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What is the remainder when 32^33^64^33 is divided by 21?

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What is the remainder when 32^33^64^33 is divided by 21? [#permalink]

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New post 24 Aug 2017, 04:31
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What is the remainder when 32^33^64^33 is divided by 21?

A) 1
B) 2
C) 8
D) 0

I know the cyclicity method but I have a problem in that.

We are asked to find remainder of (=rof) ( 32^33^64^33 ) / 21 = Rof (11^33^64^33) / 21

Now we need to look for cyclicity of 11

11^1 = 11
11^2 = 121
11^3= 1331
11^4= 14641
11^5= 161051
11^6 = 1771561

I find these last few types time consuming, although it was compartively easy to find 11^4, 11^5 , 11^6 since the multiplier was 11 but what if instead of 11 it was 17 or 18 or 19 , It would have taken a lot more time, Is there any work around to find cyclicity of 11 (or any other number) without having calculate complete 11^4, 11^5 , 11^6 etc. ?

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Re: What is the remainder when 32^33^64^33 is divided by 21? [#permalink]

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New post 03 Sep 2017, 08:10
jason201217 wrote:
What is the remainder when 32^33^64^33 is divided by 21?

A) 1
B) 2
C) 8
D) 0

I know the cyclicity method but I have a problem in that.

We are asked to find remainder of (=rof) ( 32^33^64^33 ) / 21 = Rof (11^33^64^33) / 21

Now we need to look for cyclicity of 11

11^1 = 11
11^2 = 121
11^3= 1331
11^4= 14641
11^5= 161051
11^6 = 1771561

I find these last few types time consuming, although it was compartively easy to find 11^4, 11^5 , 11^6 since the multiplier was 11 but what if instead of 11 it was 17 or 18 or 19 , It would have taken a lot more time, Is there any work around to find cyclicity of 11 (or any other number) without having calculate complete 11^4, 11^5 , 11^6 etc. ?


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Kudos [?]: 124217 [0], given: 12076

Re: What is the remainder when 32^33^64^33 is divided by 21?   [#permalink] 03 Sep 2017, 08:10
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What is the remainder when 32^33^64^33 is divided by 21?

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