2008 AIME II 第 4 题

先试着解答 2008 AIME II 第 4 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2008 AIME II 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

4.

存在唯一的一组 rr 个非负整数 n1>n2>⋯>nrn_1 \gt n_2 \gt \cdots \gt n_r,以及唯一确定的 rr 个整数 aka_k (1≤k≤r)(1 \le k \le r),其中每个 aka_k 都等于 11 或 −1-1,使得 a13n1+a23n2+⋯+ar3nr=2008。 \begin{aligned} &a_1 3^{n_1} + a_2 3^{n_2} + \cdots + a_r 3^{n_r} \\ &= 2008 \end{aligned}\text{。}求 n1+n2+⋯+nrn_1 + n_2 + \cdots + n_r。

There exist rr unique nonnegative integers n1>n2>⋯>nrn_1 \gt n_2 \gt \cdots \gt n_r and rr unique integers aka_k (1≤k≤r)(1 \le k \le r) with each aka_k either 11 or −1-1 such that a13n1+a23n2+⋯+ar3nr=2008. \begin{aligned} &a_1 3^{n_1} + a_2 3^{n_2} + \cdots + a_r 3^{n_r} \\ &= 2008. \end{aligned} Find n1+n2+⋯+nr.n_1 + n_2 + \cdots + n_r.

答案:21
知识点:进制指数
难度评级:2350
小提示:

先把 20082008 写成 33 进制

Start by writing 20082008 in base 33

大提示:

把每个数字 22 用公式 2⋅3k=3k+1−3k2 \cdot 3^k = 3^{k+1} - 3^k 替换,使每个系数都变成 11 或 −1-1

Replace each digit 22 using 2⋅3k=3k+1−3k2 \cdot 3^k = 3^{k+1} - 3^k so that every coefficient becomes 11 or −1-1

解答:

在 33 进制下,2008=220210132008 = 2202101_3,也就是说 2008=2⋅36+2⋅35+2⋅33+32+30。 \begin{aligned} 2008 &= 2 \cdot 3^6 + 2 \cdot 3^5 + 2 \cdot 3^3 \\ &\quad {}+ 3^2 + 3^0 \end{aligned}\text{。}为了把数字 22 转成系数 ±1\pm 1,使用 2⋅3k=3k+1−3k2 \cdot 3^k = 3^{k+1} - 3^k。两个相邻的数字 22 会整齐抵消:2⋅36+2⋅35=(37−36)+(36−35)=37−35, \begin{aligned} 2 \cdot 3^6 + 2 \cdot 3^5 &= (3^7 - 3^6) \\ &\quad {}+ (3^6 - 3^5) \\ &= 3^7 - 3^5 \end{aligned}\text{,}并且 2⋅33=34−332 \cdot 3^3 = 3^4 - 3^3。

因此 2008=37−35+34−33+32+30, \begin{aligned} 2008 &= 3^7 - 3^5 + 3^4 - 3^3 \\ &\quad {}+ 3^2 + 3^0 \end{aligned}\text{,}它有互不相同的指数和系数 ±1\pm 1,符合要求。指数之和为 7+5+4+3+2+0=217 + 5 + 4 + 3 + 2 + 0 = 21。

In base 3,3, 2008=22021013,2008 = 2202101_3, that is, 2008=2⋅36+2⋅35+2⋅33+32+30. \begin{aligned} 2008 &= 2 \cdot 3^6 + 2 \cdot 3^5 + 2 \cdot 3^3 \\ &\quad {}+ 3^2 + 3^0. \end{aligned} To convert the digits 22 into coefficients ±1,\pm 1, use 2⋅3k=3k+1−3k.2 \cdot 3^k = 3^{k+1} - 3^k. The two adjacent digits 22 collapse neatly: 2⋅36+2⋅35=(37−36)+(36−35)=37−35, \begin{aligned} 2 \cdot 3^6 + 2 \cdot 3^5 &= (3^7 - 3^6) \\ &\quad {}+ (3^6 - 3^5) \\ &= 3^7 - 3^5, \end{aligned} and 2⋅33=34−33.2 \cdot 3^3 = 3^4 - 3^3.

Therefore 2008=37−35+34−33+32+30, \begin{aligned} 2008 &= 3^7 - 3^5 + 3^4 - 3^3 \\ &\quad {}+ 3^2 + 3^0, \end{aligned} which has distinct exponents and coefficients ±1,\pm 1, as required. The sum of the exponents is 7+5+4+3+2+0=21.7 + 5 + 4 + 3 + 2 + 0 = 21.

第 3 题#3
完整试卷

其他年份的第 4 题