1964 AHSME Problems/Problem 16

Problem

Let $f(x)=x^2+3x+2$ and let $S$ be the set of integers $\{0, 1, 2, \dots , 25 \}$ . The number of members $s$ of $S$ such that $f(s)$ has remainder zero when divided by $6$ is:

$\textbf{(A)}\ 25\qquad \textbf{(B)}\ 22\qquad \textbf{(C)}\ 21\qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 17$

Solution

Note that for all polynomials $f(x)$ , $f(x + 6) \equiv f(x) \pmod 6$ .

Proof: If $f(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_0$ , then $f(x+6) = a_n(x+6)^n + a_{n-1}(x-6)^{n-1} +...+ a_0$ . In the second equation, we can use the binomial expansion to expand every term, and then subtract off all terms that have a factor of $6^1$ or higher, since subtracting a multiple of $6$ will not change congruence $\pmod 6$ . This leaves $a_nx^n + a_{n-1}x^{n-1} + ... + a_0$ , which is $f(x)$ , so $f(x+6) \equiv f(x) \pmod 6$ .

So, we only need to test when $f(x)$ has a remainder of $0$ for $0, 1, 2, 3, 4, 5$ . The set of numbers $6, 7, 8, 9, 10, 11$ will repeat remainders, as will all other sets. The remainders are $2, 0, 0, 2, 0, 0$ .

This means for $s=1, 2, 4, 5$ , $f(s)$ is divisible by $6$ . Since $f(1)$ is divisible, so is $f(s)$ for $s=7, 13, 19, 25$ , which is $5$ values of $s$ that work. Since $f(2)$ is divisible, so is $f(s)$ for $s=8, 14, 20$ , which is $4$ more values of $s$ that work. The values of $s=4, 5$ will also generate $4$ solutions each, just like $f(2)$ . This is a total of $17$ values of $s$ , for an answer of $\boxed{\textbf{(E)}}$

1964 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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1964 AHSME Problems/Problem 16

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