AMC10 2016 A

AMC10 2016 A · Q9

AMC10 2016 A · Q9. It mainly tests Sequences & recursion (algebra), Fractions.

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枚硬币组成的三角形排列：第一行有1枚硬币，第二行有2枚，第三行有3枚，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 $1+2+\cdots+N=\dfrac{N(N+1)}{2}$ 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 $1+2+\cdots+N=\dfrac{N(N+1)}{2}$ 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。

Topics