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0 replies
jlacosta
May 1, 2025
0 replies
2023 Japan Mathematical Olympiad Preliminary
parkjungmin   0
20 minutes ago
Please help me if I can solve the Chinese question
0 replies
parkjungmin
20 minutes ago
0 replies
Polynomial divisible by x^2+1
Miquel-point   1
N an hour ago by luutrongphuc
Source: Romanian IMO TST 1981, P1 Day 1
Consider the polynomial $P(X)=X^{p-1}+X^{p-2}+\ldots+X+1$, where $p>2$ is a prime number. Show that if $n$ is an even number, then the polynomial \[-1+\prod_{k=0}^{n-1} P\left(X^{p^k}\right)\]is divisible by $X^2+1$.

Mircea Becheanu
1 reply
Miquel-point
Apr 6, 2025
luutrongphuc
an hour ago
the same prime factors
andria   5
N an hour ago by bin_sherlo
Source: Iranian third round number theory P4
$a,b,c,d,k,l$ are positive integers such that for every natural number $n$ the set of prime factors of $n^k+a^n+c,n^l+b^n+d$ are same. prove that $k=l,a=b,c=d$.
5 replies
andria
Sep 6, 2015
bin_sherlo
an hour ago
A sharp estimation of the product
mihaig   0
an hour ago
Source: VL
Let $n\geq4$ and let $a_1,a_2,\ldots, a_n\geq0$ be reals such that $\sum_{i=1}^{n}{\frac{1}{2a_i+n-2}}=1.$
Prove
$$a_1+\cdots+a_n+2^{n-1}\geq n+2^{n-1}\cdot\prod_{i=1}^{n}{a_i}.$$
0 replies
mihaig
an hour ago
0 replies
No more topics!
IMO 2014 Problem 5
codyj   72
N Apr 15, 2025 by math-olympiad-clown
For each positive integer $n$, the Bank of Cape Town issues coins of denomination $\frac1n$. Given a finite collection of such coins (of not necessarily different denominations) with total value at most most $99+\frac12$, prove that it is possible to split this collection into $100$ or fewer groups, such that each group has total value at most $1$.
72 replies
codyj
Jul 9, 2014
math-olympiad-clown
Apr 15, 2025
IMO 2014 Problem 5
G H J
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Shreyasharma
682 posts
#61
Y by
This seems right, wouldn't mind somebody checking though :whistling:

We generalize to $k$ containers, with total sum at most $k - \frac{1}{2}$. First, we proceed with optimization.
  • For any coins that sum to 1, place them into a container thus reducing the problem to $k - 1$ containers with a total sum at most $k - \frac{3}{2}$.
  • For any two coins of denomination $\frac{1}{2d}$, combine them into one coin with denomination $\frac{1}{d}$.
Note now there is at most one coin of form $\frac{1}{2d}$ for any $d$ and $k-1$ coins of the form $\frac{1}{k}$. Now consider placing all the coins into the containers such that the coins are grouped in the following manner:
  • $\{\frac{1}{2}, \frac{1}{3}, \dots, \frac{1}{3} \}$

  • $\{ \frac{1}{4}, \frac{1}{5}, \dots, \frac{1}{5} \}$

  • $\{\frac{1}{6}, \frac{1}{7}, \dots, \frac{1}{7}\}$,
and so on such that the last container contains coins of the form $\{\frac{1}{2k}, \frac{1}{2k+1}, \dots, \frac{1}{2k+1} \}$. It is easy to see that each box always has a group of coins that sum to less than 1. Then fill in the remaining gaps with smaller denominations. At the end of this process, let the minimal coin outside of a container have weight $\frac{1}{\alpha}$. Then the sum of denominations in the $i$-th container $C_i$, must be greater than $\frac{1}{\alpha}$ giving,
\begin{align*}
    \sum C_i > k - \frac{k}{\alpha}
\end{align*}However as $k - \frac{1}{2} > C_i$, we see that
\begin{align*}
    k - \frac{1}{2} > k - \frac{k}{\alpha} \iff 2k > \alpha,
\end{align*}a clear contradiction as we have already fit coins of this form.

