Three-digit numbers

Submitted by plusadmin on May 1, 1998
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

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?

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.

Submitted by Anonymous on November 21, 2013

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

Submitted by Anonymous on August 30, 2014

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.

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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