Prove that 1000....00001 is divisible by 11 if and only if there is an even number of zeros?

Answer 1

A number with an even number of 9s, like 9,999, or 999,999, are obviously divisible by 11. That means such numbers x 10, or like 99,990 or 9,999,9990 are divisible by 11. Add 11 to those numbers, and you have your proof.

Answer 2

I think this is the easiest: if the number of zeros is even then the long number is 10^(2n + 1) + 1 for some natural n and then
(10^(2n + 1) + 1) / (10 + 1) =
=10^(2n) - 10^(2n-1) + 10^(2n-2) - . . .- 10 + 1
according the well-known formula of the sum of a geometric progression, starting with 1, ratio -10.

P.S.(EDIT) Sorry, didn't notice the 2nd part.
Since 10 ≡ -1 (mod 11), 10^(2n) ≡ (-1)^2 ≡ 1, so
10^(2n) + 1 ≡ 1 + 1 = 2 (mod 11) and is not divisible by 11.
In practice the common criterion of divisibility by 11, cited in Brian W's answer, is derived this way.

Answer 3

Wow!

I found some suggestions from Dr.Math, which is to tell that x+1 is a factor x^(2n+1) + 1.

Pretty much the same thing that Duke said, except this time I used variables for it.

x^2n - x^(2n-1) + x^(2n-2) - x^(2n-3) + x^(2n-4) ... + x^2 - x + 1.

Every "even-numbered" term we subtract.

Every "odd-numbered" term we add.

When using synthetic division we can get a clear view on how this works.

We bring the first term, which is "odd-numbered", down.

We multiply that by -1 so we subtract.

This continues on and on.

So for every "even-numbered" term we would subtract.

This is just an add-on to Duke's answer. Couldn't find any other solutions besides the hints that Dr. Math gave me.
_____________________________

Here's the shortest proof:

Prove that x + 1 is a factor of x^(2n+1) + 1

Use factor theorem

x = -1

-1^(2n+1) + 1

2n + 2 = odd

-1 + 1 = 0

Therefore, proved.

Answer 4

The answer or rule of thumb is...

If you add the odd digits up and the even digits up, and find the difference of the two, if it equals ZERO, its divisible by 11.

So, if you have an EVEN number of digits in the number, the answer would be 1 for even digits and 1 for odd digits. So the difference would be ZERO so it works.

If you have an odd number of digits, EVEN digits = 0 and ODD digits =2 so it doesn't work.

Good Luck.

Answer 5

Using method hit and try
11/11=1 (number of zeros is zero)
1001/11=91
100001/11=9091
10000001/11=909091
1000000001/11=90909091
100000000001/11=9090909091
Therefor we can say
1000....0001/11 (even number of zeroes)
the result well be true number
909090....909091