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If n is a positive integer, how many of

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by NandishSS » Wed Nov 30, 2016 9:32 pm
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

OA:E

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by MartyMurray » Wed Nov 30, 2016 9:56 pm
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.
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by DavidG@VeritasPrep » Thu Dec 01, 2016 6:45 am
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
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by Matt@VeritasPrep » Thu Dec 08, 2016 8:48 pm
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.)
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by NandishSS » Wed Dec 14, 2016 7:14 am
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.
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by Jay@ManhattanReview » Wed Dec 14, 2016 11:07 pm
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

OA:E

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Hi Nandish,

Adding some value to the already solved question. You must know the power cycles of unit digits for the first ten digits.

Power cycles of unit digits of the first ten digits:
  • 0^m: 0 for all the values of positive integer m
    1^m: 1 for all the values of positive integer m
    2^m: (2, 4, 8, 6), (2, 4, 8, 6), ... for m = 1, 2, 3, 4 => cycle of 4
    3^m: (3, 9, 7, 1), ... for m = 1, 2, 3, 4 => cycle of 4
    4^m: (4, 6), (4, 6), ... for m = 1, 2 => cycle of 2
    5^m: 5 for all the values of m
    6^m: 6 for all the values of m
    7^m: (7, 9, 3, 1), ... for m = 1, 2, 3, 4 => cycle of 4
    8^m: (8, 4, 2, 6), ... for m = 1, 2, 3, 4 => cycle of 4
    9^m: (9, 1), (9, 1), ... for m = 1, 2, 3, 4 => cycle of 4
The blue colored digits appear on the third iteration of multiplications (n^3), we see that all the ten digits have chances of their occurrence.

OA: E

-Jay

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by Jay@ManhattanReview » Wed Dec 14, 2016 11:10 pm
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

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by NandishSS » Fri Dec 16, 2016 4:59 am
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

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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!!!
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by DavidG@VeritasPrep » Fri Dec 16, 2016 9:30 am
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

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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...
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by NandishSS » Sat Dec 17, 2016 7:04 am
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

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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!!!
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by [email protected] » Sat Dec 17, 2016 10:23 am
Hi NandishSS,

Yes, IF the question is rewritten so that N is a positive single-digit integer, then your answer would be correct.

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Last edited by [email protected] on Sun Dec 18, 2016 9:27 pm, edited 1 time in total.
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by Jay@ManhattanReview » Sun Dec 18, 2016 8:22 pm
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

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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...
Thanks David. Yes, agree with you, "Details are everything...".

-Jay

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by Jeff@TargetTestPrep » Wed Dec 21, 2016 9:12 am
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

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