Oindri has$100$ chocolates. She finished eating all her chocolates in $58$ days by eating at least one chocolate each day. Prove that, in some consecutive days, she eats exactly $15$ chocolates.

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

[SPOILER!!]

[SPOILER!!]

[SPOILER!!]

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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestLet $f(n)$ be the number of chocolates eaten by the end of day

n. There are only 15 possible remainders $\pmod{15}$ 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: $f(a) \equiv f(b) \equiv f(c) \equiv f(d) \pmod{15}$ where $f(d) > f(c) > f(b) > f(a)$. Also $f(d)-f(c) \equiv f(c)-f(b) \equiv f(b)-f(a) \equiv 0 \pmod {15}$. Say that there was no set of consecutive days where Oindri ate 15 chocolates. Then we know that $f(d)-f(c), f(c)-f(b), f(b)-f(a) \ge 30$. Therefore $f(d)-f(a) \ge 90$. Since she only has 100 chocolates we know in fact that $f(d)-f(a) = 90$. But we have this fact 13 times over, ie. $f(d_{1})-f(a_{1}) = f(d_{2})-f(a_{2}) = ... = f(d_{13})-f(a_{13}) = 90$. But all $f(a_{k})$ are distinct, so $f(a_{i}) \ge 13$ for some value ofiwhere $1 \leq i \leq 13$. But that implies that $f(d_{i}) \ge 103$ 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 $f(a) \equiv f(b) \equiv f(c) \equiv f(d) \equiv f(e) \pmod{15}$ where $f(e) > f(d) > f(c) > f(b) > f(a)$. Using just the same assumption and argument as before we realise that $f(e) \ge f(a) + 120$, 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.Log in to reply

what is pigeonhole priniciple?

Log in to reply

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

Log in to reply

Log in to reply

could u pls solve it more simply?

Log in to reply

I was unable to answer this question at all :(

Log in to reply