Prove that among any 18 consecutive 3digit numbers there must be at least one which is divisible by the sum of its digits.
case 1)max sum of digits of 3digit number = 27
when number is 999
Here 18 consecutive digits are 982,983,...,999.
In this case 990 is divisible by 18.
case 2)In 18 consecutive 2 are divisible by 9.
one is an even multiple and other is odd multiple.
rule of divisibility by 9: divisible by sum of digits.
So, sum of digits can be 9,18.
The even multiple of 9 is divisible by both 9 and 18.
Hence Proved.

Posted by Praneeth
on 20100806 19:00:40 