AMC12 2016 A

AMC12 2016 A · Q6

AMC12 2016 A · Q6. It mainly tests Quadratic equations, Arithmetic sequences basics.

A triangular array of $2016$ coins has $1$ coin in the first row, $2$ coins in the second row, $3$ coins in the third row, and so on up to $N$ coins in the $N$th row. What is the sum of the digits of $N$?

由$2016$枚硬币组成的一个三角形排列：第一行有$1$枚硬币，第二行有$2$枚硬币，第三行有$3$枚硬币，依此类推，直到第$N$行有$N$枚硬币。问$N$的各位数字之和是多少？

(A) 6 6

(B) 7 7

(D) 9 9

(E) 10 10

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): There are $$ 1+2+\cdots+N=\frac{N(N+1)}{2} $$ coins in the array. Therefore $N(N+1)=2\cdot 2016=4032$. Because $N(N+1)\approx N^2$, it follows that $N\approx \sqrt{4032}\approx \sqrt{2^{12}}=2^6=64$. Indeed, $63\cdot 64=4032$, so $N=63$ and the sum of the digits of $N$ is 9.

答案（D）：该阵列中有 $$ 1+2+\cdots+N=\frac{N(N+1)}{2} $$ 枚硬币。因此 $N(N+1)=2\cdot 2016=4032$。由于 $N(N+1)\approx N^2$，可得 $N\approx \sqrt{4032}\approx \sqrt{2^{12}}=2^6=64$。确实，$63\cdot 64=4032$，所以 $N=63$，且 $N$ 的各位数字之和为 9。

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