Squares, Cubes, and Subtraction

Probability Level 3

Let the positive difference between $1^2+2^2+3^2+4^2+...+2014^2$ and $1^3+2^3+3^3+4^3+...+2014^3$ be equal to $N$ . Compute the digit sum of $N$ .

The answer is 26.

2 solutions

Saurabh Mallik
Mar 27, 2014

The value of $N$ :

$N = (1^{3}+2^{3}+3^{3}+4^{3}+...+2014^{3}) - (1^{2}+2^{2}+3^{2}+4^{2}+...+2014^{2}) = 4,114,542,013,010$

Digit sum of $N$ $= 4 + 1 + 1+ 4 + 5 + 4 + 2 + 0 + 1 + 3 + 0 + 1 + 0 = \boxed{26}$

Can somebody help me find solutions to this question?

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Saurabh Mallik - 7 years, 2 months ago

Satvik Golechha
Mar 18, 2014

I had to calculate the whole number to find the digit sum.... Please tell me to do it in a better way without finding the whole number...ThAnKs

Squares, Cubes, and Subtraction

The answer is 26.

2 solutions

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