Question

Question for Fun :)

Answer 1

\[\large \bf 19^8+18^7+17^6+16^5+15^4+14^3+13^2+12^1+11^0\]
Find the remainder when the number above is divided by 11.

Answer 2

Here are few useful facts about remainders when divided by some integer :

1) Remainder of \(x+y\) equals the remainder of the sum of remainders of \(x\) and \(y\)


For example,
\(10\) leaves a remainder of \(\color{red}{2}\) when divided by \(4\), and
\(5\) leaves a remainder of \(\color{red}{1}\) when divided by \(4\);
therefore, \(10+5\) leaves a remainder of \(\color{red}{2+1=3}\) when divided by \(4\)

Answer 3

2) Remainder of \(x^k\) equals the remainder of \((remainder ~of~ x)^k\)


For example,
\(10\) leaves a remainder of \(\color{red}{2}\) when divided by \(8\), and
therefore, \(10^2\) leaves a remainder of \(\color{red}{2}^2=4\) when divided by \(8\)

Answer 4

I think those two properties are sufficient to work this problem

Answer 5

The given sum, as far as remainders when divided by 11 are concerned, is same as
\[8^8+7^7+6^6+5^5+4^4+3^3+2^2+1^1+0^0\]

Answer 6

cool

Answer 7

but what next ?

Answer 8

I might get some ideas after lunch! brb :)

Answer 9

okay! have a good lunch

Answer 10

just a quick suggestion thing to simplify the expression
Write each term like (11+x)^n all the terms if this expression will be divisible by 11 except for the last term which would be x^n

Answer 11

@KendrickLamar2014 yo sup

Answer 12

I think ganeshie8 meant to say breakfast? xD

Answer 13

*

Answer 14

@imqwerty your method is same as @ganeshie8 's method

Answer 15

Do i post my solution now ?

Answer 16

slightly different from ganeshie's method

Answer 17

:p oops didn't read the whole post

Answer 18

You can reduce it a little bit more by writing 19^8 as (22-3)^8 and same for 18^7, 17^6

Soo the remainder obtained from the original expression will be the same as that obtained from this-

3^8-4^7+5^6+5^5+4^4+3^3+2^2+1^1+1

Answer 19

help me instead

Answer 20

I always think of the binomial theorem when I think of remainders of powers, for instance check out this term:
\[14^3\]
We can write it in terms of the remainder as:
\[(11+3)^3\]
Which when you expand the binomial becomes:
\[3^3 + 3*3^2*11+3*3*11^2+11^3\]
The point is, really every single term in the binomial expansion carries at least one multiple of our remainder \(11\), so we never even have to worry about them cause they'll evenly divide out. So we can say:

\(14^3=(3+11)^3\) has the same remainder as \(3^3\) when divided by 11. I think ganeshie might have already given this fact but I think it's pretty clear after a bit of reasoning that it should be true too!

Answer 21

ok I totally didn't read @imqwerty said the same thing pretty much but w/e can't hurt to read what I wrote anyways lol

Answer 22

Good job :)
and thanks @ganeshie8 ,@imqwerty and @Kainui

Answer 23

@imqwerty