Question

回忆：回文数是正着读和倒着读都相同的数。求小于 $1000$ 的最大整数，使它用十进制和八进制表示时都是回文数，例如 $292 = 444_8$ 。

Recall that a palindrome is a number that reads the same forward and backward. Find the greatest integer less than $1000$ that is a palindrome both when written in base ten and when written in base eight, such as $292 = 444_8.$

答案

答案：585

知识点：回文数进制系统列举

难度评级：2110

小提示：

在 $512$ 与 $999$ 之间的八进制回文数有四位，形如 $\overline{1bb1}_8 = 513 + 72b$

Any base-eight palindrome between $512$ and $999$ has four base-eight digits and looks like $\overline{1bb1}_8 = 513 + 72b$

大提示：

三位八进制数至多为 $511$ ，所以只需从 $b = 6, 5, 4, \ldots$ 开始检验它的十进制表示是否为回文数

Three-digit base-eight numbers are at most $511,$ so just test $b = 6, 5, 4, \ldots$ for a base-ten palindrome

解答：

四位八进制数在 $512$ 与 $4095$ 之间，所以小于 $1000$ 的四位八进制回文数的首位（也是末位）必须是 $1$ ：它的形式为 $\overline{1bb1}_8 = 512 + 64b$ ${}+ 8b + 1$ $= 513 + 72b$ 。要使它小于 $1000$ ，需 $b \le 6$ ，得到候选数 $513, 585, 657, 729, 801, 873, 945$ 。

从大到小检查，其中唯一在十进制下也是回文数的是 $585 = 1111_8$ 。所有至多三位的八进制回文数至多为 $777_8 = 511 \lt 585$ ，所以答案是 $585$ 。

A four-digit base-eight number lies between $512$ and $4095,$ so a base-eight palindrome less than $1000$ with four digits must have leading (and trailing) digit $1:$ it has the form $\overline{1bb1}_8 = 512 + 64b$ ${}+ 8b + 1$ $= 513 + 72b.$ Keeping this below $1000$ requires $b \le 6,$ giving the candidates $513, 585, 657, 729, 801, 873, 945.$

Checking from the top, the only one of these that is also a palindrome in base ten is $585 = 1111_8.$ Every base-eight palindrome with at most three digits is at most $777_8 = 511 \lt 585,$ so the answer is $585.$

2023 AIME II 第 2 题

答案：585

难度评级：2110

小提示：

大提示：

解答：

其他年份的第 2 题