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A student has 37 days to prepare for an examination .from past experience she  knows that she will require no more than 60 hours of study.She also wishes to study atleast 1 hour/day.

Show that no matter how she shedules her study time,there is a succession of days during which she will have studied exactly 13 hours.
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Let $y_i$  be the sum of hrs studied till the $i^{th}$ day(including). So we have the following inequality:

$1\leq y_1 < y_2 < \cdots <y_{37}\leq 60$ $-\,(1)$

Adding 13 to every element in the inequality we have:

$14\leq y_1+13 < y_2+13< \cdots <y_{37}+13\leq 73$ $-\,(2)$

So, considering the list $<y_1,y_2,\cdots,y_{37},y_1+13,y_2+13,\cdots,y_{37}+13>$ we have $74$ elements each with value as one of $\begin{Bmatrix} 1,2,\cdots 73 \end{Bmatrix}$. Thus, by Pigeonhole principle, we have 2 elements in the list with the same value such that, one element is from the first list$(1)$ and the other from the second list$(2)$ since elements in each list have an ordering as mentioned in the above inequalities.

For example we can have, $(y_{34} = y_2 + 13)$ where $y_{34}$ is from $(1)$ and $y_2 + 13$ is from $(2)$ . Thus, the condition $y_i = y_j + 13$ definitely holds. Thus, there is definitely an interval where the $\#hrs$ studied is $13$.
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