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

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.

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

EvaJager , yeah, I overlooked it

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

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.

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

Thanks EvaJager, that was smooth
Still, has anybody else found a short approach for this one? I'm still trying.

Let me know
Cheers!
J

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

Thanks!
Cheers!
J

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

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

*Removed the solution

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.

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?

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

2^100/7=2^2
2^200/7=2^4
2^300/7=2^6
2^400=2^1
2^500=2^3
2^600=2^5
All additions make it perfect div by 7
only 2^800 +2^900 will give rem of -1
-1 repeated for 9 times in cyclical order
and 2^10000 ill give rem 4
-9+4=-5
So rem is 2