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

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Director
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What is the remainder when 2^1344452457 is divided by 11?  [#permalink]

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02 Nov 2005, 23:31
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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
SVP
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02 Nov 2005, 23:39
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

A)
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02 Nov 2005, 23:45
Laxie> Khong hieu
Director
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02 Nov 2005, 23:45
laxieqv wrote:
ah, sorry, it's 7

apologies not accepted without explanation
Btw, where are my 'chocolates'?
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02 Nov 2005, 23:46
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

HEY YOU STOLE MY OLD AVATAR!!!! I WANT IT BACK!!!!!

CALL THE POLICE!!!

_________________

"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

http://www.nytimes.com/2005/11/21/international/asia/21prexy.html

Director
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02 Nov 2005, 23:51
Titleist wrote:

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!"
Director
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02 Nov 2005, 23:53
gsr wrote:
Titleist wrote:

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!"

LOL! You got all the bases covered my friend!
_________________

"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

http://www.nytimes.com/2005/11/21/international/asia/21prexy.html

SVP
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02 Nov 2005, 23:53
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

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.
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03 Nov 2005, 00:08
laxieqv wrote:
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

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.
VP
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03 Nov 2005, 00:40
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

so 2^1344452457 mod 11 = 2^7 mod 11 = 7
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03 Nov 2005, 01:21
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

so 2^1344452457 mod 11 = 2^7 mod 11 = 7

how is it ? 5,15,99995 are all mult. of 5 but 1344452457 is not mult. of 7 ? how do you deduce the pattern ?
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03 Nov 2005, 01:47
christoph wrote:
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

so 2^1344452457 mod 11 = 2^7 mod 11 = 7

how is it ? 5,15,99995 are all mult. of 5 but 1344452457 is not mult. of 7 ? how do you deduce the pattern ?

well what i meant was to show how remainders, when 2^n is divided by 11 repeates after each 10 counts(for n).

for example, 2^1 / 11 gives remainder 2
we will again get remainder 2 if we divide 2^(10+1) by 11 similarly 2^21, 2^31

so in oder to find remainder when 2^n is divided by 11, just look at last digit and find remainder 2^lastdigit / 11

its coincidence that in this case 2^7 / 11 gives remainder 7
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03 Nov 2005, 02:52
duttsit wrote:
christoph wrote:
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

so 2^1344452457 mod 11 = 2^7 mod 11 = 7

how is it ? 5,15,99995 are all mult. of 5 but 1344452457 is not mult. of 7 ? how do you deduce the pattern ?

well what i meant was to show how remainders, when 2^n is divided by 11 repeates after each 10 counts(for n).

for example, 2^1 / 11 gives remainder 2
we will again get remainder 2 if we divide 2^(10+1) by 11 similarly 2^21, 2^31

so in oder to find remainder when 2^n is divided by 11, just look at last digit and find remainder 2^lastdigit / 11

its coincidence that in this case 2^7 / 11 gives remainder 7

_________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

Intern
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03 Nov 2005, 05:36
duttsit wrote:
christoph wrote:
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

so 2^1344452457 mod 11 = 2^7 mod 11 = 7

how is it ? 5,15,99995 are all mult. of 5 but 1344452457 is not mult. of 7 ? how do you deduce the pattern ?

well what i meant was to show how remainders, when 2^n is divided by 11 repeates after each 10 counts(for n).

for example, 2^1 / 11 gives remainder 2
we will again get remainder 2 if we divide 2^(10+1) by 11 similarly 2^21, 2^31

so in oder to find remainder when 2^n is divided by 11, just look at last digit and find remainder 2^lastdigit / 11

its coincidence that in this case 2^7 / 11 gives remainder 7

Is there anything special about mod 11?
if we do the same with mod 5, the remainter is different:

2^7 mod 5 = 3
2^17 mod 5 = 2
2^23457 mod 5 = 2
Director
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03 Nov 2005, 13:46
As always, duttsit and laxie are right!!
It is 7
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03 Nov 2005, 19:38
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

so 2^1344452457 mod 11 = 2^7 mod 11 = 7
Director
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03 Nov 2005, 19:41
Did you guys all do this in under 1 min? Coz this is a 400+ question
Current Student
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03 Nov 2005, 19:50
I dont think you will see a question like this on the real exam?

gsr wrote:
Did you guys all do this in under 1 min? Coz this is a 400+ question
Director
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03 Nov 2005, 19:51
fresinha12 wrote:
I dont think you will see a question like this on the real exam?

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

Yeah...because most of us perform above 400+ level!
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03 Nov 2005, 20:06
gsr wrote:
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
03 Nov 2005, 20:06

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