What is the remainder when 2^1344452457 is divided by 11?

I wrote 1344452457= 4K+1
So it becomes 2*2^(4*k)= 2*16^k
Reminder(16^K/11) = 5
And reminder of 2/11= 2
Therefore, 5*2= 10

Where I am wrong?

\(2^5\)= -1 mod 11
\((2^5)^{odd}\)= \((-1)^{odd}\) mod 11

\((2^5)^{odd}\)*\(2^2\)= -1*4 mod 11

1344452457= 5*odd+2
\(2^{1344452457}\)=-4 mod 11=7 mod 11

2^1 % 11 = 2
2^2 % 11 = 4
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2^10 = 1024 % 11 = 1

make power a multiple of 10 so we have 2^(10K + 7 )

Remainder from 2^10K = 1
remainder from 2^7 = 7
Hence answer.

The easiest and quickest way to solve this question is by pattern by using the formula=> dividend = divisor x quotient + remainder
2^1= 11(0) + 2 =>r=2
2^2= 11(0) + 4 =>r=4
2^3= 11(0) + 8 =>r=8
2^4= 11(1) + 5 =>r=5
2^5= 11(2) + 10 =>r=10
2^6= 11(5) + 9 =>r=9
2^7= 11(11) + 7 =>r=7
2^8= 11(23) + 3=>r=3
2^9= 11(46)+6=>r=6
2^10= 11(93)+1=>r=1
2^11= 11(186)+2=>r=2
2^12= 11(372)+4=>r=4
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....
Notice the cyclicity of 10 starts at 2^11.....which means remainder for 2^(any number ending with 7) will be 7. Hence, option D

P.S: This method might seem lengthy. I am aware of the values till 2^10 which made my calculations easier. If you are quick at calculations (calculation with 2 is very quick and simple) and approximation, then this method will be the best choice (at a point I was really pissed off while calculating cyclicity....if this happened in exam, I would have chosen an remainder of 2^7 and moved on -_- )

Quick way is the use of Totient formula for finding out the remainder.

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