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Re: Find the remainder when the sum of [m]2^100 + 2^200 + 2^300+ [#permalink]
15 Oct 2012, 21:31

LalaB wrote:

my answer is D. is it right?

the thing is 2^100 has 6 as the last digit. so, we have multiple 6.as we know 6^n always has 6 as the last digit (where n>0 and n is an integer)

so 6/7 will have the remainder 6

please let me know whether this answer is ok thnx

Knowing the last digit of a number is not sufficient to determine what is the remainder when that number is divided by 7. Compare 6 and 36. Both are divisible by 6, end in 6, but give different remainders when divided by 7. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Last edited by EvaJager on 15 Oct 2012, 22:14, edited 1 time in total.

The three remainders, 2, 4, and 1, repeat cyclically for the terms in the given sum. We have 100 terms in the sum (\(10000=100\cdot{100}\)), and the last term gives again a remainder of 2.

Answer B. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Last edited by EvaJager on 15 Oct 2012, 21:53, edited 1 time in total.

Re: Find the remainder when the sum of [m]2^100 + 2^200 + 2^300+ [#permalink]
15 Oct 2012, 21:47

EvaJager wrote:

LalaB wrote:

my answer is D. is it right?

the thing is 2^100 has 6 as the last digit. so, we have multiple 6.as we know 6^n always has 6 as the last digit (where n>0 and n is an integer)

so 6/7 will have the remainder 6

please let me know whether this answer is ok thnx

Knowing the last digit of a number is not sufficient to determine whether what is the remainder when that number is divided by 7. Compare 6 and 36. Both are divisible by 6, end in 6, but give different remainders when divided by 7.

Well agreed with EvaJager.. this should not be the approach. Even my first try was based on a similar approach.

Re: Find the remainder when the sum of [m]2^100 + 2^200 + 2^300+ [#permalink]
15 Oct 2012, 21:48

EvaJager wrote:

Knowing the last digit of a number is not sufficient to determine whether what is the remainder when that number is divided by 7. Compare 6 and 36. Both are divisible by 6, end in 6, but give different remainders when divided by 7.

EvaJager , yeah, I overlooked it _________________

Happy are those who dream dreams and are ready to pay the price to make them come true

Re: Find the remainder when the sum of [m]2^100 + 2^200 + 2^300+ [#permalink]
15 Oct 2012, 21:57

The remainder when each term is divided by 7 is 1. In total there are 100 terms. So total of remainders is 100. When 100 is divided by 7, remainder is 2. So answer is 2. i.e B _________________

Did you find this post helpful?... Please let me know through the Kudos button.

The three remainders, 2, 4, and 1, repeat cyclically for the terms in the given sum. We have 100 terms in the sum (\(10000=100\cdot{100}\)), and the last term gives again a remainder of 2.

Answer B.

Amazing!!! Kudos Well, I learnt few new things from this method.

I need to write this solution on a piece of paper to understand it better.

Re: Find the remainder when the sum of [m]2^100 + 2^200 + 2^300+ [#permalink]
15 Oct 2012, 22:03

MacFauz wrote:

The remainder when each term is divided by 7 is 1. In total there are 100 terms. So total of remainders is 100. When 100 is divided by 7, remainder is 2. So answer is 2. i.e B

I am not quite sure of this.. the answer is right But the Remainder of 2^100 is 2, that of 2^200 is 4 and of 2^300 is 1 ... In cyclicity of 2,4,1

Re: Find the remainder when the sum of [m]2^100 + 2^200 + 2^300+ [#permalink]
15 Oct 2012, 22:11

mindmind wrote:

MacFauz wrote:

The remainder when each term is divided by 7 is 1. In total there are 100 terms. So total of remainders is 100. When 100 is divided by 7, remainder is 2. So answer is 2. i.e B

I am not quite sure of this.. the answer is right But the Remainder of 2^100 is 2, that of 2^200 is 4 and of 2^300 is 1 ... In cyclicity of 2,4,1

Please check EvaJager's explanation.

Oh.. Yeah.. Silly mistake by me. I took the cyclicity as 2,4,6,1. Hence my mistake.. Thanks for pointing it out _________________

Did you find this post helpful?... Please let me know through the Kudos button.

