Three-digit numbers

May 1998

A three-digit number is such that its second digit is the sum of its first and third digits.

Prove that the number must be divisible by 11.

Solution

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