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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. ?
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. ?
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
03 Sep 2017, 08:10
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jason201217
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. ?
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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Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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