Combinatorics #2

A guy eats 1\geq1 donuts a day. Given that he ate a total of 4545 donuts in 3636 days, prove that he ate exactly 2626 donuts in some period of time, where some period of time refers to a whole number of days.

#Combinatorics

Note by Victor Loh
6 years, 11 months ago

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Comments

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.

Xuming Liang - 6 years, 11 months ago

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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.

Victor Loh - 6 years, 11 months ago

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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.

Calvin Lin Staff - 6 years, 11 months ago

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@Calvin Lin Ok. But it's also possible for him to eat 2 donuts on the first day (I said 1\geq1).

Victor Loh - 6 years, 11 months ago

Got it, here's my solution:

Let aia_i denote the number of donuts eaten all together before the end of day ii. Hence we have: 0<a1<a2<...<a35<a36=450<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,mn,m with n<mn<m such that aman=26a_m-a_n=26.

Consider the 1919pigeon holes (1,27),(2,28),...,(19,45)(1,27),(2,28),...,(19,45). Since we are tossing in 3535 numbers aia_i where i=135i=1-35 and 2619=726-19=7 numbers aren't used in our pigeon holes, hence there are at least 357=2835-7=28 numbers that will end up in the holes...and since 28>1928>19 there must exists a pigeon hole that contains two numbers and we are done..

Xuming Liang - 6 years, 11 months ago
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