2016 AMC 12A Problem 6

Attempt Problem 6 of the 2016 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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?

6

7

8

9

10

1+2+\cdots+N=\dfrac{N(N+1)}{2}

Solve

N(N+1)=4032

by noting

N\approx\sqrt{4032}

The total number of coins is

\begin{gathered} 1+2+\cdots+N\\ =\dfrac{N(N+1)}{2}\\ =2016, \end{gathered}

N(N+1)=4032.

Since

63\cdot 64=4032,

we have

N=63,

and the sum of its digits is

6+3=9.

Thus, the correct answer is D.

Problem 6 in other years