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2025-08-30 18:35:01 +08:00
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回文数有两种
一种是单数位数，$abcba$，一种是双数位数，$abccba$
而双数位数是质数的只有$11$，为什么？
中间$0$的个数必然是偶数个$0$
证明： 设$f(n)$ = 1000...0001, $n$为偶数 则有
$f(n) - 11 = 999...9990$, 9的个数为n个 ，对于偶数个9，必然被11整除
偶回文数的各个数位可以分解为 n * 100...001(两个1中间若干个0)的形式 而这个 100..001 是能被11整除的
因为其余的都是11的倍数