Remark: Just something I noticed, this isn't exactly optimal. Similar to how we bounded coins of the form $\frac{1}{2k}$ we could extend to bounding coins of the form $\frac{1}{pk}$ for some prime $p$ though I doubt that it strengthens the bound by a lot.
This post has been edited 1 time. Last edited by Shreyasharma, Oct 27, 2023, 1:37 AM
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Math4Life7
1703 posts
#62
Y by
We change all pairs of the coins of the form $\frac{1}{2n}$ to one coin of value $\frac{1}{n}$. If we have $n$ coins of value $\frac{1}{n}$ we make a coin of value $1$.

We prove the more general problem where we substitute $100$ with $x$. We make it so that we make all the $\frac{1}{1}$ coins irrelevant by decreasing our $x$. Notice that we have at most $1$ of each of the coins with even denominators and we have at most $2n$ of the coins with denominators $2n+1$.

Claim 1:
We can place all the coins from value $\frac{1}{2}, \frac{1}{3} \ldots \frac{1}{2x}$

proof:
We let each of our $x$ piles contain the $x$ numbers with even denominators ($\frac{1}{2}, \frac{1}{4} \ldots \frac{1}{2x}$). We now can see that we can also place the odd coins in their "neighboring" piles. Specifically we claim that $\frac{n-1}{2n-1} + \frac{1}{2n} + \frac{n}{2n+1} \leq 1$ this is obvious by multiplying. We can see that this proves the claim.

Claim 2:
If we cannot place a coin with denominator greater than $2n$ than our total value is above $x - \frac{1}{2}$.

proof:
Let this coin have value $\frac{1}{y}$. We need the piles to each have more than $\frac{y-1}{y}$. Specifically we need the piles to sum to at least $x \frac{y-1}{y} \leq x - \frac{1}{2}$ which obviously has size issues when $y > 2x$

Combining these claims we see the result. $\blacksquare$
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OronSH
1747 posts
#63 • 2 Y
Y by GeoKing, megarnie
Solved in Cape Town.

We prove this in general, that a collection of coins with total value at most $x-\frac 12$ can be split into $x$ such groups. First notice that we may assume that there is at most one of each coin of denomination $\frac 1{2n},$ since replacing two of them with one of denomination $\frac 1n$ is a more restrictive version of the same problem. Similarly we assume that there are at most $k-1$ coins of denomination $\frac 1k$ when $k$ odd (we may do this for $k=1$ since it is equivalent to decreasing $x$ by $1,$ and the situation cannot happen if $x=1$ so this will work).

Now we place coins one at a time with nonincreasing denomination. Suppose that this fails for some coin with denomination $\frac 1{2n}.$ Then we have all groups have total value more than $1-\frac 1{2n},$ so the total value of coins is greater than $x-\frac {x-1}{2n}$ and thus $x-\frac{x-1}{2n} < x-\frac 12$ so simplifying gives $x > n+1.$ But now we can check that the total value of the coins placed before is less than or equal to $\frac 12+1-\frac 13+\frac 14+1-\frac 15+\cdots+1-\frac 1{2n-1}.$ This is less than $n-\frac 12$ so we must have $x-\frac x{2n}<n-\frac 12,$ but $x-\frac x{2n}=x\left(1-\frac 1{2n}\right) > (n+1)\left(1-\frac 1{2n}\right)=n-\frac 12+1-\frac 1{2n},$ contradiction.

Now suppose this fails for a coin with denomination $\frac 1{2n+1}.$ We have all groups have total value more than $1-\frac 1{2n+1},$ so the total value of coins is more than $x-\frac{x-1}{2n+1}$ and this is less than or equal to $x-\frac 12$ so we can simplify to get $x>n+\frac 32.$ Now we can see that the total value of coins placed before is less than or equal to $\frac 12+1-\frac 13+\frac 14+\cdots+\frac 1{2n}+1-\frac 2{2n+1}$ which is less than $n+\frac 12$ so we must have $x-\frac x{2n+1}<n+\frac 12$ but $x-\frac x{2n+1}=x\left(1-\frac 1{2n+1}\right)>\left(n+\frac 32\right)\left(1-\frac 1{2n+1}\right)=n+1-\frac 1{2n+1}>n+\frac 12,$ contradiction, so we are done.
This post has been edited 1 time. Last edited by OronSH, Dec 16, 2023, 2:40 PM
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ihatemath123
3446 posts
#64 • 2 Y
Y by OronSH, Inconsistent
If we have two coins of denomination $\frac{1}{2n}$, we merge them into one coin of denomination $\frac{1}{n}$; we can then separate the coin into two once we've allocated the piles.

