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# Combinatorics #2

A guy eats $$\geq1$$ donuts a day. Given that he ate a total of $$45$$ donuts in $$36$$ days, prove that he ate exactly $$26$$ donuts in some period of time, where some period of time refers to a whole number of days.

Note by Victor Loh
2 years, 6 months ago

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Could you elaberate on "in some period of time"? Is it that there exists two days such that the number of donuts he ate inbetween is exactly 26? or is it something else. · 2 years, 6 months ago

It can be any number of days. The question is asking to prove that there will be at least one span of days where the guy eats 26 donuts in total. · 2 years, 6 months ago

Got it, here's my solution:

Let $$a_i$$ denote the number of donuts eaten all together before the end of day $$i$$. Hence we have: $$0<a_1<a_2<...<a_{35}<a_{36}=45$$(we could imagine them as points on the number line) and we have to prove that there exists $$n,m$$ with $$n<m$$ such that $$a_m-a_n=26$$.

Consider the $$19$$pigeon holes $$(1,27),(2,28),...,(19,45)$$. Since we are tossing in $$35$$ numbers $$a_i$$ where $$i=1-35$$ and $$26-19=7$$ numbers aren't used in our pigeon holes, hence there are at least $$35-7=28$$ numbers that will end up in the holes...and since $$28>19$$ there must exists a pigeon hole that contains two numbers and we are done.. · 2 years, 6 months ago

You should clarify that "some period of time" refers to a whole number of days. Otherwise, from the time that he ate the first donut, to the time that the finished the 26th donut, clearly he has eaten exactly 26 donuts during that period. Staff · 2 years, 6 months ago

Ok. But it's also possible for him to eat 2 donuts on the first day (I said $$\geq1$$). · 2 years, 6 months ago