Question

求最小的质数，使它是一个递增等差数列的第五项，并且前四项也都是质数。

Find the smallest prime that is the fifth term of an increasing arithmetic sequence, all four preceding terms also being prime.

答案

答案：29

知识点：质数等差数列整除性

难度评级：1890

小提示：

如果公差不是 $3$ 的倍数，那么前三项中有一项能被 $3$ 整除

If the common difference is not a multiple of $3,$ one of the first three terms is divisible by $3$

大提示：

公差必须是 $6$ 的倍数，所以从小质数开始尝试公差为 $6$ 的五项等差数列

The difference must be a multiple of $6,$ so try five-term progressions with difference $6$ starting from small primes

解答：

设这些项为 $p$ 、 $p + d$ 、 $\ldots$ 、 $p + 4d$ 。偶质数不能作为第一项：若公差为偶数，第二项为偶数；若公差为奇数，第三项为偶数。如果 $d$ 是奇数，相邻两项奇偶性相反，因此除了第一项以外会有某一项是大于 $2$ 的偶数，不可能是质数。如果 $d$ 不是 $3$ 的倍数，那么 $p$ 、 $p + d$ 、 $p + 2d$ 会覆盖模 $3$ 的所有余数，所以其中一项能被 $3$ 整除；这一项只能等于 $3$ ，从而迫使 $p = 3$ ，但此时 $p + 3d = 3(1 + d)$ 是合数。因此 $d$ 能被 $6$ 整除。

因为 $d \ge 6$ ，第五项至少为 $p + 24$ 。取 $p = 5$ 且 $d = 6$ ，得到 $5, 11, 17, 23, 29$ ，全部为质数；又因为 $p \ge 5$ （以 $p = 2$ 和 $p = 3$ 开头都会失败），不可能有更小的第五项。答案是 $29$ 。

Let the terms be $p,$ $p + d,$ $\ldots,$ $p + 4d.$ The even prime cannot be the first term: an even difference makes the second term even, while an odd difference makes the third term even. If $d$ were odd, consecutive terms would have opposite parity, so some term other than the first would be even and greater than $2$ — impossible. If $d$ were not a multiple of $3,$ then $p,$ $p + d,$ $p + 2d$ would cover all residues mod $3,$ so some term would be divisible by $3;$ that term would have to be $3$ itself, forcing $p = 3,$ but then $p + 3d = 3(1 + d)$ is composite. Hence $d$ is divisible by $6.$

With $d \ge 6$ the fifth term is at least $p + 24.$ Trying $p = 5$ and $d = 6$ gives $5, 11, 17, 23, 29,$ all prime, and no smaller fifth term is possible since $p \ge 5$ (the starts $p = 2$ and $p = 3$ fail as above). The answer is $29.$

1999 AIME 第 1 题

答案：29

难度评级：1890

小提示：

大提示：

解答：

其他年份的第 1 题