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Cyclicity and remainders

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Cyclicity and remainders [#permalink] New post 18 Sep 2013, 12:38
Hey guys, and calling for the legend Bunuel!

So I am learning the cyclicity and finding a units digit trick. Very neat. Only thing I am a bit confused on is if the remainder is zero.

For instance, 17^12 is 7^12. 7 has a cyclicity of 4. 12 divided by 4 yields no remainder. So does that mean the unit digit is just the 4th in the cycle, which in this case is 1?

Thanks!
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Re: Cyclicity and remainders [#permalink] New post 18 Sep 2013, 16:04
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TheLostOne wrote:
Hey guys, and calling for the legend Bunuel!

So I am learning the cyclicity and finding a units digit trick. Very neat. Only thing I am a bit confused on is if the remainder is zero.

For instance, 17^12 is 7^12. 7 has a cyclicity of 4. 12 divided by 4 yields no remainder. So does that mean the unit digit is just the 4th in the cycle, which in this case is 1?

Thanks!

Dear TheLostOne,
I'm happy to help with this. :-)

First of all, here's a blog you may find informative.
http://magoosh.com/gmat/2013/gmat-quant ... questions/

Let's think about this. The powers of 7 indeed have a cycle of 4 --- this means
7^1 has a units digit of 7
7^2 has a units digit of 9
7^3 has a units digit of 3
7^4 has a units digit of 1
7^5 has a units digit of 7
7^6 has a units digit of 9
7^7 has a units digit of 3
7^8 has a units digit of 1
That's a cycle of 4. Notice, at every exponent that's a multiple of four, the unit digit is 1.

When you divide by 4 and get no remainder, you are at a multiple of four. Therefore, the units digit it 1. Therefore, 7^12 (or 17^12 or 1037^12) would have to have a units digit of 1. You are perfectly correct. :-)

If you think about it, when the power is a multiple of the cycle, the units digit would have to be 1 for any base, because having a units digit of 1 allows the next power to have the same units digit as the base.

Does all this make sense?
Mike :-)
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Mike McGarry
Magoosh Test Prep

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Re: Cyclicity and remainders [#permalink] New post 19 Sep 2013, 05:50
mikemcgarry wrote:
TheLostOne wrote:
Hey guys, and calling for the legend Bunuel!

So I am learning the cyclicity and finding a units digit trick. Very neat. Only thing I am a bit confused on is if the remainder is zero.

For instance, 17^12 is 7^12. 7 has a cyclicity of 4. 12 divided by 4 yields no remainder. So does that mean the unit digit is just the 4th in the cycle, which in this case is 1?

Thanks!

Dear TheLostOne,
I'm happy to help with this. :-)

First of all, here's a blog you may find informative.
http://magoosh.com/gmat/2013/gmat-quant ... questions/

Let's think about this. The powers of 7 indeed have a cycle of 4 --- this means
7^1 has a units digit of 7
7^2 has a units digit of 9
7^3 has a units digit of 3
7^4 has a units digit of 1
7^5 has a units digit of 7
7^6 has a units digit of 9
7^7 has a units digit of 3
7^8 has a units digit of 1
That's a cycle of 4. Notice, at every exponent that's a multiple of four, the unit digit is 1.

When you divide by 4 and get no remainder, you are at a multiple of four. Therefore, the units digit it 1. Therefore, 7^12 (or 17^12 or 1037^12) would have to have a units digit of 1. You are perfectly correct. :-)

If you think about it, when the power is a multiple of the cycle, the units digit would have to be 1 for any base, because having a units digit of 1 allows the next power to have the same units digit as the base.

Does all this make sense?
Mike :-)


Indeed it does!

Thanks.
Re: Cyclicity and remainders   [#permalink] 19 Sep 2013, 05:50
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