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Re: If n is a positive integer, how many of the ten digits from [#permalink]
08 Feb 2014, 11:43

Im also not completely sure about this one. But if you simply plug in values for ''n'' from 0 tot 9. This will show you that units digit can be 0,1,2,3,4,5,6,7,8,9 = 10. 0^3=0 , 1^3=1 , 2^3=8 , 3^3=27, 4^3=64, 5^3=125, 6^3=216, 7^3=343, 8^3=512, 9^3=729

These possibilities keep repeating themselves. _________________

Structural persistence is the key to succes . Party hard, study harder.

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Last edited by CarloCjm on 09 Feb 2014, 02:15, edited 1 time in total.

Re: If n is a positive integer, how many of the ten digits from [#permalink]
08 Feb 2014, 12:06

I did it exactly like carlocjm explained, in around 2 minutes. Keep in mind that you don't have to go through the whole multiplication process, just get the units digits and stop, since that's what we're concerned about. Also, for some numbers, you actually don't have to do manual stuff (5^3 is by heart 125, for example).

Re: If n is a positive integer, how many of the ten digits from [#permalink]
02 Apr 2014, 07:31

1

This post received KUDOS

Expert's post

Qoofi wrote:

Any one with a better solution for this?

If n is a positive integer, how many of the ten digits from 0 through 9 could be the units digits of n^3 ?

A. Three B. Four C. Six D. Nine E. Ten

The question is fairly straightforward, it basically asks: which digits can be the units digit of a perfect cube.

Can it be 0? Yes, 10^3=1,000. Can it be 1? Yes, 1^3=1. Can it be 2? Yes, 8^3=512. Can it be 3? Yes, 7^3=343. Can it be 4? Yes, 4^3=64. Can it be 5? Yes, 5^3=125. Can it be 6? Yes, 6^3=...6. Can it be 7? Yes, 3^3=27. Can it be 8? Yes, 2^3=8. Can it be 9? Yes, 9^3=...9.