Bunuel wrote:
If x is a positive integer greater than 3 and \(x^4 - 10x^2 + 9 = y\), is y a multiple of 30 ?
(1) x is a prime number such that when x divided by 5 gives the remainder of 1
(2) x is a prime number such that when x divided by 6 gives the remainder of 5
Official Solution:If \(x\) is a positive integer greater than 3 and \(x^4 - 10x^2 + 9 = y\), is \(y\) a multiple of 30 ? Work on the expression in the stem: \(x^4 - 10x^2 + 9 = y\);
\(x^4 - 9x^2 -x^2 + 9 = y\);
\(x^2(x^2 - 9) -(x^2 - 9) = y\);
\((x^2 - 9)(x^2 -1) = y\);
\((x - 3)(x -1)(x+1)(x+3) = y\);
(1) \(x\) is a prime number such that when \(x\) divided by 5 gives the remainder of 1.
First of all, since \(x\) is prime greater than 3, then \(x\) is odd. Now, if \(x\) is odd, then \((x - 3)(x -1)(x+1)(x+3) \) is the product of four consecutive even integer, so even. Plus, out of four consecutive even integers, at least one will for sure be divisible by 3. Thus, \((x - 3)(x -1)(x+1)(x+3) \) is divisible by \(2*3=6\).
Next, since \(x = 5q + 1\) (\(x\) divided by 5 gives the remainder of 1), then \((x - 3)(x -1)(x+1)(x+3) =(5q + 1 - 3)(5q + 1 -1)(5q + 1+1)(5q + 1+3) =(...)(5q)(...)(...)\) is a multiple of 5.
Therefore, \(x^4 - 10x^2 + 9 = y\) is a multiple of \(6*5=30\). Sufficient.
(2) \(x\) is a prime number such that when \(x\) divided by 6 gives the remainder of 5
Test numbers:
If \(x=5\), then \((x - 3)(x -1)(x+1)(x+3) = y\) won't be a multiple of 5, so won't be a multiple of 30 either.
If \(x=11\), then \((x - 3)(x -1)(x+1)(x+3) = 8*10*12*14\), which is a multiple of 2, 3, and 5, so a multiple of 30.
Two different answers. Not sufficient.
Answer: A
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