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Great help....Thanx

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khanna_sumit Saysactually 2^20 will do instead of 2^44

when a power of 2 is divided by 17, then 2^4 gives a remainder -1 and hence 2^8 gives a remainder +1....and that will be true for any power of 2^8, or any number of the form 2^8k will give remainder 1 when divided by 17...

also when divided by 9, 2^6a gives a remainder 1......or when a number is divided by 9*17, 2^{LCM(8,6)*b} will give a remainder 1......means 2^24b will give a remainder 1 when divided by 153........3500 = 24k + 20......

so 2^3500 = 2^(24k + 20) will give same remainder as is given by 2^20.....

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noops you will take 10 min to solve in a cat paper going by your approach...(no offence meant)

MY approach.

2^3500 mod 153

= 2^44 mod 153

= 59^4 mod 153

=67

I follow this method because i believe in some approaches to be followed in certain problems. People may not agree.

-Maverick

actually 2^20 will do instead of 2^44

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noops you will take 10 min to solve in a cat paper going by your approach...(no offence meant)

MY approach.

2^3500 mod 153

= 2^44 mod 153

= 59^4 mod 153

=67

I follow this method because i believe in some approaches to be followed in certain problems. People may not agree.

-Maverick

well can u elaborate more on ur approach... how u reached to 2^44 from 2^3500.

then probably it wud no longer take 10 min!!

well, all the questions on remainders that i ve solved in mock tests, didnt take this long... they were quite easy n direct.

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well just a little more patience n u cud have reached the end. ( same goes for ur name - suicidal just read the few lines in my signature... )

lemme try n explain sumit's solution in the manner i did it.

128^500 = 2^3500

= 16^875

now 16 = -1(mod 17) ----------> { i.e. remainder when 16 is divided by 17.}

=> 16^2n = 1(mod 17) and 16^(2n+1) = -1(mod 17)

i.e. for odd powers, remainder is -1

therefore 2^3500 = 17x -1

similarly, 2^3500 = 4 * 8^1166

=> 8^2n = 1(mod 9) and 8^(2n+1) = -1(mod 9)

i.e. for even powers, remainder is 4

therefore 2^3500 = 9y + 4

equating, 17x-1=9y+4

=> x = (9y+5)/17

for x to be integral, y must be of form (17z+7)

substituting in 2^3500 = 9y+4

=> 2^3500 = 9(17z+7) + 4

= 153z + 67

hence remainder of (2^3500)/153 will be 67. :)

noops you will take 10 min to solve in a cat paper going by your approach...(no offence meant)

MY approach.

2^3500 mod 153

= 2^44 mod 153

= 59^4 mod 153

=67

I follow this method because i believe in some approaches to be followed in certain problems. People may not agree.

-Maverick

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Hi Noops, Thanks for taking the trouble to help me out......but, you were right - i spent a little more time with this problem, and saw that life is pretty simple after all!

tke cre

tke cre

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Well, join the club madguy!

Me being an Arts student, Sumit's soln. was a little too smart for me too, to be honest.

However, I did make an attempt to decipher it and this is what i came up with:

128^500 = 2^3500

= 2^(4*875) = 16^875

= (17 - 1)^85.

Therefore 128^500 = 17a - 1 (- since 85 is odd)

Simmilarly, 128^500 = 2^3500

=2^(3*1166 + 2) = (8^1166)*(2^2)

= ((9 - 1)^1166) * 4.

Therefore, 128^500 = 9b + 4 (+ since 1166 is even)

sumit's steps after this is understandable.

maverick, pls post ure soln for tyro's like me !!!

well just a little more patience n u cud have reached the end. ( same goes for ur name - suicidal just read the few lines in my signature... )

lemme try n explain sumit's solution in the manner i did it.

