Prove that among any 18 consecutive 3-digit numbers there must be at least one which is divisible by the sum of its digits.
case 1)max sum of digits of 3-digit 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.
Posted by Praneeth
on 2010-08-06 19:00:40