Digit sum

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What is the digit sum of 1 3 + 2 3 + 3 3 + . . . + 99 8 3 + 99 9 3 + 100 0 3 ? 1^{3}+2^{3}+3^{3}+...+998^{3}+999^{3}+1000^{3}?


The answer is 19.

Victor Loh by Victor Loh , 19, Singapore

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2 solutions

Prasun Biswas
Feb 9, 2014

This is an example of special series case where the series is in the form 1 3 + 2 3 + 3 + . . . + n 3 1^3+2^3+3^+...+n^3 . For the sum, we have an identity as follows----

1 3 + 2 3 + 3 3 + . . . . . + ( n − 1 ) 3 + n 3 = ( 1 2 n ( n + 1 ) ) 2 1^3+2^3+3^3+.....+(n-1)^3+n^3=(\frac{1}{2}n(n+1))^2 . So, we have ----

1 3 + 2 3 + 3 3 + . . . . + 100 0 3 1^3+2^3+3^3+....+1000^3

= ( 1 2 × 1000 × 1001 ) 2 = ( 500500 ) 2 = ( 5005 ) 2 × ( 100 ) 2 = 25050025 × 10000 = 250500250000 =(\frac{1}{2}\times 1000\times 1001)^2 = (500500)^2=(5005)^2\times (100)^2=25050025\times 10000 = \boxed{250500250000}

Now, digit sum = 2 + 5 + 0 + 5 + 0 + 0 + 2 + 5 + 0 + 0 + 0 + 0 = 19 =2+5+0+5+0+0+2+5+0+0+0+0=\boxed{19}

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Lorenc Bushi
Feb 1, 2014

We will use here this identity:

1 3 + 2 3 + 3 3 + . . . . . . . . + n 3 = ( n ( n + 1 ) 2 ) 2 1^3+2^3+3^3+........+n^3=(\dfrac{n(n+1)}{2})^2 .For our problem we have

1 3 + 2 3 + 3 3 + . . . . . . . . + 100 0 3 = ( 1000 ∗ 1001 2 ) 2 = 250500250000 1^3+2^3+3^3+........+1000^3=(\dfrac{1000*1001}{2})^2=250500250000 , thus the digit sum is 2 + 5 + 5 + 2 + 5 = 19 2+5+5+2+5=\boxed{19}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

why is the square?

Partho Kunda - 7 years, 4 months ago

I will give you the whole proof. First look at this pattern:

1 3 = 1 1^3=1

1 3 + 2 3 = ( 1 + 2 ) 2 = 9 1^3+2^3=(1+2)^2=9

1 3 + 2 3 + 3 3 = ( 1 + 2 + 3 ) 2 = 36 1^3+2^3+3^3=(1+2+3)^2=36

1 3 + 2 3 + 3 3 + 4 3 = ( 1 + 2 + 3 + 4 ) 2 = 100 1^3+2^3+3^3+4^3=(1+2+3+4)^2=100

1 3 + 2 3 + 4 3 + 5 3 = ( 1 + 2 + 3 + 4 + 5 ) 2 = 225 1^3+2^3+4^3+5^3=(1+2+3+4+5)^2=225 . Isn't math wonderful?Thus,from this we can conclude that the sum of the form:

1 3 + 2 3 + 3 3 + . . . . . . . + n 3 1^3+2^3+3^3+.......+n^3 will always be a square and it will be the square of 1 + 2 + 3 + . . . . . . . + n = n ( n + 1 ) 2 1+2+3+.......+n=\dfrac{n(n+1)}{2} .

Lorenc Bushi - 7 years, 4 months ago

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thanks!

Partho Kunda - 7 years, 4 months ago

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