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Find the remainder when the sum of 2^100 + 2^200 + 2^300+ [#permalink]
 15 Oct 2012, 21:23
Question Stats:
100% (02:13) correct
0% (00:00) wrong based on 2 sessions
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
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Re: Find the remainder when the sum of [m]2^100 + 2^200 + 2^300+ [#permalink]
15 Oct 2012, 21:47
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
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Re: Find the remainder when the sum of [m]2^100 + 2^200 + 2^300+ [#permalink]
15 Oct 2012, 22:31
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.
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Last edited by EvaJager on 15 Oct 2012, 23:14, edited 1 time in total.
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Re: Find the remainder when the sum of [m]2^100 + 2^200 + 2^300+ [#permalink]
15 Oct 2012, 22:42
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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
Please help me out with this. 2^{100}=(2^3)^{33}\cdot{2}=(M7+1)\cdot{2}=M7+2 ( M7 denotes multiple of 7). 2^3=8, which is a M7+1. 2^{200}=(2^{100})^2=(M7+2)^{2}=M7+4. 2^{300}=(2^{100})^3=(M7+2)^{3}=M7+8=M7+1. Therefore, 2^{100}+2^{200}+2^{300}=M7+2+4+1=M7. 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.
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Last edited by EvaJager on 15 Oct 2012, 22:53, edited 1 time in total.
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Re: Find the remainder when the sum of [m]2^100 + 2^200 + 2^300+ [#permalink]
15 Oct 2012, 22:47
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.
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Re: Find the remainder when the sum of [m]2^100 + 2^200 + 2^300+ [#permalink]
15 Oct 2012, 22:48
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 
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Re: Find the remainder when the sum of [m]2^100 + 2^200 + 2^300+ [#permalink]
15 Oct 2012, 22:57
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
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Re: Find the remainder when the sum of [m]2^100 + 2^200 + 2^300+ [#permalink]
15 Oct 2012, 22:59
EvaJager 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
Please help me out with this. 2^{100}=(2^3)^{33}\cdot{2}=(M7+1)\cdot{2}=M7+2 ( M7 denotes multiple of 7). 2^3=8, which is a M7+1. 2^{200}=(2^{100})^2=(M7+2)^{2}=M7+4. 2^{300}=(2^{100})^3=(M7+2)^{3}=M7+8=M7+1. Therefore, 2^{100}+2^{200}+2^{300}=M7+2+4+1=M7. 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.
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Re: Find the remainder when the sum of [m]2^100 + 2^200 + 2^300+ [#permalink]
15 Oct 2012, 23:03
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 Please check EvaJager's explanation.
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Re: Find the remainder when the sum of [m]2^100 + 2^200 + 2^300+ [#permalink]
15 Oct 2012, 23:11
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
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Re: Find the remainder when the sum of [m]2^100 + 2^200 + 2^300+
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15 Oct 2012, 23:11
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