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2016 AMC 12A Problem 6

Problem 6 of 25EasierAlgebra

A triangular array of 20162016 coins has 11 coin in the first row, 22 coins in the second row, 33 coins in the third row, and so on up to NN coins in the NNth row. What is the sum of the digits of N?N?

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Solution

The total number of coins is 1+2+⋯+N=N(N+1)2=2016, \begin{gathered} 1+2+\cdots+N\\ =\dfrac{N(N+1)}{2}\\ =2016, \end{gathered} so N(N+1)=4032.N(N+1)=4032. Since 63⋅64=4032,63\cdot 64=4032, we have N=63,N=63, and the sum of its digits is 6+3=9.6+3=9. Thus, the correct answer is D.

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Concepts: triangular number · estimation

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.