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PS: Exponents

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Source: — Problem Solving |

Re: PS: Exponents

by Stuart@KaplanGMAT » Wed Sep 17, 2008 11:30 am
beater wrote:If n is a positive integer, what is the remainder when 3^8n+3 + 2 is divided by 5?
A. 0
B. 1
C. 2
D. 3
E. 4
From the answer choices, we know that there's one and only one answer to the question no matter which value we choose for n. So, let's pick n=1 and see what happens.

If n=1, then we have:

3^(8+3) + 2
3^11 + 2

Let's look at powers of 3, focusing on the units digit (since we're ultimately dividing by 5, knowing the units digit will give us our remainder):

3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243

We can stop right there, since we've hit a units digit of 3 again. We now know that powers of 3 go in a 4 term cycle.

We want to know the units digit of 3^11. The 11th term in our sequence will have a units digit of 7, since we could map our our sequence (1,2,3,4), (5,6,7,8), (9,10,11,12) and see that 11 falls in the same spot as "3" for units digits. (If you're comfortable with modular arithmetic, that's another way you could look at it.)

So, the units digit of 3^11 is 7. Therefore, the units digit of 3^11 + 2 will be 7 + 2 = 9.

When we divide a number ending in "9" by 5, we get a remainder of 4: choose (e).
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by beater » Wed Sep 17, 2008 11:49 am
Stuart - Appreciate the very detailed explanation.

Could you please elaborate on this "since we're ultimately dividing by 5, knowing the units digit will give us our remainder". Thanks!
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by Stuart@KaplanGMAT » Wed Sep 17, 2008 11:53 am
beater wrote:Stuart - Appreciate the very detailed explanation.

Could you please elaborate on this "since we're ultimately dividing by 5, knowing the units digit will give us our remainder". Thanks!
5 is a multiple of all numbers ending in 0 or 5.

So, any number ending in "1" will have a remainder of 1 when you divide by 5. Any number ending in "2" will have a remainder of 2... any number ending in "3" will have a remainder of 3... and so on.

Therefore, if our number were 8.432.679 and we wanted to know the remainder when we divide by 5, we'd just say "our number ends in 9 and 9/5 will always have 4 left over, so the remainder is 4".
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by beater » Wed Sep 17, 2008 11:56 am
Perfect. You rock. Thanks a lot!
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