We know that the cycle repeats after every 3rd term - 1,2,4,1,2,4.... So, divide the power by 3 to find the number of complete cycles and the remaining powers. 2^100 = (2^99)*2 = R1*2 = 2. 2/7 => R=2 ------- (99/3 complete cycles and one 2 left.) 2^200 = (2^198)*(2^2) = R1*4 = 4. /7 => R=4 ------- (198/3 complete cycles and two 2 left.) 2^300 = R1. 1/7 => R=1 ------- (100 complete cycles and no 2 left.) Similarly, 2^400 => R=2, 2^500 => R=4, 2^600 => R=1 and the cycle repeats. 2^10000 = (2^9999)*2 => R2.

So, \(2^{100} + 2^ {200} + 2^ {300}+........+ 2^ {10000}\) in terms of remainders is equal to 2 + 4 + 1 + 2 + 4 + 1 +....... 2 Number of terms in the sequence = 10000/100 = 100 = 99 + 1. Each remainder cycle consists of 3 digits - 2,4,1. There are 99/3 = 33 complete cycles and 1 left out 100th term Therefore, 2 + 4 + 1 + 2 + 4 + 1 +....... 2 = (2+4+1)*33 + 2 = > 7*33 + 2. On dividing this result by 7, the remainder is 2.

Re: Find the remainder when the sum of [m]2^100 + 2^200 + 2^300+ [#permalink]
13 Jan 2014, 13:28

MacFauz wrote:

mindmind wrote:

MacFauz wrote:

The remainder when each term is divided by 7 is 1. In total there are 100 terms. So total of remainders is 100. When 100 is divided by 7, remainder is 2. So answer is 2. i.e B

I am not quite sure of this.. the answer is right But the Remainder of 2^100 is 2, that of 2^200 is 4 and of 2^300 is 1 ... In cyclicity of 2,4,1

Please check EvaJager's explanation.

Oh.. Yeah.. Silly mistake by me. I took the cyclicity as 2,4,6,1. Hence my mistake.. Thanks for pointing it out

MacFauz what were you trying to do here? Could you please elaborate more?

Re: Find the remainder when the sum of 2^100 + 2^200 + 2^300+ [#permalink]
15 Jan 2014, 01:10

mindmind wrote:

Find the remainder when the sum of \(2^{100} + 2^ {200} + 2^ {300}+........+ 2^ {10000}\) is divided by 7

A. 0 B. 2 C. 1 D. 6 E. 5

Hello Bunuel, Karishma

Let me know what you think of this approach

Consider : 2^100+2^200+2^300 at first

What is the remainder when (2^100+2^200+2^300)/7

The expression can be written as (2* 2^99+2^2*2^198+2^3*2*297)/7--------> {2*(7+1)^33+ 2^2*(7+1)^66+2^3*(7+1)^99}/7

The remainder for the above expression will be (2*1^33+ 4*(1^66)+8*(1^99)/7 is 0

when you consider the next 3 terms ie. 2^400+2^500+2^600 and simplify the expression the Remainder is still 0 because the sum of the remainder of these terms will (2^4+2^5+2^6 )/7 OR 112/7 AND THUS REMAINDER IS 0

Notice that the remainder in each case is power of 2 and follow the pattern: (2,4,8),(16,32,64)......

Since there are 100 terms in the Original expression therefore the sum of remainder from Term 1 i.e 2^100 to 2^9900 will be divisible by 7 and the remainder for the last term that 2^10000 will be 2^100-------> 2(2^99)-----> 2(7+1)^99/7

Ans is 2

Definitely not a Sub 600 level Q _________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Re: Find the remainder when the sum of 2^100 + 2^200 + 2^300+ [#permalink]
18 Apr 2014, 09:46

VeritasPrepKarishma wrote:

WoundedTiger wrote:

mindmind wrote:

Find the remainder when the sum of \(2^{100} + 2^ {200} + 2^ {300}+........+ 2^ {10000}\) is divided by 7

A. 0 B. 2 C. 1 D. 6 E. 5

Hello Bunuel, Karishma

Let me know what you think of this approach

Consider : 2^100+2^200+2^300 at first

What is the remainder when (2^100+2^200+2^300)/7

The expression can be written as (2* 2^99+2^2*2^198+2^3*2*297)/7--------> {2*(7+1)^33+ 2^2*(7+1)^66+2^3*(7+1)^99}/7

The remainder for the above expression will be (2*1^33+ 4*(1^66)+8*(1^99)/7 is 0

when you consider the next 3 terms ie. 2^400+2^500+2^600 and simplify the expression the Remainder is still 0 because the sum of the remainder of these terms will (2^4+2^5+2^6 )/7 OR 112/7 AND THUS REMAINDER IS 0

Notice that the remainder in each case is power of 2 and follow the pattern: (2,4,8),(16,32,64)......

