Dost thou seek the answer to yesterday’s Challenge Problem? As promised, find it below!
Question
If p, x, and y are positive integers, y is odd, and p = x2 + y2, is x divisible by 4?
(1) When p is divided by 8, the remainder is 5.
(2) x – y = 3
(A) Statement (1) ALONE is sufficient to answer the question, but statement (2) alone is not.
(B) Statement (2) ALONE is sufficient to answer the question, but statement (1) alone is not.
(C) Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient to answer the question.
(E) Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question.
Solution
Statement (1): SUFFICIENT. We know that p is odd. We know from the problem stem that y is odd, which means that y2 is odd. Therefore, x2 must be even (because O = E + O), so x must be even.
However from all of this, we can infer something else—specifically, that x is NOT a multiple of 4. Here’s why:
y2 = (y2 — 1) + 1, which using the quadratic property, yields: y = (y+1)(y—1) +1. Because y is odd, then (y+1)(y—1) is even times even, which is a multiple of 4. However we ALSO know that either y+1 or y—1 is a multiple of 4, because they are consecutive multiples of 2. Therefore, y2 is 1 greater than a multiple of 8. (You can confirm this by thinking about all squared odd numbers: 12 = 1, 32 = 9, 52 = 25, 72 = 49, etc.)
Since y2 divided by 8 yields a remainder of 1, from statement (1) we need x2 divided by 8 to yield a remainder of 4. This means that x must be even, but NOT be a multiple of 4, because if it were a multiple of 4, then x2 would be a multiple of 16, and therefore a multiple of 8.
Statement (2): INSUFFICIENT. We know that x is even. However, we can come up with two different examples which lead to different answers. For example, if x = 8 and y = 5, then x is a multiple of 4, but if x = 6 and y = 3, then x is not a multiple of 4.
The answer is A: Statement (1) is sufficient to answer the question, but statement (2) is insufficient.












