# Pigeonhole principle problem - avoiding a sum in consecutive sets. We have 48 golf balls and 30 gol

Pigeonhole principle problem - avoiding a sum in consecutive sets.
We have 48 golf balls and 30 golf holes, the holes are labeled from 1 to 30. Prove whether or not it is possible to distribute these balls into the holes while satisfying the following conditions:
1.) There must be at least one ball in each hole.
2.) The sum of balls in any number of consecutive holes cannot be 11 nor 18. (So if the fourth hole had 5 balls, fifth hole 3 balls, and sixth hole 3 balls, this would not satisfy the rule.)
I tried many different approaches (and I think that it is not possible to distribute the balls that way). First, I realized that I only need to distribute 18 balls as the 30 mandatory balls are already determined to be in the their respective holes. If I distributed 18 balls each into a different hole, at least 12 of the 30 holes would only have one ball inside. That means that in any case, there will be at least 12 holes with one ball only. But this is where I got stuck. I am not asking you to solve this problem, but I think I need a nudge towards the right way.
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Step 1
Hint: Let ${x}_{k}$ be the sum of the number of balls in the first k holes. Since at least one ball is placed in each hole, $\left({x}_{1},{x}_{2},{x}_{3},\dots ,{x}_{30}\right)$ is a strictly increasing sequence satisfying
$1\le {x}_{1}<{x}_{2}<{x}_{3}<\cdots <{x}_{30}=48$
Step 2
Let ${y}_{k}$ be the sequence defined by ${y}_{k}={x}_{k}+11$. Notice that $\left({y}_{1},{y}_{2},{y}_{3},\dots ,{y}_{30}\right)$ is also a strictly increasing sequence. Moreover,
$12\le {y}_{1}<{y}_{2}<{y}_{3}<\cdots <{y}_{30}=59$