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If n is a positive integer, how many of the ten digits from

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

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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

Can anyone help me with this question? Thanks!
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Re: If n is a positive integer, how many of the ten digits from [#permalink]

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New post 08 Feb 2014, 11:43
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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.

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

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New post 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).

I'd say (E), too. Waiting for other solutions.

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

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New post 02 Apr 2014, 07:18
Any one with a better solution for this?
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Re: If n is a positive integer, how many of the ten digits from [#permalink]

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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.

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

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For a number to be a cube, any digit from 0 to 9 may be in the units place,

however for a number to be a square, there are only six possibilities in units place which are

0, 1, 4, 5, 6, 9
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Re: If n is a positive integer, how many of the ten digits from [#permalink]

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New post 03 Apr 2014, 11:36
I'm confused. I got D) 9, because I read "n is a positive integer" and immediately wrote down on my paper n>0. So I did not consider 0^3.

I thought zero was not positive or or negative. Can someone confirm??

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

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macey15 wrote:
I'm confused. I got D) 9, because I read "n is a positive integer" and immediately wrote down on my paper n>0. So I did not consider 0^3.

I thought zero was not positive or or negative. Can someone confirm??


Yes, zero is neither positive nor negative.

But 0 still can be the units digit of a perfect cube, consider 10^3=1,000.
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Re: If n is a positive integer, how many of the ten digits from [#permalink]

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New post 03 Apr 2014, 12:37
Bunuel wrote:
macey15 wrote:
I'm confused. I got D) 9, because I read "n is a positive integer" and immediately wrote down on my paper n>0. So I did not consider 0^3.

I thought zero was not positive or or negative. Can someone confirm??


Yes, zero is neither positive nor negative.

But 0 still can be the units digit of a perfect cube, consider 10^3=1,000.


Of course!! For some reason I got stuck on n = 1-9, which looking back, the problem definitely does NOT say. Thanks!

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

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New post 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

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

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New post 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


Hence all ten digits are possible.

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

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New post 15 Jun 2017, 09:16
ok im not the only one who didn't read thoroughly :lol:
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Re: If n is a positive integer, how many of the ten digits from [#permalink]

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New post 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.

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

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New post 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

For instance: 5^2 = 25 --> 5^3 = 5*5 = xx5
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Re: If n is a positive integer, how many of the ten digits from [#permalink]

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Pimenton wrote:
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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New post 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?"

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

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New post 17 Dec 2017, 08:31
sunilrawat wrote:
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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