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Re: Remainder problem. [#permalink]
16 Oct 2012, 20:11

Actually the cyclicity is 3:

(2^0)/7 = 0 R 1 (2^1)/7 = 0 R 2 (2^2)/7 = 0 R 4 (2^3)/7 = 1 R 1 (2^4)/7 = 2 R 2 (2^5)/7 = 4 R 4 ...

The pattern is such that the remainder is 4 for every third one and we know there are 201 numbers between 0 and 200. Since 201 is a multiple of 3, D is in fact correct.

The remainder of 4^40 would be same as that of 4 . Is my approach wrong? Can you provide similar examples, where it is not.

In this case, 4^{40} is a M7+4 : \,\,4^{40}=2^{80} and 80=M3+2 (the cycle is 3, because 2^{3}=8=M7+1). 4^{40}=2^{80}=(2^3)^{26}\cdot{2^2}=(M7+1)\cdot{4}=M7+4. Or - 4^3=64=M7+1, therefore 4^{40}=(4^3)^{13}\cdot{4}=(M7+1)^{13}\cdot{4}=(M7+1)\cdot{4}=M7+4.

Consider for example 5^{40}: 5=7\cdot{0}+5=M7+5 and 5^3=125=126-1=7\cdot{18}-1=M7-1=M7+6. 5^{40}=(5^3)^{13}\cdot{5}=(M7-1)^{13}\cdot{5}=(M7-1)\cdot{5}=M7-5=M7+2 and not M7+5. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Last edited by EvaJager on 17 Oct 2012, 02:48, edited 1 time in total.

Re: Remainder problem. [#permalink]
17 Oct 2012, 02:43

Please help me understand better :

The remainder of 4^40 would be same as that of 4 . Is my approach wrong? Can you provide similar examples, where it is not.[/quote]

In this case, 4^{40} is a M7+4 : \,\,4^{40}=2^{80} and 80=M3+2 (the cycle is 3, because 2^{3}=8=M7+1). 4^{40}=2^{80}=(2^3)^{26}\cdot{2^2}=(M7+1)\cdot{4}=M7+4.

Consider for example 5^{40}: 5=7\cdot{0}+5=M7+5 and 5^3=125=126-1=7\cdot{18}-1=M7-1=M7+6. 5^{40}=(5^3)^{13}\cdot{5}=(M7-1)^{13}\cdot{5}=(M7-1)\cdot{5}=M7-5=M7+2 and not M7+5.[/quote]

Yes, Agreed So I should consider : Something near to Remainder 1

Any Trend where, (number or prime )^n divided by (other no, prime) gives the same remainder as the original no. or prime eg : 4^40 divided by 7 give a remainder of 4.

Re: Remainder problem. [#permalink]
17 Oct 2012, 03:01

mindmind wrote:

Please help me understand better :

The remainder of 4^40 would be same as that of 4 . Is my approach wrong? Can you provide similar examples, where it is not.

In this case, 4^{40} is a M7+4 : \,\,4^{40}=2^{80} and 80=M3+2 (the cycle is 3, because 2^{3}=8=M7+1). 4^{40}=2^{80}=(2^3)^{26}\cdot{2^2}=(M7+1)\cdot{4}=M7+4.

Consider for example 5^{40}: 5=7\cdot{0}+5=M7+5 and 5^3=125=126-1=7\cdot{18}-1=M7-1=M7+6. 5^{40}=(5^3)^{13}\cdot{5}=(M7-1)^{13}\cdot{5}=(M7-1)\cdot{5}=M7-5=M7+2 and not M7+5.[/quote]

Yes, Agreed So I should consider : Something near to Remainder 1

Any Trend where, (number or prime )^n divided by (other no, prime) gives the same remainder as the original no. or prime eg : 4^40 divided by 7 give a remainder of 4.[/quote]

I am not sure what you mean here: Any Trend where, (number or prime )^n divided by (other no, prime) gives the same remainder as the original no. or prime eg : 4^40 divided by 7 give a remainder of 4.

3^4=81=11\cdot{7}+4. 7 is prime, but the remainder is neither 3, nor a prime. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Remainder problem. [#permalink]
17 Oct 2012, 03:09

EvaJager wrote:

mindmind wrote:

Please help me understand better :

The remainder of 4^40 would be same as that of 4 . Is my approach wrong? Can you provide similar examples, where it is not.

In this case, 4^{40} is a M7+4 : \,\,4^{40}=2^{80} and 80=M3+2 (the cycle is 3, because 2^{3}=8=M7+1). 4^{40}=2^{80}=(2^3)^{26}\cdot{2^2}=(M7+1)\cdot{4}=M7+4.

Consider for example 5^{40}: 5=7\cdot{0}+5=M7+5 and 5^3=125=126-1=7\cdot{18}-1=M7-1=M7+6. 5^{40}=(5^3)^{13}\cdot{5}=(M7-1)^{13}\cdot{5}=(M7-1)\cdot{5}=M7-5=M7+2 and not M7+5.

Yes, Agreed So I should consider : Something near to Remainder 1

Any Trend where, (number or prime )^n divided by (other no, prime) gives the same remainder as the original no. or prime eg : 4^40 divided by 7 give a remainder of 4.[/quote]

I am not sure what you mean here: Any Trend where, (number or prime )^n divided by (other no, prime) gives the same remainder as the original no. or prime eg : 4^40 divided by 7 give a remainder of 4.

3^4=81=11\cdot{7}+4. 7 is prime, but the remainder is neither 3, nor a prime.[/quote]

Yes.. I was referring to 3.. and not the remainder 4.. But I got your point.. thanks..

Re: What is the remainder when you divide 2^200 by 7? [#permalink]
26 Oct 2012, 10:34

Another approach is the Theorem of Remainder 2^200 = 4*(2^3)^66;7 = 2^3 -1 Theorem of Remainder of f(X)/X-a is f(a) 2^200 = 4*f(X)/X-a X=2^3 a = 1 ==> Remainder of f(X)/X-a = 1 Remainder of 2^200 divided by7 is 4*1 = 4 Brother Karamazov

Re: What is the remainder when you divide 2^200 by 7? [#permalink]
08 Aug 2013, 22:39

Easiest and the smallest possible solution ever:

Keep your mind open while dealing with such question.It is not that you have to be math pro for that.

REM(2^200/7) [ REM(x/y) means remainder when x is divided by y]

We know that : Rem when 8 is divided by 7 is '1'.

Also by powers of 2 we can reach to 8.Using this concept:

[ (2^3)^198 * 2^2] /7

[ (8)^198 * 2^2] /7

Since REM(8/7) =1

We are left with REM(4/7) = 4 _________________

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