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\(2^{1927}=(2^5)^{385}\cdot2^2=(33-1)^{385}\cdot{4}=(M33-1)\cdot{4}=M33-4=M33-33+29=M33+29\). (\(M33\) designates multiple of \(33\).)

\(2^{1929}=2^{1927}\cdot2^2=(M33+29)\cdot{4}=M132+29\cdot{4}=M132+116\). \(M33+29\) means \(33n+29\), for some integer \(n\). Then \((M33+29)\cdot{4}=132n+29\cdot{4}=M132+116\).

Hence, the remainder is 116. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

\(2^{1929}=2^{1927}\cdot2^2=(M33+29)\cdot{4}=M132+29\cdot{4}=M132+116\). \(M33+29\) means \(33n+29\), for some integer \(n\). Then \((M33+29)\cdot{4}=132n+29\cdot{4}=M132+116\).

Hence, the remainder is 116.

What is wrong if i cancel out 2^2 in the numerator and divide by 33 . I still have to get the same reminder right?

\(2^{1927}=(2^5)^{385}\cdot2^2=(33-1)^{385}\cdot{4}=(M33-1)\cdot{4}=M33-4=M33-33+29=M33+29\). (\(M33\) designates multiple of \(33\).)

\(2^{1929}=2^{1927}\cdot2^2=(M33+29)\cdot{4}=M132+29\cdot{4}=M132+116\). \(M33+29\) means \(33n+29\), for some integer \(n\). Then \((M33+29)\cdot{4}=132n+29\cdot{4}=M132+116\).

\(2^{1929}=2^{1927}\cdot2^2=(M33+29)\cdot{4}=M132+29\cdot{4}=M132+116\). \(M33+29\) means \(33n+29\), for some integer \(n\). Then \((M33+29)\cdot{4}=132n+29\cdot{4}=M132+116\).

Hence, the remainder is 116.

What is wrong if i cancel out 2^2 in the numerator and divide by 33 . I still have to get the same reminder right?

Hi g3kr,

Both of the last 2 solutions fetching 29 and 116 are correct processwise, but for a logical mistake in former.

Consider the case, 16/10 will have a remainder of 6, but its not same as 8/5 which has a remainder of 3 only.

The difference? notice that for any y/x remainder can not be greater than x, hence if you have factored out a number from numerator and denominator, you have reduced remainder by that times. In our example of 16/10: if we make it 8/5 by dividing each term by 2 and get 3 as remainder. we should multiply result by 2 to get correct ans 6.

hence, once you arrived in first solution (M33+29)

you should remember to multiply it by 4, the one which was factored out earlier. Or better dont factor out and cancel a term in a remainder problem.

Re: Remainder problem. [#permalink]
16 Oct 2012, 19:39

What is wrong if i cancel out 2^2 in the numerator and divide by 33 . I still have to get the same reminder right?[/quote]

Hi g3kr,

Both of the last 2 solutions fetching 29 and 116 are correct processwise, but for a logical mistake in former.

Consider the case, 16/10 will have a remainder of 6, but its not same as 8/5 which has a remainder of 3 only.

The difference? notice that for any y/x remainder can not be greater than x, hence if you have factored out a number from numerator and denominator, you have reduced remainder by that times. In our example of 16/10: if we make it 8/5 by dividing each term by 2 and get 3 as remainder. we should multiply result by 2 to get correct ans 6.

hence, once you arrived in first solution (M33+29)

you should remember to multiply it by 4, the one which was factored out earlier. Or better dont factor out and cancel a term in a remainder problem.

Hope it helps.[/quote]

Thanks,

Even I had the same ambiguity.. But just to be sure.. there wont be both options in the answer : 29 & 116 ?

Re: Remainder problem. [#permalink]
16 Oct 2012, 20:33

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mindmind wrote:

Even I had the same ambiguity.. But just to be sure.. there wont be both options in the answer : 29 & 116 ?

Both 29 and 116 can be in the options because the answer is only 116. In fact, it is quite likely that both will be there since if there is a way to trip you, GMAT will not let it go. You can choose to cancel off terms in the numerator and denominator or you can choose to keep them as it is. In either case, answer will be 116.

\(\frac{8^{643}}{132} = \frac{8*8^{642}}{132}\)

We get \(2*\frac{8^{642}}{33} = 2*\frac{2^{1926}}{33}\)

Re: Remainder problem. [#permalink]
16 Oct 2012, 20:45

1

This post received KUDOS

mindmind wrote:

What is wrong if i cancel out 2^2 in the numerator and divide by 33 . I still have to get the same reminder right?

Hi g3kr,

Both of the last 2 solutions fetching 29 and 116 are correct processwise, but for a logical mistake in former.

Consider the case, 16/10 will have a remainder of 6, but its not same as 8/5 which has a remainder of 3 only.

The difference? notice that for any y/x remainder can not be greater than x, hence if you have factored out a number from numerator and denominator, you have reduced remainder by that times. In our example of 16/10: if we make it 8/5 by dividing each term by 2 and get 3 as remainder. we should multiply result by 2 to get correct ans 6.

hence, once you arrived in first solution (M33+29)

you should remember to multiply it by 4, the one which was factored out earlier. Or better dont factor out and cancel a term in a remainder problem.

Hope it helps.

Thanks,

Even I had the same ambiguity.. But just to be sure.. there wont be both options in the answer : 29 & 116 ?

Hi mindmind (too many minds :D),

Exactly what I explained with example of 16/10 and 8/5. if i ask you remainder of 16/10 and give u option of 6 and 3, will you be confused? I dont think so

If u factor out a number, the "fraction" remains same but "remainder" does not. please re-read the post. Also this is why it is important not to factor out in remainder problem. Also try to check one more way of understanding it: a remainder is shown as below: if n = qm +r then kn = kqm +kr where Kn is a multiple of n and km is multiple of m. But this changes remainder from r to kr.

Re: What is the remainder when 8^643/132? [#permalink]
12 Feb 2014, 03:29

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