2023 AIME II 第 2 题

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2.

回忆:回文数是正着读和倒着读都相同的数。求小于 10001000 的最大整数,使它用通常写法和以八为底表示时都是回文数, 例如 292=444eight292 = 444_{\text{eight}}

Recall that a palindrome is a number that reads the same forward and backward. Find the greatest integer less than 10001000 that is a palindrome both when written in base ten and when written in base eight, such as 292=444eight.292 = 444_{\text{eight}}.

答案:585
知识点:回文数进制系统列举
难度评级:2070
解答:

四位的以八为底的数在 51251240954095 之间,所以小于 10001000 的四位以八为底的回文数的首位(也是末位)必须是 11:它的形式为 1bb1eight=512+64b\overline{1bb1}_{\text{eight}} = 512 + 64b +8b+1{}+ 8b + 1 =513+72b= 513 + 72b。要使它小于 10001000,需 b6b \le 6,得到候选数 513,585,657,729,801,873,945513, 585, 657, 729, 801, 873, 945

从大到小检查,其中唯一按通常写法也是回文数的是 585=1111eight585 = 1111_{\text{eight}}。所有至多三位的以八为底的回文数至多为 777eight=511<585777_{\text{eight}} = 511 \lt 585,所以答案是 585585

A four-digit base-eight number lies between 512512 and 4095,4095, so a base-eight palindrome less than 10001000 with four digits must have leading (and trailing) digit 1:1: it has the form 1bb1eight=512+64b\overline{1bb1}_{\text{eight}} = 512 + 64b +8b+1{}+ 8b + 1 =513+72b.= 513 + 72b. Keeping this below 10001000 requires b6,b \le 6, giving the candidates 513,585,657,729,801,873,945.513, 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=1111eight.585 = 1111_{\text{eight}}. Every base-eight palindrome with at most three digits is at most 777eight=511<585,777_{\text{eight}} = 511 \lt 585, so the answer is 585.585.

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