AMC8 2012

AMC8 2012 · Q15

AMC8 2012 · Q15. It mainly tests GCD & LCM, Remainders & modular arithmetic.

The smallest number greater than 2 that leaves a remainder of 2 when divided by 3, 4, 5, or 6 lies between what numbers?

大于 2 的最小数，除以 3、4、5 或 6 余数均为 2，它位于什么数之间？

(A) 40 and 50 40 和 50

(B) 51 and 55 51 和 55

(D) 61 and 65 61 和 65

(E) 66 and 99 66 和 99

Answer

Correct choice: (D)

正确答案：（D）

Solution

To find the answer to this problem, we need to find the least common multiple of $3$, $4$, $5$, $6$ and add $2$ to the result. To calculate the least common multiple, we need to find the prime factorization for each of them. $3 = 3^{1}$, $4 = 2^{2}$, $5 = 5^{1}$, and $6 = 2^{1}*{3^{1}}$. So the least common multiple of the four numbers is $2^{2}*{3^{1}*{5^{1}}} = 60$, and by adding $2$, we find that that such number is $62$. Now we need to find the only given range that contains $62$. The only such range is answer $\textbf{(D)}$, and so our final answer is $\boxed{\textbf{(D)}\ 61\text{ and }65}$.

求 3、4、5、6 的最小公倍数并加 2。先求质因数分解：$3 = 3^{1}$，$4 = 2^{2}$，$5 = 5^{1}$，$6 = 2^{1}\times{3^{1}}$。最小公倍数为 $2^{2}\times{3^{1}\times{5^{1}}} = 60$，加 2 得 62。包含 62 的唯一范围是选项 $\textbf{(D)}$，最终答案 $\boxed{\textbf{(D)}\ 61\text{ and }65}$。

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