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Find the remainder when the sum of 2^100 + 2^200 + 2^300+

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Re: Find the remainder when the sum of 2^100 + 2^200 + 2^300+  [#permalink]

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New post 15 Feb 2016, 00:12
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


Solved this in one minute.

Just remember to use the concept explained @ quarter-wit-quarter-wisdom-a-remainders-post-for-the-geek-in-you and half of the remainder Qs become a piece of cake :)

All the terms in the expression (a+b)^z are divisible by a except b^z

(8=7+1) Make the modifications in the expression so that it is in some relation with the given divisor (usually plus or minus 1)

2^100 = 2* 8^33 = 2* (7+1)^33
2^200 = 4* 8^66 = 4* (7+1)^66
2^300 = 1* 8^100 = 1* (7+1)^100
2^400 = 2* 8^133 = 2* (7+1)^133
2^500 = 4* 8^166 = 4* (7+1)^166
2^600 = 1* 8^200 = 1* (7+1)^200
Observer that this it is a repeat cycle of 2, 4 and 1

Make a note of total number of terms = 100
remainder from first term = 2*1
remainder from second term = 4*1
remainder from third term = 1*1

Sum of first three remainders = 7. Remainder =0 .Sum of first 99 remainders = 7*33 Remainder =0 . Remainder on the 100th term = 2 (from the cycle)

Answer B

Though I sound pretty confident I need someone to validate the method :oops: Thanks!
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Re: Find the remainder when the sum of 2^100 + 2^200 + 2^300+  [#permalink]

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New post 04 Mar 2016, 13:18
jns wrote:
mindmind wrote:
Find the remainder when the sum of \(2^{100} + 2^ {200} + 2^ {300}+........+ 2^ {10000}\) is divided by 7

2^0 = 1. 1/7 => R=1, R = remainder
2^1 = 2. 2/7 => R=2
2^2 = 4. 4/7 => R=4
2^3 = 8. 8/7 => R=1
2^4 = 16. 16/7 => R=2 .... and the cycle repeats.

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.


Why do you start the cycle (2^0; 2^1; 2^2 etc) from 2^0? Is it a special case? I've always started from n^1 and the answer was right. Thank you!
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Re: Find the remainder when the sum of 2^100 + 2^200 + 2^300+  [#permalink]

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New post 05 Mar 2016, 03:41
1
Viktoriaa wrote:
jns wrote:
mindmind wrote:
Find the remainder when the sum of \(2^{100} + 2^ {200} + 2^ {300}+........+ 2^ {10000}\) is divided by 7

2^0 = 1. 1/7 => R=1, R = remainder
2^1 = 2. 2/7 => R=2
2^2 = 4. 4/7 => R=4
2^3 = 8. 8/7 => R=1
2^4 = 16. 16/7 => R=2 .... and the cycle repeats.

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.


Why do you start the cycle (2^0; 2^1; 2^2 etc) from 2^0? Is it a special case? I've always started from n^1 and the answer was right. Thank you!


hi,
you are CORRECT in your approach.
Always start with 1, whenever you are finding such answers..
although the person has calculated for 2^0, but has not used in his calculations for finding remainder of 2^100...
You should maintain the approach you have been following

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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: Find the remainder when the sum of 2^100 + 2^200 + 2^300+  [#permalink]

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New post 21 Mar 2016, 04:31
Bunuel
Need Some Input..
I solved the Question using this approach=>
there are 100 terms in the series.
First term =2^100 => 2*[7+1]^33 => using the Binomial expansion => 2* [7K +1] = 7p+2
2nd term => 2^200 => 4*[7+1]^66 => 7p+4
3rd term => 2^300 => 7p+1
and the next terms are all following the same pattern ..
hence the series can be written as => 7p+2 + 7t+4 + 7s+1 +....
now i noticed that there are 100 terms => sum of the first 99 terms => 7c for some c
hence the last term = 7r+2 => sum of all the terms => 7z+2 hence 2 is the remainder..
I look forward to a method other than this too..

also is this approach right ?
thanks and regards
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Re: Find the remainder when the sum of 2^100 + 2^200 + 2^300+  [#permalink]

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New post 01 Dec 2016, 06:17
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


Cyclicity is a good approach but modular arithmetic will help us to solve this question a little bit faster.