Since there are 100 terms in the Original expression therefore the sum of remainder from Term 1 i.e 2^100 to 2^9900 will be divisible by 7 and the remainder for the last term that 2^10000 will be 2^100-------> 2(2^99)-----> 2(7+1)^99/7

Ans is 2

Definitely not a Sub 600 level Q

You seem to have used two methods here: both are correct. But let me segregate them.

For first three terms, \(2^{100} + 2^{200} + 2^{300}\), you have used the method used above by EvaJager. For the next three terms, you have used a different method. Let me do the whole question using that.

Taking 3 terms at a time (2 + 4+ 8 = 14), they are divisible by 7. There are 100 terms so we will form 33 groups of 3 terms each and last term will be left i.e. \(2^{100}\)

\(2^{100} = 2 * 8^{33} = 2 * (7 + 1)^{33}\) Remainder when divided by 7 is 2.

So you are left with following remainders

\([0 + 0 ... + 2]^{100}\)

Again, \(2^{100}\) gives a remainder of 2.

What is the relationship between \(2^{100} + 2^{200} + 2^{300} + 2^{400} + ... + 2^{10000}\) and \([2^{1} + 2^{2} + 2^{3} + 2^{4} + ... + 2^{100}]^{100}\) ?

Definitely, the two expressions are not equal. As in general, \(a^n+b^n\neq{(a+b)}^n\), and not for sums with more than two terms. Are they giving the same remainder when divided by 7? Why? It isn't obvious to me. So, what is that different method that you are mentioning?

In the above solution (WoundedTiger), in the second step, we can write \(2^{400}+2^{500}+2^{600}=2^{300}(2^{100}+2^{200}+2^{300})\). The expression in the parenthesis is divisible by 7 (was proven in the first step), so the remainder is 0. The remainders cannot be greater than 7, therefore is not correct to say they are 16, 32, 64. In fact, they are 2, 4 and 1, as \(16 = 2 * 7 + 2\), \(32 = 4 * 7 + 4\), \(64 = 9 * 7 + 1\). The process can be continued by grouping three terms each time and taking out an appropriate factor (\(2^{600},2^{900}...)\) We are left to determine the remainder given by the last term when divided by 7. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Last edited by EvaJager on 18 Apr 2014, 10:18, edited 1 time in total.

Re: Find the remainder when the sum of 2^100 + 2^200 + 2^300+ [#permalink]
18 Apr 2014, 09:58

jlgdr wrote:

Here's what I did

We have that the remainders of powers of 2 divided by 7 follow the pattern: 2,4,1

Therefore for 2^100 remainder 2 For 2^200 remainder 4 For 2^200 remainder 1 and so on...

We have 2+4+1= 7 so 7*3 = 21 + 2 = 23

So 23/7 remainder is 6

Answer is thus 6

Hope this helps Cheers! J

Wrong answer!

What is this We have 2+4+1= 7 so 7*3 = 21 + 2 = 23? And how 23 divide by 7 gives remainder 6? Isn't it 23 = 3 * 7 + 2, meaning a remainder of 2? _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: Find the remainder when the sum of 2^100 + 2^200 + 2^300+ [#permalink]
18 Apr 2014, 18:04

Expert's post

EvaJager wrote:

What is the relationship between \(2^{100} + 2^{200} + 2^{300} + 2^{400} + ... + 2^{10000}\) and \([2^{1} + 2^{2} + 2^{3} + 2^{4} + ... + 2^{100}]^{100}\) ?

Definitely, the two expressions are not equal. As in general, \(a^n+b^n\neq{(a+b)}^n\), and not for sums with more than two terms. Are they giving the same remainder when divided by 7? Why? It isn't obvious to me. So, what is that different method that you are mentioning?

In the above solution (WoundedTiger), in the second step, we can write \(2^{400}+2^{500}+2^{600}=2^{300}(2^{100}+2^{200}+2^{300})\). The expression in the parenthesis is divisible by 7 (was proven in the first step), so the remainder is 0. The remainders cannot be greater than 7, therefore is not correct to say they are 16, 32, 64. In fact, they are 2, 4 and 1, as \(16 = 2 * 7 + 2\), \(32 = 4 * 7 + 4\), \(64 = 9 * 7 + 1\). The process can be continued by grouping three terms each time and taking out an appropriate factor (\(2^{600},2^{900}...)\) We are left to determine the remainder given by the last term when divided by 7.

Yes, you are right. I am surprised I made such a conceptual mistake! For some reason I was thinking of them as multiplication signs! _________________

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