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

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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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

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.

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

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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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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why are we taking 0 here? 0 does not come under a positive integer right?

Your doubt is already addressed here.
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Your doubt is already addressed


I am sorry, I didn't quite understand why we are taking n as 0 when it can only be a positive integer. Pls explain

n cannot be 0 but the units digit of n can be. Please re-read the question and the solutions provided above more carefully.
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Hi bunuel,
this question appeared on my gmat prep 3 yesterday, (i scored Q49, V33) i got it wrong, because i was not able to figure out what does it mean by tens digit. I thought tens digit of intger n.