Question

求正整数 $m$ 的个数，使得存在非负整数 $x_0, x_1, \ldots, x_{2011}$ ，满足 $m^{x_0} = \sum_{k=1}^{2011} m^{x_k}.$

Find the number of positive integers $m$ for which there exist nonnegative integers $x_0, x_1, \ldots, x_{2011},$ such that $m^{x_0} = \sum_{k=1}^{2011} m^{x_k}.$

答案

答案：16

知识点：模运算因数个数

难度评级：2710

解答：

$m = 1$ 不成立，因为右边会是 $2011$ ，而左边是 $1$ 。对 $m \ge 2$ ，模 $m - 1$ 考察两边： $m$ 的每个幂都 $\equiv 1$ ，所以方程迫使 $1 \equiv 2011 \pmod{m - 1}$ ，也就是 $m - 1$ 整除 $2010$ 。

反过来，假设 $2010 = (m - 1)n$ 。取 $x_0 = n$ ，令 $m$ 个 $x_k$ 等于 $0$ ，并且对每个 $r = 1, 2, \ldots, n - 1$ ，令 $m - 1$ 个 $x_k$ 等于 $r$ 。这共用了 $m + (m - 1)(n - 1)$ $= n(m - 1) + 1$ $= 2011$ 项，且和会望远镜相消： $\begin{aligned} &m \\ &\quad {}+ (m - 1) \\ &\quad {}\cdot (m + m^2 + \cdots + m^{n-1}) \\ &\quad {}= m + (m^n - m) \\ &\quad {}= m^n = m^{x_0}. \end{aligned}$

所以方程有解当且仅当 $m - 1$ 整除 $2010 = 2 \cdot 3 \cdot 5 \cdot 67$ ，而它有 $2^4 = 16$ 个因数。因此这样的 $m$ 有 $16$ 个。

The value $m = 1$ fails, since the right side would be $2011$ while the left side is $1.$ For $m \ge 2,$ reduce mod $m - 1:$ every power of $m$ is $\equiv 1,$ so the equation forces $1 \equiv 2011 \pmod{m - 1},$ that is, $m - 1$ divides $2010.$

Conversely, suppose $2010 = (m - 1)n.$ Take $x_0 = n,$ let $m$ of the $x_k$ equal $0,$ and for each $r = 1, 2, \ldots, n - 1$ let $m - 1$ of the $x_k$ equal $r.$ This uses $m + (m - 1)(n - 1)$ $= n(m - 1) + 1$ $= 2011$ terms, and the sum telescopes: $\begin{aligned} &m \\ &\quad {}+ (m - 1) \\ &\quad {}\cdot (m + m^2 + \cdots + m^{n-1}) \\ &\quad {}= m + (m^n - m) \\ &\quad {}= m^n = m^{x_0}. \end{aligned}$

So the equation is solvable exactly when $m - 1$ divides $2010 = 2 \cdot 3 \cdot 5 \cdot 67,$ which has $2^4 = 16$ divisors. There are $16$ such $m.$

2011 AIME I 第 7 题

答案：16

难度评级：2710

解答：

其他年份的第 7 题