2018 AMC 8 第 7 题

先试着解答 2018 AMC 8 第 7 题,然后核对你的答案与视频与精心整理的解答,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 2018 AMC 8 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。

7.

55 位数 2‾\underline{2} 0‾\underline{0} 1‾\underline{1} 8‾\underline{8} U‾\underline{U} 能被 99 整除。这个数除以 88 的余数是多少?
The 55-digit number 2‾\underline{2} 0‾\underline{0} 1‾\underline{1} 8‾\underline{8} U‾\underline{U} is divisible by 9.9. What is the remainder when this number is divided by 8?8?
11
33
55
66
77
答案:B
知识点:整除性模运算
小提示:
先用 99 的整除规则确定 UU
Use the divisibility-by-99 rule to determine UU.
大提示:
UU 确定后,求模 88 的余数只需要看最后三位
Once UU is known, only the last three digits matter for the remainder modulo 88.
视频讲解:
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文字解答:
注意,一个数能被 99 整除,当且仅当它的各位数字之和也能被 99 整除。
题中这个 55 位数的数字和为 2+0+1+8+U=11+U。2+0+1+8+U= 11+U\text{。}因为 2‾\underline{2} 0‾\underline{0} 1‾\underline{1} 8‾\underline{8} U‾\underline{U} 能被 99 整除,11+U11+U 也必须被 99 整除。又因为 UU 是一位数字,0≤U≤90\le U\le 9,所以 UU 只能是 77。
现在可知这个 55 位数是 2018720187,而我们要求 2018720187 除以 88 的余数。长除法给出 20187=2523⋅8+320187=2523\cdot 8 + 3,所以余数是 33。
所以正确答案是 B。
Notice that a number is divisible by 99 if and only if the sum of its digits is also divisible by 9.9.
The sum of the digits of the 55-digit number in the problem is: 2+0+1+8+U=11+U.2+0+1+8+U= 11+U. As 2‾\underline{2} 0‾\underline{0} 1‾\underline{1} 8‾\underline{8} U‾\underline{U} is divisible by 9,9, 11+U11+U must also be divisible by 9.9. Also, as UU is a digit, we know that 0≤U≤9.0\le U\le 9. This means that UU can only be 7.7.
Now we know that the 55-digit number in question is 20187,20187, and we want to find the remainder when we divide 2018720187 by 8.8. To solve this, simply use long division to see that 20187=2523⋅8+3.20187=2523\cdot 8 + 3. Therefore, the remainder is 3.3.
Thus, the correct answer is B.
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