Similarly, if we have $2n-1$ coins of denomination $\frac{1}{2n-1}$, we just place them all into one pile since that's extremely efficient (no leftover space).

So we may now assume we have at most one coin of denomination $\frac{1}{2n}$ and at most $2n-2$ coins of denomination $\frac{1}{2n-1}$. For $1 \leq i \leq 100$, we place all coins of denomination $\frac{1}{2i}$ and $\frac{1}{2i-1}$ in pile $i$.

We now fill in all spaces with coins of denomination $\frac{1}{201}$, $\frac{1}{202}$, etc. When we can't fill any more spaces, each pile must be at least $\frac{200}{201} > \frac{199}{200}$ in value, for a total of $\frac{199}{2} =99.5$ spent; we've used all our coins.
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naonaoaz
332 posts
#65 • 1 Y
Y by ihatemath123
Generalize $100$ to $n$. First, do the following reductions:
1. If there are two coins of denomination $\frac{1}{2k}$ for any $k \ge 1$, replace them with one coin of denomination
2. If there are $2k+1$ coins of value $\frac{1}{2k+1}$, replace them with a coin of value $1$.
If we can find a valid grouping after these reductions then we can also find one before since we can split the coins in both bullet points.

If we have a coin of denomination $1$, then we're done by induction, so assume we don't.

Now place the one (if we one) $\frac{1}{2}$ value coin into one group. Next, for all $1 \le k \le n-1$ place all $\frac{1}{2k+1}$ and $\frac{1}{2k+2}$ coins into one group. By our restrictions, this group has value at most
\[\frac{2n}{2n+1} + \frac{1}{2n+2} < 1\]Now we can just shove the rest of the coins that have value $\le \frac{1}{2n+1}$ into any group that still has room. FTSoC, assume we can't do this. Then the total value is at least
\[n \cdot \left(1-\frac{1}{2n+1}\right) > n-\frac{1}{2}\]a contradiction, which finishes the problem.
This post has been edited 1 time. Last edited by naonaoaz, Feb 25, 2024, 5:58 AM
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dolphinday
1325 posts
#66
Y by
Replace $100$ with general $k$.
Now we can make the following optimization. Assume that the Bank of Cape Town allows us to trade $2m - 1$ coins of denomination $\frac{1}{2m - 1}$ for a coin of denomination $1$ and $2$ coins of denomination $\frac{1}{2m}$ for a single coin of denomination $\frac{1}{m}$. We make this optimization repeatedly, and are left with at most $2m - 2$ coins of denomination $\frac{1}{2m - 1}$ and at most $1$ coin of denomination $\frac{1}{2m}$. Now we can set aside all coins that have denomination $1$. Place all coins of denomination $\frac{1}{2m - 1}$ and $\frac{1}{2m}$ in pile $m$ which is valid since the total value of pile $m$ is at most $\frac{2m - 2}{2m - 1} + \frac{1}{2m} < 1$.
Notice that all coins of denomination $< \frac{1}{2k}$ can be placed into groups.
If not, then we can create pre-existing groups and show that it is impossible that the coins of denomination $< \frac{1}{2k}$ cannot fit anywhere.. Then this implies that all of the groups have remaining space that is $< \frac 1 2n$ which is a contradiction since the sum of the preexisting groups would be $> n - \frac 12$, so we are done.
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Pyramix
419 posts
#68
Y by
Incorrect solution
This post has been edited 1 time. Last edited by Pyramix, Apr 15, 2024, 2:03 PM
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Pyramix
419 posts
#69
Y by
Replace 100 with $n$ and $99.5$ with $n-0.5$. We induct on $n$.
Replace two coins $\frac1{2m}$ with one coin $\frac1m$. Replace $2m+1$ coins $\frac1{2m+1}$ with $1$ coin, and make separate group (this is just induction hypothesis).
Keep doing this until we cannot. At this point we have at most 1 coin with denominator is even and at most $2t$ coins if denominator = $2t+1$. Consider set $S_0$ containing the $\frac12$ coin, and $S_m$ containing all coins $\frac1{2m}$ and $\frac1{2m+1}$ for every $m<100$. Then, each set has sum at most 1. We are now left with denominations $\frac1{201},\frac1{202},$ etc. Distribute them into the sets, we can do this because the sets have sum less than 1. Suppose we are still left with some "extra" denominations. So, the "sets" each have sum as at least $1-\frac1{201}=\frac{200}{201}$. Since there are 100 such sets, we already have a total sum of $\frac{20000}{201}>\frac{199}2$, a contradiction.
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joshualiu315
2534 posts
#70
Y by
Replace $100$ with a general $n$. We will perform an algorithm on the coins as follows:
  • Exchange $2$ coins of denomination $\tfrac{1}{2k}$ for a coin of denomination $\tfrac{1}{k}$.
  • Exchange $k$ coins of denomination $\tfrac{1}{k}$ for a coin $1$.

