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AMC10 2018 B

AMC10 2018 B · Q11

AMC10 2018 B · Q11. It mainly tests Primes & prime factorization, Remainders & modular arithmetic.

Which of the following expressions is never a prime number when \(p\) is a prime number?
当 \(p\) 是质数时,下列哪个表达式永远不是质数?
(A) \(p^2 + 16\) \(p^2 + 16\)
(B) \(p^2 + 24\) \(p^2 + 24\)
(C) \(p^2 + 26\) \(p^2 + 26\)
(D) \(p^2 + 46\) \(p^2 + 46\)
(E) \(p^2 + 96\) \(p^2 + 96\)
Answer
Correct choice: (C)
正确答案:(C)
Solution
Answer (C): If $p=3$, then $p^2+26=35=5\cdot 7$. If $p$ is a prime number other than 3, then $p=3k\pm 1$ for some positive integer $k$. In that case $$ p^2+26=(3k\pm 1)^2+26=9k^2\pm 6k+27=3(3k^2\pm 2k+9) $$ is a multiple of 3 and is not prime. The smallest counterexamples for the other choices are $5^2+16=41$, $7^2+24=73$, $5^2+46=71$, and $19^2+96=457$.
答案(C):如果$p=3$,则$p^2+26=35=5\cdot 7$。如果$p$是除3以外的素数,则对某个正整数$k$有$p=3k\pm 1$。在这种情况下 $$ p^2+26=(3k\pm 1)^2+26=9k^2\pm 6k+27=3(3k^2\pm 2k+9) $$ 是3的倍数,因此不是素数。其他选项的最小反例如下:$5^2+16=41$、$7^2+24=73$、$5^2+46=71$、以及$19^2+96=457$。
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