128^500 = 2^3500

= 16^875

now 16 = -1(mod 17) ----------> { i.e. remainder when 16 is divided by 17.}

=> 16^2n = 1(mod 17) and 16^(2n+1) = -1(mod 17)

i.e. for odd powers, remainder is -1

therefore 2^3500 = 17x -1

similarly, 2^3500 = 4 * 8^1166

=> 8^2n = 1(mod 9) and 8^(2n+1) = -1(mod 9)

i.e. for even powers, remainder is 4

therefore 2^3500 = 9y + 4

equating, 17x-1=9y+4

=> x = (9y+5)/17

for x to be integral, y must be of form (17z+7)

substituting in 2^3500 = 9y+4

=> 2^3500 = 9(17z+7) + 4

= 153z + 67

hence remainder of (2^3500)/153 will be 67. 😃

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Well, join the club madguy!

Me being an Arts student, Sumit's soln. was a little too smart for me too, to be honest.

However, I did make an attempt to decipher it and this is what i came up with:

128^500 = 2^3500

= 2^(4*875) = 16^875

= (17 - 1)^85.

Therefore 128^500 = 17a - 1 (- since 85 is odd)

Simmilarly, 128^500 = 2^3500

=2^(3*1166 + 2) = (8^1166)*(2^2)

= ((9 - 1)^1166) * 4.

Therefore, 128^500 = 9b + 4 (+ since 1166 is even)

sumit's steps after this is understandable.

maverick, pls post ure soln for tyro's like me !!!

Me being an Arts student, Sumit's soln. was a little too smart for me too, to be honest.

However, I did make an attempt to decipher it and this is what i came up with:

128^500 = 2^3500

= 2^(4*875) = 16^875

= (17 - 1)^85.

Therefore 128^500 = 17a - 1 (- since 85 is odd)

Simmilarly, 128^500 = 2^3500

=2^(3*1166 + 2) = (8^1166)*(2^2)

= ((9 - 1)^1166) * 4.

Therefore, 128^500 = 9b + 4 (+ since 1166 is even)

sumit's steps after this is understandable.

maverick, pls post ure soln for tyro's like me !!!

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thanx sumit, for the detailed explanation u provided.

actually, i was not able to make it out.

thanx to maverick too.

answer is 67......

(12^500 = 2^3500,153=17*9

when any number a is coprime with 153....a^24 gives a remainder 1 when divided by 153(eular's extension)....it can be proved like this also that 2^4 divided by 17 gives remainder -1 therefore 2^8 gives a remainder 1 when divided by 17.....3500=24x+20...therefor 2^3500 will give the same remainder from 17 as is given by 2^20 i.e. -1........2^6 when divided by 9 gives remainer 1..

3500=2^6*y+2.....so remiander given by 2^3500 when divided by 9 is 2^2 = 4..

2^3500 can be therefore be written in the form 17a-1 and 9b+4....

17a-1=9b+4

or b=(17a-5)/9

b will be integral when a is of the form 9c-5(remainder thm)

our number is 17a-1

putting value of a in it

17(9c-5)-1=153c-86...so when this number is divided by 153, remainder is -86 or 67...

rgds

Sumit

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hi ppl,

can anyone pls solve

the question?

wats the remaionder

128^500/153

answer is 67......

(12^500 = 2^3500,153=17*9

when any number a is coprime with 153....a^24 gives a remainder 1 when divided by 153(eular's extension)....it can be proved like this also that 2^4 divided by 17 gives remainder -1 therefore 2^8 gives a remainder 1 when divided by 17.....3500=24x+20...therefor 2^3500 will give the same remainder from 17 as is given by 2^20 i.e. -1........2^6 when divided by 9 gives remainer 1..

3500=2^6*y+2.....so remiander given by 2^3500 when divided by 9 is 2^2 = 4..

2^3500 can be therefore be written in the form 17a-1 and 9b+4....

17a-1=9b+4

or b=(17a-5)/9

b will be integral when a is of the form 9c-5(remainder thm)

our number is 17a-1

putting value of a in it

17(9c-5)-1=153c-86...so when this number is divided by 153, remainder is -86 or 67...

rgds

Sumit

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