As per formula of cube series

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

Ans is 0

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
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