Mo2men wrote:

Method 3:

(2^1) / 5 will have remainder 2

(2^2) / 5 will have remainder 4

(2^3) / 5 will have remainder 3

(2^4) / 5 will have remainder 1

(2^5) / 5 will have remainder 2

I welcome your comments to help me figure where my problem is

Thanks

(2^1) / 5 will have remainder 2

(2^2) / 5 will have remainder 4

(2^3) / 5 will have remainder 3

(2^4) / 5 will have remainder 1

(2^5) / 5 will have remainder 2

(2^6) / 5 will have remainder 4

(2^7) / 5 will have remainder 3

(2^8) / 5 will have remainder 1

So, Cyclicity will be 4

\(2^{586}\) =

\(2^{4*146}\) *

\(2^2\)The blue part will have remainder 1 and

the red part will always have the remainder 4So IMHO result will be 4Perfect as you stated.....Mo2men wrote:

Method 2:

\(2^{586} = (2^2)^{293} =(5-1)^{195} * 2\)

\(2^{586}\) = \(2^{2*293}\) =>4^{293}

\(4^{293}\) = \((5 - 1)^{293}\)

Now 5^{293} divided by 5 will have remainder as 5

-1^{293} divided by 5 will have remainder as -1 { Since 293 is an odd power -1^Odd power = -1 }

Now you have 5 - 1 = 4
_________________

Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only )