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

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08 Feb 2014, 12:43

2

This post received KUDOS

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

Still bashing, will continue to do so , although it's important to chill aswell ; ) STUDY+CHILL=VICTORY

Last edited by CarloCjm on 09 Feb 2014, 03:15, edited 1 time in total.

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

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08 Feb 2014, 13:06

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).

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.

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

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13 Jun 2015, 14:13

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

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02 Oct 2015, 14:44

CarloCjm wrote:

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.

You got to the right conclusion, but be careful: the question states n has to be a POSITIVE integer, therefore 0 is excluded. The 0 units digit comes from 10^3

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

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03 Oct 2016, 21:07

Hello from the GMAT Club BumpBot!

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

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24 Nov 2016, 01:50

Outstanding Question. Here we need to get the units digit of n^3 Consider the unit digit of n being 0,1,2... 0=> 0 1=> 1 2=>8 3=>7 4=>4 5=>5 6=>6 7=>3 8=>2 9=>9

Concentration: Entrepreneurship, General Management

WE: Engineering (Energy and Utilities)

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

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15 Jun 2017, 10:20

I am expecting a quicker way than below.. 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.

If n is a positive integer, how many of the ten digits from [#permalink]

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15 Jun 2017, 10:36

hotcool030 wrote:

I am expecting a quicker way than below.. 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.

what do you mean by "I am expecting a quicker way"

Thats a very fast way

you don't have to compute x^3 if that is your questions