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Re: What is the units digit of 1^3 + 2^3 + 3^3 + ... + 99^3 ? [#permalink]
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Bunuel wrote:
What is the units digit of \(1^3 + 2^3 + 3^3 + ... + 99^3\) ?

A. 0
B. 1
C. 2
D. 3
E. 4

­If you know the formula it is straight forward...
1) Sum of first n positive integers: \(1+2+...+n = \frac{n(n+1)}{2}\)
2) Sum of square of first n positive integers: \(1^2+2^2+...+n^2 =\frac{ n(n+1)(2n+1)}{6}\)
3) Sum of cube of first n positive integers: \(1^3+2^3+...+n^3 = (\frac{n(n+1)}{2})^2\)

So, sum of cube of first 99 positive integers: \(1^3+2^3+...+99^3 = (\frac{99(99+1)}{2})^2 = (\frac{99(100)}{2})^2=99^2*50^2\). Thus answer is 0.

Or, you should know the property of units digit of cube of different digits.
Each digit cubed gives a different units digit, so all digits from 1 to 10 or 11 to 20 .... till 91 to 99 will give set of units digit as 1+2+3+4+5+6+7+8+9+0 or 45. Thus 45*10 as 1 to 99 will make 10 sets. Again we can say 0 as the answer.

Further, even if you do not know the above property, knowing that the sum would be 10 sets (1-9,11-19,21-29,31-39,41-49,...81-89 and 91-99), what ever be the digits of each set, when you multiply it by 10, you will have 0 in the end.

 
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Re: What is the units digit of 1^3 + 2^3 + 3^3 + ... + 99^3 ? [#permalink]
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