Prove that among any 18 consecutive 3-digit numbers there must be at least one which is divisible by the sum of its digits.
(In reply to
solution by Praneeth)
Absolutely true, but case 1 is included in case 2.
Erasing the first four rows and the words "case 2" does not affect the validity of your proof.
Your solution boils down to a shorter text:
In a row of 18 consecutive numbers two are divisible by 9:
one is an even multiple and the other is odd.
Both 2 and 9 divide the even one.
q.e.d.
re: solution