We have:

\(\frac{2^{100} + 2^{200} + 2^{300} + 2^{400} + … + 2^{10000}}{7}\)

\(2^3 = 1 (mod _7)\)

\(2^{100} = 2^{99}*2 = (2^3)^{33}*2 = 1^{33}*2 = 2\)

\(2^{100} + 2^{200} + 2^{300} + 2^{400} + … + 2^{10000} = (2^{100})^1 + (2^{100})^2 + (2^{100})^3 + … + (2^{100})^{100}\)

In mod 7 this is the same as:

\(\frac{2 + 2^2 + 2^3 + 2^4 + 2^5 + … + 2^{100}}{7}\) where the nominator is the sum of geometric progression.

\(\frac{2*(2^{100}-1)}{(2-1)*7} = \frac{2*(2^{100}-1)}{7}\)

But we already know that \(2^{100} = 2 (mod_7)\)

And our expression boils down to:

\(\frac{2*(2-1)}{7} = \frac{2}{7}\)

Our remainder is \(2\).

Answer B.
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Re: Find the remainder when the sum of 2^100 + 2^200 + 2^300+  [#permalink]

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New post 24 Dec 2017, 19:14
chetan2u Bunuel niks18 VeritasPrepKarishma

Hi,
I was wondering if one of you experts could please check my work? I was able to get to the right answer in under 2 mins but I am not entirely positive if my way is correct. I simply took the first term and last term and added them together to get the final equation as:

(2^100+2^10000)/7 - (2^100(1*2^9900))/7 - 2^100/7= Remainder of 2 & 2^9900/7= Remainder of 1. I then multiplied the remainders together (2*1)/7 to finally get a remainder of 2.

Please let me know if my way was correct? Would greatly appreciate it!
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Find the remainder when the sum of 2^100 + 2^200 + 2^300+  [#permalink]

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New post 24 Dec 2017, 20:32
csaluja wrote:
chetan2u Bunuel niks18 VeritasPrepKarishma

Hi,
I was wondering if one of you experts could please check my work? I was able to get to the right answer in under 2 mins but I am not entirely positive if my way is correct. I simply took the first term and last term and added them together to get the final equation as:

(2^100+2^10000)/7 - (2^100(1*2^9900))/7 - 2^100/7= Remainder of 2 & 2^9900/7= Remainder of 1. I then multiplied the remainders together (2*1)/7 to finally get a remainder of 2.

Please let me know if my way was correct? Would greatly appreciate it!


Hi csaluja,

At first I could not understand your logic for adding the two extremes. Also the highlighted portion is incorrect. if you take 2^100 out then the expression will be 2^100(1+something)

Note that this is a GP(geometric series) and formula to find sum of GP series \(=\frac{a_1(r^n-1)}{(r-1)}\), where \(a_1=\)first term, \(r=\) common ration\(=\frac{a_2}{a_1}\) and \(n=\)number of terms

Also note that \(\frac{2^{100}}{7}=\frac{(2^3)^{33}*2}{7}=\frac{8^{33}*2}{7}\), Now 8 leaves a remainder of 1 when divided by 7 and 2 leaves a remainder of 2

\(=>remainder = 1^{33}*2=2\)

Similarly \(2^{200}=(2^{100})^2\), remainder for this term will be \((2)^2\)

So we will have new G.P series of REMAINDERS

\(2 + 2^2 + 2^3 +......+ 2^{100}\)

Sum of the remainders: \(S_n = \frac{a(r^n - 1)}{r-1}\), where \(a=2; r=2, n=100\)

\(S_{100}=\frac{2(2^{100} - 1)}{(2-1)} =>2(2^{100} - 1)=2^{101}-2\)

Hence \(Remainder = \frac{(2^{101}-2)}{7}=\frac{[(2^3)^{33}*2^2]}{7}-\frac{2}{7}\)

or, \(remainder= \frac{8^{33}*4}{7}-\frac{2}{7} =1^{33}*4-2=4-2=2\)
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Find the remainder when the sum of 2^100 + 2^200 + 2^300+ &nbs [#permalink] 24 Dec 2017, 20:32

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