AMC8 2018
AMC8 2018 · Q21
AMC8 2018 · Q21. It mainly tests Chinese remainder / simultaneous congruences (basic).
How many positive three-digit integers have a remainder of 2 when divided by 6, a remainder of 5 when divided by 9, and a remainder of 7 when divided by 11?
有有多少个正的三位数，除以6余2，除以9余5，除以11余7？
(A)
1
1
(B)
2
2
(C)
3
3
(D)
4
4
(E)
5
5
Answer
Correct choice: (E)
正确答案：（E）
Solution
Answer (E): Let $n$ be a positive three-digit integer that satisfies the conditions stated in the problem. Note that $n$ has a remainder of $2$ when divided by $6$ and $5$ when divided by $9$. Similarly, it has a remainder of $7$ when divided by $11$. Hence $n+4$ is divisible by the least common multiple of $6$, $9$, and $11$, which is $198$. Thus $n+4=198k$ where $k=1,2,3,4,$ or $5$, so $n=194,392,590,788, or\ 986$. The five numbers satisfy the conditions in the problem, so the answer is $5$.
解答 (E)：设 $n$ 为满足题意的三位正整数。注意到 $n$ 被 $6$ 除余 $2$、被 $9$ 除余 $5$，并且被 $11$ 除余 $7$。因此 $n+4$ 能被 $6,9,11$ 的最小公倍数 $198$ 整除。于是有 $n+4=198k$，其中 $k=1,2,3,4,5$，所以 $n=194,392,590,788,986$。这五个数都满足条件，所以答案为 $5$。
Topics
Related Questions
Practice full AMC exams on amcdrill.
Try full-length practice and diagnostics at www.amcdrill.com.