It is obvious that if we can split the collection into $n$ groups after the algorithm, we can make the split before. Hence, we can assume that the algorithm is performed as many times as possible.

We are left with at most $2k-2$ coins of denomination $\tfrac{1}{2k-1}$ and at most $1$ coin of denomination $\tfrac{1}{2k}$. Group the above coins together, which is possible since the sum of each group can equal at most

\[\frac{2k-2}{2k-1}+\frac{1}{2k} = \frac{4k^2-2k-1}{4k^2-2k} < 1.\]
Then, for all coins with denomination less than $\tfrac{1}{2n}$, we can simply add them to groups that have room. If there is not room for such a coin, then each group has sum greater than $1-\tfrac{1}{2n}$, which implies a total sum greater than $n-\tfrac{1}{2}$, a contradiction.

Hence, we have a desired grouping. $\blacksquare$
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Mr.Sharkman
500 posts
#71
Y by
This was massively overestimated in difficulty.

We can use induction, so that, if the coin denomination of value $\frac{1}{k}$ is repeated more than $k$ times, we can simply put a whole bucket just for that coin. Also, we can take any even valued coin which is repeated more than twice, and group two of them together, so that $2$ coins of denomination $\frac{1}{2k}$ become one of denomination $\frac{1}{k}.$ Now, we have reduced the problem to showing that we can group coins with total value at most $\frac{2n+1}{2}$ into at most $n+1$ groups, where the coins of even denomination appear at most once, and the coins of denomination $2i+1$ occur at most $2i$ times.

Put the coins of denomination $2$ into the first bucket (we do not have any coins of denomination $1$). For the other buckets, put one coin of denomination $2i$ in, and all of the other coins of denomination $2i-1$ as well. Then, notice that we are left over with denominations at least $2n+1$ which cannot be stored in its own bucket, since then there would be $1$ too many buckets. These denominations can be tossed into any buckets with leftover room. Then, we can see that it is always possible to put in these last coins, since otherwise we would have more than $(k-1)+\frac{1}{2}$ in denominations. Hence, we are done.
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N3bula
276 posts
#72
Y by
I will prove that if we have $2n+\frac{1}{2}$ as the total value we can split it into $n$ groups of value at most $1$. Suppose we have $2$ elements $\frac{1}{2n}$, we can
treat this as one element of $\frac{1}{n}$. We can also say that we don't have $n$ elements of $\frac{1}{n}$ because we can induct from previous case that this works. Now
consider $2n$ categories labled with the odd values from $1$ to $2n$. Clearly we can put any elements of the form $m2^k$ into the category labled $m$ without exceeding $100$
elements total. Now consider the elements smaller than $\frac{1}{4n}$, suppose an element smaller than that can not be put in any category, this means the total sum of
all categories is more than $2n+\frac{1}{2}$ which is a contradicition.
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Saucepan_man02
1342 posts
#73
Y by
(Replace $100$ with $N$)
Consider the following step: If sum of multiple coins sum to a number of the form $\frac{1}{n}$, then we remove all of them and add a $\frac{1}{n}$ coin.
In the current coin collection, assume that there are $d$ coins with value $1$, then we make single-ton groups with them and thus, we replace $N$ with $N-d$.

We make the number $\frac{1}{2}$ into a single-ton group.
For every $2 \le k \le N-1$, we have at-most one occurrence of $\frac{1}{2k}$ and at-most $2k$ occurrences of $\frac{1}{2k+1}$ and therefore, they sum to: $$\frac{1}{2k}+\frac{2k}{2k+1} < 1.$$Now, we claim that, for all $t \ge 2N-1$, we can freely skip it: If not, then each group has sum of at-least $1-\frac{1}{t}$ and therefore: total sum is at-least: $$\frac{N} - \frac{N-1}{t} \le N-\frac{1}{2}$$But this implies: $t \le 2N-2$, contradiction. Thus, we can place denominations of the form $\frac{1}{t}$ for all $t \ge 2N-1$ into any group which has space for it.
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cursed_tangent1434
632 posts
#74
Y by
Solved this with a concerning amount of hints. But then again, I'm bad at combinatorics and especially at local arguments in particular.

