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.
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
Anonymous
All right, but how can you prove that a number like 62345678987654555548 (or even a bigger number) is divisible bij 11?
Anonymous
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.
Anonymous
It can be done with 121 1+1=2 and 121 is divisble by 11
Anonymous
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.
Anonymous
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
opksalu@gmail.com