Let digits are a,b,c
Let b=a+c because second digit is sum of last and first
So number=a*100+b*10+c*1
=a*100+(a+c)*10+c*1
=a*100+a*10+c*10+c*1
=110*a+11*c
=11*10*a+11*c
=11*(10*a+c)
That is the number is multiple of 11 so it is divisible by 11

yes.
1 way: Use calculator for windows 7, which is totally capable of numbers larger than that.(12 digits longer than 62345678987654555548)
the other speedy way:
What is:
6 - 2 + 3 - 4 ...?
If it is 0 or 11 or -11, it is divisible...
if it isn't, no.

## Comments

## proof 3digit

Let digits are a,b,c

Let b=a+c because second digit is sum of last and first

So number=a*100+b*10+c*1

=a*100+(a+c)*10+c*1

=a*100+a*10+c*10+c*1

=110*a+11*c

=11*10*a+11*c

=11*(10*a+c)

That is the number is multiple of 11 so it is divisible by 11

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## three-digit numbers

All right, but how can you prove that a number like 62345678987654555548 (or even a bigger number) is divisible bij 11?

## You can, in this way: First

You can, in this way:

First (from the first digit), divide the no: into groups of 2.

Eg: 62 34 56 78 98 76 54 55 55 48

Then, add them up.

Eg: 62 + 34 + 56 + 78 + 98 + 76 + 54 + 55 + 55 + 48 = 616

If the sum is more than 100, repeat step 1 and 2.

Eg: 6 16

6 + 16 = 22

If the sum obtained is divisible by 11, then the initial no: is divisible by 11.

## yes...

yes.

1 way: Use calculator for windows 7, which is totally capable of numbers larger than that.(12 digits longer than 62345678987654555548)

the other speedy way:

What is:

6 - 2 + 3 - 4 ...?

If it is 0 or 11 or -11, it is divisible...

if it isn't, no.

## Yes

It can be done with 121 1+1=2 and 121 is divisble by 11