# Pigeon hole principle — an example for RMO preparation

Problem:

A student has 6 weeks (that is, 42 days) to prepare for her examination and she has decided that during this period she will put in a total of 70 hours towards her preparation for the examination. She decides to study in full hours every day, studying at least one hour each day. We have to prove that no matter how she schedules her studying pattern, she will study for exactly 13 hours during some consecutive days.

Solution:

If $a_{i}$ denotes the number of hours she studies on the ith day, then we are given that each $a_{i}$ is a positive integer and the sum $a_{1}+a_{2}+ \ldots + a_{42}=70$. To get the required succession of days, we must find some m and j such that $m \leq j$ and $a_{m}+ \ldots + a_{j}=13$. Trying out all the possibilities, is close to impossible as well as dumb. Let $b_{i}$ denote the partial sum $a_{1}+ \ldots + a_{i}$ which is the number of hours she studies upto the ith day. Our given data then translates into $1 \leq b_{1} < b_{2} < \ldots < b_{41} < b_{42}=70$

and we have to find some $i such that $b_{i}+13=b_{j}$, that is, $b_{j}-b_{i}=a_{i+1}+ \ldots + a_{j}=13$ (clearly, then $i. Hence, besides the 42 numbers in $B= \{ b_{1}, b_{2}, \ldots , b_{42}\}$ we also look at 42 more numbers in $B^{'}= \{ b_{1}+13, b_{2}+13, \ldots, b_{42}+13\}$ which are also 42 different numbers and the largest among them is $b_{42}+13=70+13=83$. Hence, the $2 \times 42$ are actually among the positive integers from 1 to 83 and hence, by the pigeonhole principle, we see that two numbers in $B \bigcup B^{'}$ must be equal. As we already saw the numbers in B are all distinct and so are the numbers in $B^{'}$. Hence, we must have for some i and j, $b_{i}=b_{j}+13$ giving the required succession of days when she studied exactly for 13 hours a day.

More on pigeonhole principle,

Later,

Till then,

Nalin Pithwa

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