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

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

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.

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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