What is the units digit of 1^3 + 2^3 + 3^3 + ... + 99^3 ?

Question

What is the units digit of  1^3 + 2^3 + 3^3 + ... + 99^3  ?

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

chetan2u · Accepted Answer

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.

HinalAmin24 · Answer

As per formula of cube series

= (n(n+1)/2)^ 2

Ans is 0

CrackverbalGMAT · Answer

We need to have the units place from  1^3  to  9^3 . We do not need to include the numbers ending with 0 as the last digit will be 0.

This value needs to be multiplied 10 times as it will be the same for every 9 numbers.

Last digit of  1^3 = 1 ,  2^3 = 8 ,  3^3 = 7 ,  4^3 = 4 ,  5^3 = 5 ,  6^3 = 6 ,  7^3 = 3 ,  8^3 = 2  and  9^3 = 9

So the sum of the last digits = 1 + 8 + 7 + 4 + 5 + 6 + 3 + 2 + 9 = 45. The last digit is 5.

The last digit for the sum of  1^3 \space to \space 99^3  = 10 * 5 = 50. The last digit is 0.

Option A

Arun Kumar

Yosemite98 · Answer

there is no need for knowing formulas:

we can group the sum terms like   +   +  ........

We know that   for all n  odd  has a factor of (a+b), thus we can extract common factor 100 on the sum above.

the resulting sum will be = 100* (some sum), since the "some sum" factor is an integer (as it is the multiplication and sum of integers), we know that multiplying an integer by 100 we will have 0 in the units digit as we will be "adding two 0's" to that number.

IMO A

hope it helps!

What is the units digit of 1^3 + 2^3 + 3^3 + ... + 99^3 ?

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What is the units digit of 1^3 + 2^3 + 3^3 + ... + 99^3 ?

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