Three-digit numbers

Submitted by plusadmin on April 30, 1998
in
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

Submitted by Anonymous on July 13, 2011.

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

Submitted by Anonymous on June 13, 2011.

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

yes...

Submitted by Anonymous on August 18, 2011.

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.