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# Find the remainder when number 10111213....9899 is divided by 33 ?plz tell with full explanation .

Yuvraj Singh
42 Points
2 years ago

SOLUTION :The no. 1011121314........9596979899 is completely divisible by 33.

From 10 to 99 there are 90 two digit no...
from 1 to 9 every digit no. is 20 times …
to show divisibility with 33 we need to show that its divisible by 3 and 11..
To show divisibility by 3 the sum should be divisible by 3..
If from 1 to 9 every no. is 20 times so using Arithmetic Progression sum formula..

a=20
d=20
l=180
n=9

Sum =n/2(a+l)
=900
And 900 is divisible by 3 so the whole number is divisible by 3.
Now to show divisibility by 11 we need to show that the difference  of odd terms and even terms is divisible by 11..

In odd terms the no. from 1 to 9 would be 10 times .
So again using A. P. Sum formula..

a=10
d=10
l=90
n=9

sum =450

Similarly sum of even terms would be 450..
So their difference would be 450-450=0 and 0 is divisible by 11...
So the whole no. is divisible by 11..
Therefore the number is divisible by 33...