Question

This question was on the American Mathematics Contest and it made 0 sense.
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 Nth row. What is the sum of the digits of N?

Answer 1

So the rows sum in this way,\[\large\rm \sum_{i=1}^n i=1+2+...+n\]And we know that\[\large\rm \sum_{i=1}^n i=\frac{n(n+1)}{2}\]

So hmm what can we do...
2016 is this total number of coins, right?\[\large\rm 2016=\frac{n(n+1)}{2}\]And then solve the quadratic that is formed from this equation.

Does that work maybe?
Hmm thinking...

Answer 2

That question is worded rather poorly.
"What is the sum of the digits of N?"
Since there are "N" coins in the "Nth" row and n=63, then the "sum of the digits is 9???
(6+3=9)

Answer 3

ya seems that way :o

Answer 4

Wow... one of the answer choices was 9, so that was probably the answer. Very poorly made question.

Answer 5

But thanks guys for your effort.

Answer 6

Okay

Answer 7

I don't understand why you think it was poorly worded...

You need to find the number of rows,
then the question is asking for another simple step beyond that,
add up the digits of this number that you found.