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.
Let digits are a,b,c
Let b=a+c because second digit is sum of last and first
That is the number is multiple of 11 so it is divisible by 11
All right, but how can you prove that a number like 62345678987654555548 (or even a bigger number) is divisible bij 11?
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:
6 - 2 + 3 - 4 ...?
If it is 0 or 11 or -11, it is divisible...
if it isn't, no.
It can be done with 121 1+1=2 and 121 is divisble by 11