Prove that in every 100 consecutive integers there is an integer whose digits sum to a number divisible by 14

Question

Prove that in every 100 consecutive integers there is an integer whose digits sum to a number divisible by 14

How would one go about proving this? Many thanks!

Answer 1

One of these 100 consecutive integers $a$ will end in 00. Say that the remainder of $a$ divided by 14 is $v\in\{0,1,\ldots,13\}$.

Case 1. There exist among these 100 consecutive integers at least 49 larger than $a$. In such case the remainders of the sum of the digits of $a,a+1,\ldots,a+49$ will range between $$ a\!\!\!\!\!\!\mod\!\! 14,\,\, a+1\!\!\!\!\!\!\mod\!\! 14,\ldots, a+49\!\!\!\!\!\!\mod\!\! 14, $$ and hence will cover the whole of $\{0,1,\ldots,13\}$.

Case 2. There exist among these 100 consecutive integers at least 50 smaller than $a$. Then we shall 50 numbers with two last digits ranging for 50 to 99 and all the other digits identical, and if the remainder the sum of the digits of the one ending in 50 is $w$ then we shall have as remainders $$ w\!\!\!\!\!\!\mod\! 14, w+1\!\!\!\!\!\!\mod 14,\ldots, w+49\!\!\!\!\!\!\mod 14 $$ which again will cover the whole of $\{0,1,\ldots,13\}$.