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Remainders - Please help!

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Remainders - Please help!

by melifox » Sat Apr 24, 2010 9:03 pm
If k is a positive integer, what is the remainder when 13^(4k+2) + 8 is divided by 10?
(A) 7
(B) 4
(C) 2
(D) 1
(E) 0

Don't have an OA. This is way to complicated for me or maybe I'm missing something. The fact that 13 is prime should mean something? Can't figure out how to make 13^(4k+2) a multiple of 10... I'm lost.
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by eaakbari » Sat Apr 24, 2010 9:36 pm
IMO A

In such questions, because the questions implies that the remained will be same for all n , just substitute n = 1.

We get (13^6 + 8)/10

We can find out what 13's unit digit will be finding a pattern.

13^1=13
13^2=169
13^3=___7
13^4=____1
13^5=_____3

Hence ^6 will have a units digit of 9
so ____9+ 8 = _____7
Hence for the number 10, 7 will be remainder

Hence A
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by melifox » Sat Apr 24, 2010 9:52 pm
eaakbari wrote:IMO A

In such questions, because the questions implies that the remained will be same for all n , just substitute n = 1.

We get (13^6 + 8)/10

We can find out what 13's unit digit will be finding a pattern.

13^1=13
13^2=169
13^3=___7
13^4=____1
13^5=_____3

Hence ^6 will have a units digit of 9
so ____9+ 8 = _____7
Hence for the number 10, 7 will be remainder

Hence A
Thank you very much, but it's still not clear for me. Why do you stop at 6? How is the fact that we divide the number by 10 helpful?
Sorry, remainder problems are so hard to understand...
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by this_time_i_will » Sat Apr 24, 2010 10:56 pm
melifox wrote:If k is a positive integer, what is the remainder when 13^(4k+2) + 8 is divided by 10?
(A) 7
(B) 4
(C) 2
(D) 1
(E) 0

Don't have an OA. This is way to complicated for me or maybe I'm missing something. The fact that 13 is prime should mean something? Can't figure out how to make 13^(4k+2) a multiple of 10... I'm lost.
for k=1,2,3,4.....
we get: 13^6+8,13^10+8,13^14+8.........
All these 13^x will have 9 as their unit digit. (since 13 follows a cycle of 4 and here x = 4k+2)

so in all cases we have 9+8 = 17
so unit digit = 7
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by Testluv » Sun Apr 25, 2010 2:22 am
Thank you very much, but it's still not clear for me. Why do you stop at 6? How is the fact that we divide the number by 10 helpful?
Sorry, remainder problems are so hard to understand...
For many remainder problems, you should focus on the units digit.

The units digits of all powers follow cyclic patterns.

This question is testing your ability to use the above two facts.

As eaakbari showed, the cycle for 13 is: 3, 9, 7, 1.

This means that 13^1, 13^5, 13^9, 13 ^13 will all end in 3. And 13^2, 13 ^6, 13 ^10 will all end in 9. And so on.

When k is a positive integer, each "4k + 2" is 4 apart from the next (because we are multplying k by 4).

So, if k is 1, then 4k+2 is 6. If k is 2, then 4k+2 is 10. The 2nd, 6th, 10th (...and so on) powers of 13 will always have the second number in the cycle (ie, 9) as its units digit....So, 13^(4k+2) will actually never be a multiple of 10 (just as we can't divide 9 by 10).

Because 9 + 8 is 17, when we add 8 to 13^(4k+2), the units digit becomes 7. Thus, when we divide the entire expression by 10, there will be 7 left over--that's our remainder.
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