A guy eats ≥1 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.
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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.
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.
Let ai denote the number of donuts eaten all together before the end of day i. Hence we have: 0<a1<a2<...<a35<a36=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 am−an=26.
Consider the 19pigeon holes (1,27),(2,28),...,(19,45). Since we are tossing in 35 numbers ai 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..
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2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
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.
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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.
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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.
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≥1).
Ok. But it's also possible for him to eat 2 donuts on the first day (I saidGot it, here's my solution:
Let ai denote the number of donuts eaten all together before the end of day i. Hence we have: 0<a1<a2<...<a35<a36=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 am−an=26.
Consider the 19pigeon holes (1,27),(2,28),...,(19,45). Since we are tossing in 35 numbers ai 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..