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Senior Manager  Joined: 23 Jul 2013
Posts: 302
Cyclicity and remainders  [#permalink]

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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!
Magoosh GMAT Instructor G
Joined: 28 Dec 2011
Posts: 4473
Re: Cyclicity and remainders  [#permalink]

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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 _________________
Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
Senior Manager  Joined: 23 Jul 2013
Posts: 302
Re: Cyclicity and remainders  [#permalink]

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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.
Non-Human User Joined: 09 Sep 2013
Posts: 13136
Re: Cyclicity and remainders  [#permalink]

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_________________ Re: Cyclicity and remainders   [#permalink] 02 Mar 2019, 02:49
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