Re M31-31
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14 Jun 2015, 14:27
Official Solution:
If \(x\) is an integer, what is the remainder when \(|1 - x^2|\) is divided by 4?
We want to find the remainder when \(|1-x^2|\) is divided by 4, which can only be 0, 1, 2, or 3.
• If \(x\) is odd, then \(|1-x^2| = |1-odd^2| = |(1-odd)(1+odd)|=|even*even|\), which is a multiple of 4 and hence \(|1-x^2|\) leaves a remainder of 0 when divided by 4.
• If \(x\) is even, then \(|1-x^2| = |1-even^2| =|1-(2k)^2|=|1-4k^2|=|4k^2-1|\), which is 1 less than a multiple of 4 and hence \(|1-x^2|\) leaves a remainder of 3 when divided by 4.
Therefore, the remainder when \(|1-x^2|\) is divided by 4 is 0 if \(x\) is odd, and 3 if \(x\) is even.
(1) The sum of any two factors of \(x\) is even.
In order for the sum of ANY two factors of \(x\) to be even, all factors of \(x\) must be odd (if \(x\) has at least one even factor, we can pair it with 1, which is a factor of every integer, and the sum would be odd). Consequently, \(x\) having only odd factors means that it's an odd number. We know that if \(x\) is odd, the remainder when \(|1-x^2|\) is divided by 4 is 0. Sufficient.
(2) The product of any two factors of \(x\) is odd.
Basically the same here: in order for the product of ANY two factors of \(x\) to be odd, all factors of \(x\) must be odd (if \(x\) has at least one even factor, we can pair it with 1, which is a factor of every integer, and the product would be even). Consequently, \(x\) having only odd factors means that it's an odd number. We know that if \(x\) is odd, the remainder when \(|1-x^2|\) is divided by 4 is 0. Sufficient.
Answer: D