Bangladesh National Math Olympiad 2014 : Junior level Problem 10

Oindri has100 100 chocolates. She finished eating all her chocolates in 5858 days by eating at least one chocolate each day. Prove that, in some consecutive days, she eats exactly 1515 chocolates.

(It was probably the hardest problem for the Junior level this year. Hope you guys have fun solving it.)

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In case you did not get the statement, here's a better explanation. Define di=d_i = the number of chocolates she ate on the ithi^{th} day. You have to prove that, for some ii and jj, k=ijdk=15\sum_{k=i}^j d_k = 15 where 1ij581\leq i\leq j\leq 58.

Note by Labib Rashid
5 years, 8 months ago

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Let f(n) f(n) be the number of chocolates eaten by the end of day n. There are only 15 possible remainders (mod15) \pmod{15} so by the pigeonhole principle we have 2 cases: The first case is that there at least 13 sets of 4 which satisfy the following congruence: f(a)f(b)f(c)f(d)(mod15) f(a) \equiv f(b) \equiv f(c) \equiv f(d) \pmod{15} where f(d)>f(c)>f(b)>f(a) f(d) > f(c) > f(b) > f(a) . Also f(d)f(c)f(c)f(b)f(b)f(a)0(mod15) f(d)-f(c) \equiv f(c)-f(b) \equiv f(b)-f(a) \equiv 0 \pmod {15} . Say that there was no set of consecutive days where Oindri ate 15 chocolates. Then we know that f(d)f(c),f(c)f(b),f(b)f(a)30 f(d)-f(c), f(c)-f(b), f(b)-f(a) \ge 30 . Therefore f(d)f(a)90 f(d)-f(a) \ge 90 . Since she only has 100 chocolates we know in fact that f(d)f(a)=90 f(d)-f(a) = 90 . But we have this fact 13 times over, ie. f(d1)f(a1)=f(d2)f(a2)=...=f(d13)f(a13)=90 f(d_{1})-f(a_{1}) = f(d_{2})-f(a_{2}) = ... = f(d_{13})-f(a_{13}) = 90 . But all f(ak) f(a_{k}) are distinct, so f(ai)13 f(a_{i}) \ge 13 for some value of i where 1i13 1 \leq i \leq 13 . But that implies that f(di)103 f(d_{i}) \ge 103 which is more than the number of chocolates, so we have a contradiction. Therefore our assumption was wrong and there must be some set of consecutive days where Oindri ate 15 chocolates. The second case is that there is one set of at least 5 terms which satisfies the following congruence f(a)f(b)f(c)f(d)f(e)(mod15) f(a) \equiv f(b) \equiv f(c) \equiv f(d) \equiv f(e) \pmod{15} where f(e)>f(d)>f(c)>f(b)>f(a)f(e) > f(d) > f(c) > f(b) > f(a) . Using just the same assumption and argument as before we realise that f(e)f(a)+120 f(e) \ge f(a) + 120 , but this again will be more than the number of chocolates, again meaning that our assumption was wrong and there must be some set of consecutive days where Oindri ate 15 chocolates.

Josh Rowley - 5 years, 8 months ago

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what is pigeonhole priniciple?

Ruhan Habib - 5 years, 5 months ago

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It's like when you have more pigeons than holes, you need more holes.

Jubayer Nirjhor - 2 years, 8 months ago

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@Jubayer Nirjhor thanks for this invaluable information :3

Ruhan Habib - 2 years, 8 months ago

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could u pls solve it more simply?

Promise Cadet - 3 years, 11 months ago

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I was unable to answer this question at all :(

Ruhan Habib - 5 years, 5 months ago

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