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What is the remainder when 2^1344452457 is divided by 11? A) 4 B) 6 C) 8 D) None of the above (then what else? ) E) 12

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HEY YOU STOLE MY OLD AVATAR!!!! I WANT IT BACK!!!!! CALL THE POLICE!!!

Everything is being done legally as this avatar was handed over to me 'officially' on 10/25/05. Quote Titleist "Anyhow, you can find my old avatar in the Upload bank. You can copy his image for prosterity!"

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"Wow! Brazil is big." â€”George W. Bush, after being shown a map of Brazil by Brazilian president Luiz Inacio Lula da Silva, Brasilia, Brazil, Nov. 6, 2005

What is the remainder when 2^1344452457 is divided by 11? A) 4 B) 6 C) 8 D) None of the above (then what else? ) E) 12

E is crossed right away

1344452457 divided by 5 has remainder of 2 ( for sure!)
---->we can write 2^1344452457= 2^(5x)* 2^2 ( x is an integer)
2^1344452457= (2^5)^x * 2^2 +2^2 - 2^2 = 2^2 ( 32^x +1^x) - 2^2
= 4 * (32+1)* A -2^2 ( A is the exponential expression gained by expressing 33^x+1^x and for sure A is an integer!)
= 4*33*A - 4
We have 4*33*A is divided by 11 so 4*33*A -4 divided by 11 has remainder of 7.

What is the remainder when 2^1344452457 is divided by 11? A) 4 B) 6 C) 8 D) None of the above (then what else? ) E) 12

E is crossed right away

1344452457 divided by 5 has remainder of 2 ( for sure!) ---->we can write 2^1344452457= 2^(5x)* 2^2 ( x is an integer) 2^1344452457= (2^5)^x * 2^2 +2^2 - 2^2 = 2^2 ( 32^x +1^x) - 2^2 = 4 * (32+1)* A -2^2 ( A is the exponential expression gained by expressing 33^x+1^x and for sure A is an integer!) = 4*33*A - 4 We have 4*33*A is divided by 11 so 4*33*A -4 divided by 11 has remainder of 7.

Let me elaborate some ,we are sure to have formula for a^x+b^x only when x is odd. Since the result of 1344452457 dividing by 5 must be an odd number ---> x is odd ----> the formula works here.

why is this the long way...i do it this way...its the simpilest and fastest way to do...basically with 2 you have to recognize that the remainder repeats every 5 counts...

duttsit wrote:

Good job lexi. 7 seems correct.

if we go the hard way of checking 2^n mod 11, we find remainders repeats after 10 counts.

so, 2^5 mod 11 is same as 2^15 mod 11 or 2^99995 mod 11

Did you guys all do this in under 1 min? Coz this is a 400+ question

this question is solved in under 1(0) second since the question asks "what is the reminder" and the reminder must be an odd integer because any power of 2 is even and any even integer divided by odd integer gives odd integer as reminder and there is no odd integer in the ACs. So it is D.

if there were only one odd integer given as answer in D, then it must also be D as OA.

You all guys went in wrong way because solving this question and finding the reminder when 2^x is divided by 11 is something different.

gsr wrote:

What is the remainder when 2^1344452457 is divided by 11? A) 4 B) 6 C) 8 D) None of the above (then what else? ) E) 12

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