that is divisible by 4 AND divisible by 3; SUFFICIENT!

S2::

(n + 1) = 12, 15, 18, ...

which means

(n - 1) = 10, 13, 16, ...

10 * 12 is divisible by 12, but 15 * 13 is not, so this is NOT sufficient.