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What is the units digit of 9^19 - 7^15? a) 2 b) 4 c) 5 d) 6

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by regor60 » Mon Oct 10, 2016 12:17 pm
This can be solved by identifying the repeating digits for each exponential.
9^1 has 9 as the units digit.
9^2 has 1,and repeats. So the units digit of 9^19 has 9 as its units digit.

Similarly, 7^X has 7,9,3 and 1,repeating. So, 7^15 has 3 as its units digit.

9-3=6
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by Brent@GMATPrepNow » Mon Oct 10, 2016 2:09 pm
alanforde800Maximus wrote:What is the units digit of 9^19 - 7^15?

a) 2
b) 4
c) 5
d) 6
e) 7

Please assist with above problem.
If you're not sure how to proceed with this question, you can eliminate 2 answer choices in a matter of seconds by recognizing two things:
9^19 = some ODD integer, since 9^19 is a product consisted entirely of odd integers
Likewise, 7^15 = some ODD integer, since 7^15 is a product consisted entirely of odd integers

So, 9^19 - 7^15 = (some ODD integer) - (some ODD integer) = EVEN

So, the correct answer is EVEN.
Eliminate C and E, and make a guess.

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by Matt@VeritasPrep » Fri Oct 14, 2016 1:58 am
One nice thing to remember is that the units digit pattern for ANY units digit goes in a block of AT MOST 4. For instance:

1, 1, 1, 1
2, 4, 8, 16, 32, 64, 128, 256 {2, 4, 8, 6, 2, 4, 8, 6, ...}
3, 9, 27, 81, 243, 729, 2187, 6561 {3, 9, 7, 1, 3, 9, 7, 1, ...}
etc.

So you can always consider the remainder when your exponents are divided by 4. In this case, since 19/4 has remainder 3 and 15/4 has remainder 3, that gives us
What's the units digit of 9³ - 7³?
9³ = 729 and 7³ = 343, so we've got 9 - 3, or 6.
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by Matt@VeritasPrep » Fri Oct 14, 2016 1:59 am
Also, to piggyback on Brent's approach:

9 has a clear pattern: 9, 1, 9, 1, ...

7² ends in 9, so that has a clear pattern too. (7²)� = 9� = ends in 9, so 7¹� = (7²)� * 7 = (...9) * 7 = (...3)

That means we've got 9 - 3 again, or 6.
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by alanforde800Maximus » Fri Oct 14, 2016 5:49 am
Thanks Guys for amazing solutions.
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by Scott@TargetTestPrep » Sun Oct 16, 2016 5:22 pm
alanforde800Maximus wrote:What is the units digit of 9^19 - 7^15?

a) 2
b) 4
c) 5
d) 6
e) 7

Please assist with above problem.
We need to separately determine the units digit of 9^19 and 7^15.

Since all odd powers of 9 have a units digit of 9, and all even powers of 9 have a units digit of 1, the units digit of 9^19 is 9.

To determine the units digit of 7^15, we look at the pattern of the units digits of 7^n for positive integer values of n. That is, let's look at the pattern of the units digits of powers of 7.

7^1 = 7

7^2 = 9

7^3 = 3

7^4 = 1

7^5 = 7

The pattern of any base of 7 repeats every 4 exponents. The pattern is 7-9-3-1. In this pattern, all positive exponents that are multiples of 4 will produce a 1 as its units digit. The closest multiple of 4 to 15 is 16. Thus, 7^16 will have a 1 as its units digit and, thus, 7^15 will have a 3 as its units digit.

Thus, the units digit of 9^19 - 7^15 = 9 - 3 = 6

Answer: D

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