2010 AIME I 第 10 题

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10.

NN 为将 20102010 写成如下形式的方法数: 2010=a3103+a2102+a110+a0 \begin{aligned} 2010 &= a_3 \cdot 10^3 + a_2 \cdot 10^2 \\ &\quad {}+ a_1 \cdot 10 + a_0 \end{aligned}\text{,}其中 aia_i 都是整数,且 0ai990 \le a_i \le 99。这样一种表示的例子是 11031 \cdot 10^3 +3102+ 3 \cdot 10^2 +67101+ 67 \cdot 10^1 +40100+ 40 \cdot 10^0。求 NN

Let NN be the number of ways to write 20102010 in the form 2010=a3103+a2102+a110+a0, \begin{aligned} 2010 &= a_3 \cdot 10^3 + a_2 \cdot 10^2 \\ &\quad {}+ a_1 \cdot 10 + a_0, \end{aligned} where the aia_i’s are integers, and 0ai99.0 \le a_i \le 99. An example of such a representation is 11031 \cdot 10^3 +3102+ 3 \cdot 10^2 +67101+ 67 \cdot 10^1 +40100.+ 40 \cdot 10^0. Find N.N.

答案:202
知识点:进制数字双射
难度评级:2840
小提示:

将每个系数拆成数字:ai=10bi+cia_i = 10 b_i + c_i,其中 bi,ci{0,1,,9}b_i, c_i \in \{0, 1, \ldots, 9\}

Split each coefficient into digits: ai=10bi+cia_i = 10 b_i + c_i with bi,ci{0,1,,9}b_i, c_i \in \{0, 1, \ldots, 9\}

大提示:

bib_i 是一个数 mm 的各位数字,cic_i 是一个数 nn 的各位数字,条件正好是 10m+n=201010m + n = 2010;数可能的 mm

The bib_i are the digits of a number mm and the cic_i of a number n,n, and the condition is exactly 10m+n=2010;10m + n = 2010; count the possible mm

解答:

将每个系数写成 ai=10bi+cia_i = 10b_i + c_i,其中数字 bi,ci{0,1,,9}b_i, c_i \in \{0, 1, \ldots, 9\};每个整数 0ai990 \le a_i \le 99 都能唯一这样拆分。令 m=b3b2b1b0m = b_3 b_2 b_1 b_0n=c3c2c1c0n = c_3 c_2 c_1 c_0(按 1010 进制数读取),条件变为 2010=10m+n2010 = 10m + n\text{。}

反过来,任何非负整数 m,nm, n 若满足 10m+n=201010m + n = 2010,就有 m201m \le 201n2010n \le 2010,所以二者至多四位;这些数字可恢复 bib_icic_i,从而恢复 aia_i。因此表示方法与 m{0,1,,201}m \in \{0, 1, \ldots, 201\} 的选择一一对应,其中 n=201010mn = 2010 - 10m,所以 N=202N = 202

Write each coefficient as ai=10bi+cia_i = 10b_i + c_i with digits bi,ci{0,1,,9};b_i, c_i \in \{0, 1, \ldots, 9\}; every integer 0ai990 \le a_i \le 99 splits this way uniquely. Setting m=b3b2b1b0m = b_3 b_2 b_1 b_0 and n=c3c2c1c0n = c_3 c_2 c_1 c_0 (read as base-1010 numbers), the condition becomes 2010=10m+n.2010 = 10m + n.

Conversely, any nonnegative integers m,nm, n with 10m+n=201010m + n = 2010 satisfy m201m \le 201 and n2010,n \le 2010, so each has at most four digits; those digits recover the bib_i and ci,c_i, hence the ai.a_i. So representations correspond exactly to choices of m{0,1,,201}m \in \{0, 1, \ldots, 201\} with n=201010m,n = 2010 - 10m, and N=202.N = 202.

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