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11^704 divided by 17
11^2 = 121 which when divided by 17 gives 2 as remainder
2^(704/2) = 2^352. Now 2^4 = 16 and 16/17 = -1
so we can write as
(-1)^(352/4) = (-1)^88 = 1
Therefore, the remainder is 1
Answer A
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sivakumarm786
11^704 divided by 17
11^2 = 121 which when divided by 17 gives 2 as remainder
2^(704/2) = 2^352. Now 2^4 = 16 and 16/17 = -1
so we can write as
(-1)^(352/4) = (-1)^88 = 1
Therefore, the remainder is 1
Answer A
Could you please explain the last part where 16/17=-1 ???
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suddu
sivakumarm786
11^704 divided by 17
11^2 = 121 which when divided by 17 gives 2 as remainder
2^(704/2) = 2^352. Now 2^4 = 16 and 16/17 = -1
so we can write as
(-1)^(352/4) = (-1)^88 = 1
Therefore, the remainder is 1
Answer A
Could you please explain the last part where 16/17=-1 ???

Hi suddu

this is negative remainder. In the ibid case, 16/17 gives 16 as remainder or it can be seen as -1 (-1 + 17 = 16) as well.
For details please refer this link from Bunuel
https://gmatclub.com/forum/all-about-ne ... l#p1472077

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Siva kumar
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There is a theorem I read on this platform somewhere: Fermat's little theorem, which states that any number when raised to the power of (p-1), where p is a prime number, will leave remainder of 1 when divided by p.

Here, observe that 17 is a prime number, hence 11^(17-1) will leave a remainder of 1 when divided by 17.
Therefore, 11^16 can be written as 17x+1

Now, 11^704 = 11^(16*44) = (17x+1)^44

Expanding by binomial theorem,
(17x+1)^44 = 44C0*[(17x)^44]*[1^0] + 44C1*[(17x)^43]*[1^1] +........+ 44C43*[(17x)^1]*[1^43] + 44C44*[(17x)^0]*[1^44]

Now, the final term of the expansion is the only term which does not have 17 as a factor, hence, the remainder is the remainder of final term when it is divided by 17, and since the final term is 1, the answer is 1.
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