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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, 11:43

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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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]

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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).

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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02 Oct 2015, 13:44

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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24 Nov 2016, 00:50

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

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

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15 Jun 2017, 09:20

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, 09:36

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

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

To solve, we need to raise each digit of 0 through 9 to the third power to determine how many unique units digits we can produce.

0^3 = 0

1^3 = 1

2^3 = 8

3^3 = 27 (units digit of 7)

4^3 = 64 (units digit of 4)

5^3 = 125 (units digit of 5)

Since after the base of 5 the number starts getting fairly large, we can rely on our knowledge of units digit patterns of a number raised to a power to determine the units digits of the remaining numbers.

6^3 = units digit of 6

We should recall that 6 raised to any whole number exponent will always have a units digit of 6.

7^3 = units digit of 7

We should recall that the repeating pattern for the units digits when the base of 7 is raised to an exponent is 3-9-7-1.

8^3 = units digit of 2

We should recall that the repeating pattern for the units digits when the base of 8 is raised to an exponent is 8-4-2-6.

9^3 = units digit of 9

We should recall that the pattern for the units digits when the base of 9 is raised to an exponent is 9-1.

Thus, there are 10 possible units digits for n^3 for the integers 0 through 9.

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

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17 Dec 2017, 05:52

Can someone please explain how "how many of the ten digits from 0 through 9 could be the units digits of n^3 ?" translates to "which digits can be the units digit of a perfect cube?"

Can someone please explain how "how many of the ten digits from 0 through 9 could be the units digits of n^3 ?" translates to "which digits can be the units digit of a perfect cube?"

n^3 is a prefect cube, a cube of an integer. The question asks how many values can the units digit of n^3 take.
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