A triangular array of 2016 coins has 1 coin in the first row, 2 coins

Question

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?

(A) 6

(B) 7

(C) 8

(D) 9

(E) 10

Aboyhasnoname · Accepted Answer

We'll I was not able to deduce that 4032 is 63 X 64...I took a longer route..... 
n^2 + n = 4032..... 
So N should be between 60 and 70...because 60 X 60 = 3600 .. and 70 X 70 = 4900... So n should be between these two.... 
From this...I can say that A can't be the answer...as from 61 the sum of digits will be more than 6...

Going to Option B.. 7...we have 61 only as our option for n...61 X 61 + 61....need not to solve it...since 60 X 60 = 3600..it will be somewhere around 3800... to be exact.. .61 X 61 + 61 = 3782....

Now Option C... which is 8...the only option that fits is 62... This cant be the answer as..62 X 62..will give units digit as 4 and then adding 62 will give unit's digit 6...

Option D...which is 9.... 63 X 63 + 63.... Check it..it gives 4032....Hence D is the answer....
Now... I Know its a little lengthy ...but this was is what struck me under time pressure...

Chethan92 · Answer

1st row, 1 coin
2nd row, 2 coin
So, 1+2+3+4+5+.....+N
Sum of consecutive numbers =  \frac{n(n+1)}{2} = 2016 
 n^2+n-4032 = 0 
 n^2+64n-63n-4032 = 0 
 n(n+64)-63(n+64) = 0 
 n = 63 or -64 
n is positive.
n = 63
Sum of the digits of N = 6+3 = 9

D is the answer.

CAMANISHPARMAR · Answer

Applying the formula for sum of first N natural numbers we can find the some of digits on N.
 Option D is the correct Ans.

xhaltar · Answer

Hi,

I just wanted to add, for anybody that like me cannot "guess" that 4032 is 64*63 "on-the-fly", that you need to decompose as prime factor.

I just posed :
4032
2016
1008
504
252
126
63

And then, we can deduce that 4032 = 63 * 2 * 2 * 2 * 2 * 2 * 2 = 63 * 64

Then, we are back to equation : n² + n - 4032

Note :
- We know that a polynom like this one :
- "x² + sx + p" can be written as (x+j)(x-k)
- where :
 - j + k = s
 - j * k = p

We have :
- p = 63 * 64 (as we deduced previously)
- So, we need to find s => Obviously, 64 - 63 = 1, which is our s
- So, we can deduce that j = 64 and k = 63

Which gives : n² + sn + p = (n+64)(n-64)

To be equal to 0, etc.

KarishmaB · Answer

Total number of coins = 2016 = 1 + 2 + 3 + 4 + ... + N = N*(N+1)/2

N*(N+1) = 8*252*2 = 2^6 * 7 * 9 = 63 * 64

N = 63
Sum of the digits of N = 6+3 = 9

Answer (D)

A triangular array of 2016 coins has 1 coin in the first row, 2 coins

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A triangular array of 2016 coins has 1 coin in the first row, 2 coins

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