2013 AMC 10B 第 25 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

25.

Bernardo 选择一个三位正整数 NN,并把它的五进制表示和六进制表示都写在黑板上。后来 LeRoy 看到这两个数,把它们当作以十为底的整数相加,得到整数 SS

例如,若 N=749N = 749,Bernardo 写下 10, ⁣44410,\!4443, ⁣2453,\!245,LeRoy 得到 S=13, ⁣689S = 13,\!689。有多少个 NN 的选择,使得 SS 的最右两位数字按顺序与 2N2N 的最右两位数字相同?

Bernardo chooses a three-digit positive integer NN and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer S.S.

For example, if N=749,N = 749, Bernardo writes the numbers 10, ⁣44410,\!444 and 3, ⁣245,3,\!245, and LeRoy obtains the sum S=13, ⁣689.S = 13,\!689. For how many choices of NN are the two rightmost digits of S,S, in order, the same as those of 2N?2N?

55

1010

1515

2020

2525

答案:E
知识点:进制模运算中国剩余定理
难度评级:2440
解答:

只需在模 900=lcm(25,36,100)900=\operatorname{lcm}(25,36,100) 下分析,因为 55 进制、66 进制和以十为底表示时的末两位都以此为周期。

55 进制末两位为 a1a0a_1a_066 进制末两位为 b1b0b_1b_0。个位条件给出 a0+b02N2a0(mod10)a_0+b_0\equiv2N\equiv2a_0\pmod{10},所以 a0=b0a_0=b_0

写作 N=150N3+30a1+a0N=150N_3+30a_1+a_066 进制十位条件给出 N3a1+b1(mod6)N_3\equiv a_1+b_1\pmod6,所以 N=900N4+180a1N=900N_4+180a_1 +150b1+a0+150b_1+a_0

比较 SS2N2N 的十位条件,化简得 5a1b1(mod10)5a_1\equiv b_1\pmod{10}。在 0a140\le a_1\le40b150\le b_1\le5 下,可行对为 (0,0),(2,0),(4,0),(1,5),(3,5)(0,0),(2,0),(4,0),(1,5),(3,5)

a0=b0a_0=b_055 种选择,所以共有 55=255\cdot5=25NN

所以正确答案是 E

It is enough to work modulo 900=lcm(25,36,100)900=\operatorname{lcm}(25,36,100), because the last two base-55, base-66, and decimal digits repeat with that period.

Let the last two base-55 digits be a1a0a_1a_0 and the last two base-66 digits be b1b0b_1b_0. The units digit condition gives a0+b02N2a0(mod10)a_0+b_0\equiv2N\equiv2a_0\pmod{10}, so a0=b0a_0=b_0.

Writing N=150N3+30a1+a0N=150N_3+30a_1+a_0, the base-66 tens digit condition gives N3a1+b1(mod6)N_3\equiv a_1+b_1\pmod6, so N=900N4+180a1N=900N_4+180a_1 +150b1+a0+150b_1+a_0.

The tens digit condition for SS and 2N2N reduces to 5a1b1(mod10)5a_1\equiv b_1\pmod{10}. With 0a140\le a_1\le4 and 0b150\le b_1\le5, the valid pairs are (0,0),(2,0),(4,0),(1,5),(3,5)(0,0),(2,0),(4,0),(1,5),(3,5).

There are 55 choices for a0=b0a_0=b_0, so there are 55=255\cdot5=25 choices of NN.

Thus, the correct answer is E .

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