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.
\(2^{100} + 2^{200} + 2^{300} + 2^{400} + ... + 2^{10000}\)
\([2^{1} + 2^{2} + 2^{3} + 2^{4} + ... + 2^{100}]^{100}\)
\([
2 + 4 + 8 +
16 + 32 + 64 ... + 2^{100}]^{100}\)
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.
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