AMC10 2018 B

AMC10 2018 B · Q16

AMC10 2018 B · Q16. It mainly tests Remainders & modular arithmetic, Powers & residues.

Let $a_1, a_2, \dots, a_{2018}$ be a strictly increasing sequence of positive integers such that \[a_1 + a_2 + \cdots + a_{2018} = 2018^{2018}.\] What is the remainder when $a_1^3 + a_2^3 + \cdots + a_{2018}^3$ is divided by 6?

设 $a_1, a_2, \dots, a_{2018}$ 是一个严格递增的正整数序列，使得 \[a_1 + a_2 + \cdots + a_{2018} = 2018^{2018}.\] 当 $a_1^3 + a_2^3 + \cdots + a_{2018}^3$ 除以 6 的余数是多少？

(A) 0 0

(B) 1 1

(D) 3 3

(E) 4 4

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): Let $n$ be an integer. Because $n^3-n=(n-1)n(n+1)$, it follows that $n^3-n$ has at least one prime factor of $2$ and one prime factor of $3$ and therefore is divisible by $6$. Thus $n^3\equiv n\pmod 6$. Then $$ a_1^3+a_2^3+\cdots+a_{2018}^3\equiv a_1+a_2+\cdots+a_{2018}\equiv 2018^{2018}\pmod 6. $$ Because $2018\equiv 2\pmod 6$, the powers of $2018$ modulo $6$ are alternately $2,4,2,4,\ldots$, so $2018^{2018}\equiv 4\pmod 6$. Therefore the remainder when $a_1^3+a_2^3+\cdots+a_{2018}^3$ is divided by $6$ is $4$.

答案（E）：设 $n$ 为整数。因为 $n^3-n=(n-1)n(n+1)$，可知 $n^3-n$ 至少含有一个素因子 $2$ 和一个素因子 $3$，因此能被 $6$ 整除。于是 $n^3\equiv n\pmod 6$。因此 $$ a_1^3+a_2^3+\cdots+a_{2018}^3\equiv a_1+a_2+\cdots+a_{2018}\equiv 2018^{2018}\pmod 6. $$ 又因为 $2018\equiv 2\pmod 6$，所以 $2018$ 在模 $6$ 下的幂交替为 $2,4,2,4,\ldots$，从而 $2018^{2018}\equiv 4\pmod 6$。因此 $a_1^3+a_2^3+\cdots+a_{2018}^3$ 除以 $6$ 的余数是 $4$。

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