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[数论] 同余

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cty_zhl posted 2019-2-21 17:28 |Read mode
已知数列{an}满足:an=(根号下(2)+1)^n-(根号下(2)-1)^n(n∈N),
用[x]表示不超过实数x的最大整数,则[a2017]的个位数字是

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kuing posted 2019-2-21 21:13
发现循环后暴力递推来一发

首先易知:当 `n` 为奇数时 `a_n` 为整数,当 `n` 为偶数时 `\sqrt2a_n` 为整数。

由特征方程理论可知,`a_n` 满足递推关系\[a_{n+2}=2\sqrt2a_{n+1}-a_n,\]通过不断迭代计算,可得\[a_{n+13}=47321a_{n+1}-13860\sqrt2a_n,\]那么,当 `n` 为偶数时,由于 `a_{n+1}` 以及 `\sqrt2a_n` 均为整数,因此,对上式模 `10` 即得\[a_{n+13}\equiv a_{n+1}\pmod{10},\]此式对 `n` 为偶数恒成立,所以\[a_{2017}=a_{168\times12+1}\equiv a_1=2\pmod{10},\]所以它的个位数字就是 `2`。
业余的业余 posted 2019-2-21 21:54
厉害

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