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PS question- Beat the GMAT

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PS question- Beat the GMAT

by lavinia » Thu Oct 21, 2010 6:43 am
What is the units digit of (18)to the power of 47 ?
A.0
B.2
C.4
D.6
E.8

Because we are asked for units digit of (18) to the power of 47, we should pay attention to (8) to the power of 7, right? But to simplify this, we can rewrite as [(2) to the power of 3] to the power of 7, which means (2) to the power of 21, and because we are looking for the units digit we should look at (2) to the power of 1, which equals 2. The correct answer is B.

Beat the GMAT video explanation is very good, but I feel this one is more easier. Could you please confirm that this approach is correct?

Thanks!
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by shovan85 » Thu Oct 21, 2010 7:13 am
lavinia wrote:What is the units digit of (18)to the power of 47 ?
A.0
B.2
C.4
D.6
E.8

Because we are asked for units digit of (18) to the power of 47, we should pay attention to (8) to the power of 7, right? But to simplify this, we can rewrite as [(2) to the power of 3] to the power of 7, which means (2) to the power of 21, and because we are looking for the units digit we should look at (2) to the power of 1, which equals 2. The correct answer is B.

Beat the GMAT video explanation is very good, but I feel this one is more easier. Could you please confirm that this approach is correct?

Thanks!
How about using cyclicity .

8^1 = 8 --> Unit digit 8
8^2 = 64 --> Unit digit 4
8^3 = 512 --> Unit digit 2
8^4 = 4096 --> Unit digit 6
8^5 = 32768 --> Unit digit 8

So the unit digit sequence repeats as 8,4,2,6,8,4,.....

Each 4 consecutive terms' unit digit repeat, Hence the cyclicity = 4.

Now 18^47
If u divide 47 by 4 we will get a remainder 3

And we know 8^3 unit digit is 2 from the cyclicity table.
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by Rahul@gurome » Thu Oct 21, 2010 8:17 pm
There is one more approach.
(18)^47 = (3^47)*(6^47).
Now 6 raised to a power which is a positive integer always ends in 6.
So 6^47 ends in 6.
Next consider 3^47.
3^1 ends in 3.
3^2 ends in 9.
3^3 ends in 7.
3^4 ends in 1.
47 = 4*11 + 3.
So 3^47 ends in the same last digit in which 3^3 ends which is 7.
Taking into account that 6^47 ends in 6 and 3^47 ends in 7 and since 7*6 = 42 ends in 2, we conclude that (3^47)*(6^47) = 18^47 ends in 2.
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