You could see whether for each digit there is a number that when cubed has that digit in the ones place in the cube.

The key thing for doing this quickly and efficiently is realizing that only the units digit matters. So, for instance, when considering 9³, we don't need to figure out the actual value, only that 81 x 9 will have a units digit of 9, because 1 x 9 = 9.

0 -- 10³ works.
1 -- 1³ works.
2 -- No odd number works. 2³ = 8, no. 4³ = 64, no. All powers of 6 end in 6, no. 8 x 8 = 64 8 x 4 = 32 8 works.
3 -- No even number works. 3³ = 27, no. All powers of 5 end in 5, no. 7 x 7 = 49 7 x 9 = 63 7 works.
4 -- From above, 4³ = 64.
5 -- All powers of 5 work.
6 -- All powers of 6 work.
8 -- From above, 2³ = 8.
9 -- No even number works. From above 3, 5, and 7 do not work. 9 x 9 = 81 9 x 1 = 9 9 works.

Every digit could be the units digit of n³.

The correct answer is E.

You could also just cube all the digits from 0 to 9 and see what happens.

0^3: ends in 0
1^3: ends in 1
2^3: ends in 8
3^3: ends in 7
4^3: ends in 4
5^3: ends in 5
6^3: ends in 6
7^3: ends in 3
8^3: ends in 2
9^3: ends in 9

All 10 produced unique results! The answer is E

Useful FYI: you can get all any units digit from 0 to 9 at the end of a cube, but remember that you CANNOT get any units digit at the end of a square (and this is a property the GMAT is more likely to test!) For instance, you can say that 1447 isn't a square simply because it ends in 7, which no square can do. (No number from 0 to 9, multiplied by itself, gives a result ending in 7.)

Matt@VeritasPrep wrote:Useful FYI: you can get all any units digit from 0 to 9 at the end of a cube, but remember that you CANNOT get any units digit at the end of a square (and this is a property the GMAT is more likely to test!) For instance, you can say that 1447 isn't a square simply because it ends in 7, which no square can do. (No number from 0 to 9, multiplied by itself, gives a result ending in 7.)

To Add on- No square ends with unit digit 2, 3, 7, 8.

Jay@ManhattanReview wrote:Nandish,

What could be the answer to above question had the question is modified to:

If n is a positive unit digit 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

-Jay

_________________
Manhattan Review GMAT Prep

Locations: New York | Tokyo | Manchester | Geneva | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.

The Ans still remains E

Because 0^3: ends in 0
1^3: ends in 1
2^3: ends in 8
3^3: ends in 7
4^3: ends in 4
5^3: ends in 5
6^3: ends in 6
7^3: ends in 3
8^3: ends in 2
9^3: ends in 9

Let me know if i'm wrong!!!

NandishSS wrote:

Jay@ManhattanReview wrote:Nandish,

What could be the answer to above question had the question is modified to:

If n is a positive unit digit 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

-Jay

_________________
Manhattan Review GMAT Prep

Locations: New York | Tokyo | Manchester | Geneva | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.

The Ans still remains E

Because 0^3: ends in 0
1^3: ends in 1
2^3: ends in 8
3^3: ends in 7
4^3: ends in 4
5^3: ends in 5
6^3: ends in 6
7^3: ends in 3
8^3: ends in 2
9^3: ends in 9

Let me know if i'm wrong!!!

Re-read Jay's question stem. If n is a positive unit digit integer, how many of the ten digits from 0 through 9 could be the units digits of n^3 ? Details are everything...

DavidG@VeritasPrep wrote:

NandishSS wrote:

Jay@ManhattanReview wrote:Nandish,

What could be the answer to above question had the question is modified to:

If n is a positive unit digit 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

-Jay

_________________
Manhattan Review GMAT Prep

Locations: New York | Tokyo | Manchester | Geneva | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.

The Ans still remains E

Because 0^3: ends in 0
1^3: ends in 1
2^3: ends in 8
3^3: ends in 7
4^3: ends in 4
5^3: ends in 5
6^3: ends in 6
7^3: ends in 3
8^3: ends in 2
9^3: ends in 9

Let me know if i'm wrong!!!

Re-read Jay's question stem. If n is a positive unit digit integer, how many of the ten digits from 0 through 9 could be the units digits of n^3 ? Details are everything...

@David And Jay

"The devil is in the detail"
0 is neither + nor -
So the ans would be 9
Correct me if I'm wrong!!!

NandishSS 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