2017 AIME II 第 4 题

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4.

求不超过 20172017 的正整数中,有多少个的三元表示不含数字 00

Find the number of positive integers less than or equal to 20172017 whose base-three representation contains no digit equal to 0.0.

答案:222
知识点:进制数字分类讨论
难度评级:2230
解答:

一个正整数的三元表示没有 00,当且仅当每一位都是 1122。对于 k=1,2,,6k = 1, 2, \ldots, 6,这样的 kk 位数有 2k2^k 个,并且它们全都不超过 2222223=728<2017222222_3 = 728 \lt 2017

因为 2017=220220132017 = 2202201_3,一个由 1122 组成的七位字符串不超过 20172017,当且仅当它以 111112122121 开头:任何以 2222 开头的字符串都会在第三位超过 220220132202201_3,因为它的各位都非零。因此有 325=963 \cdot 2^5 = 96 个七位数。

总数为 2+4+8+16+32+642 + 4 + 8 + 16 + 32 + 64 +96=222+ 96 = 222

A positive integer has no 00 in base three exactly when every digit is 11 or 2.2. For k=1,2,,6k = 1, 2, \ldots, 6 there are 2k2^k such kk-digit numbers, and all of them are at most 2222223=728<2017.222222_3 = 728 \lt 2017.

Since 2017=22022013,2017 = 2202201_3, a seven-digit string of 11s and 22s is at most 20172017 exactly when it begins with 11,11, 12,12, or 21:21: any string beginning 2222 already beats 220220132202201_3 at the third digit, since its digits are nonzero. That gives 325=963 \cdot 2^5 = 96 seven-digit numbers.

The total is 2+4+8+16+32+642 + 4 + 8 + 16 + 32 + 64 +96=222.+ 96 = 222.

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