We approach this via induction. We prove the more general statement that for all $k\geq 1$, we are able to split a collection of coins of total value at most $(k-1)+\frac{1}{2}$ into $k$ groups, such that each group has a total value at most 1, for all $k\geq 0$. For $k=0$, the claim is obvious as we can simply chuck all the coins in one group.

We first conduct the following optimization. Whenever we have 2 coins of denomination $\frac{1}{2r}$ we exchange it for a coin of value $\frac{1}{r}$ and whenever we have $r$ coins of denomination $\frac{1}{r}$, we exchange it for a coin of value 1. It is easy to see that if we are able to achieve our goal after this exchange, we are most obviously able to achieve it before (simply place the smaller coins where u placed the equivalent larger coin).

Now, we place all the coins of size 1 (say there are $k$ of them) in separate boxes. Thus, we have coins of total value $(n-k)+\frac{1}{2}$ to be separated into $(n-k+1)$ groups which is guaranteed to be possible by the induction hypothesis. So, assume we have no coins of size 1 after this optimization.

In this setup, we have at most 1 coin of value $\frac{1}{2k}$ for each $k \geq 1$ and at most $2k$ coins of value $\frac{1}{2k+1}$ for each $k\geq 1$.

Algorithm : We place all coins of value $\frac{1}{2}$ in box $B_1$. For all $k\geq 2$, we then place all coins of value $\frac{1}{2k}$ and $\frac{1}{2k-1}$ in Box $B_k$.

Proof : If $S_k$ is the total value of $B_k$ in such an assignment of coins,
\[S_k \leq \frac{2k-2}{2k-1} +\frac{1}{2k} = \frac{4k^2-4k+2k-1}{4k^2-2k} = \frac{4k^2-2k-1}{4k^2-2k}<1\]Thus, we can place all coins of weights $\frac{1}{r}$ for $r\leq 2n$ using this algorithm.

Now, we have successfully placed all the coins of denominations greater than $\frac{1}{2n}$. For the rest, we perform the following greedy algorithm.

Algorithm : We pick the unplaced coin with the largest denomination ($\frac{1}{i}$-for smallest $i$) and place it in the left-most box which it can occupy (boxes are placed in the order $B_1,B_2,\dots,B_n$ from left to right).

Proof : We claim that this successfully disposes of all the other coins. Assume that we eventually get stuck. That is, there exists a coin with denomination $\frac{1}{r}$ which cannot be placed in any of the boxes. Thus, each box has total value greater than $1-\frac{1}{r}$. Thus, the sum of total assigned coins $S$ is,
\[(n-1)+\frac{1}{2}> S > n\left(1-\frac{1}{r}\right) = n - \frac{n}{r}\]which rearranges to $\frac{n}{r} < \frac{1}{2} \implies r < 2n$. But, we have already packed in coins of size $\frac{1}{r}$ for $r<2n$ and are dealing with smaller coins. Thus, this algorithm never breaks and we can reach our goal.

This means the induction is complete. Thus, it is always possible to split a finite collection of coins with total value at most $99+\frac{1}{2}$ into 100 or fewer groups, such that each group has total value at most 1, as desired.
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PEKKA
1848 posts
#75
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Used an embarrassingly high number of hints. Darn.

Replace $100$ by $n.$
First, we optimize as follows: Given two coins of type $\frac{1}{2k}$ we "combine" them into a coin of type $\frac 1k.$ Given $2k+1$ coins of type $\frac{1}{2k+1}$ we combine them into one coin of type $\frac{1}{1}.$
Now we ignore all of the coins that have been combined as integers. Let the sum of the remaining coins be at most $\frac{2i-1}{i}.$ We will fit them into $i$ boxes.
In the first box we put all coins of type $\frac 12$, of which there are $1$ or less.
In the $j$th box we put the coins of type $\frac{1}{2j-1}$ and $\frac{1}{2j}.$ Of the former type there are at most $2j-2$ and of the latter type there is at most $1$ so the box has value at most $$1-\frac{1}{2j-1}+\frac{1}{2j}<1.$$Now all that remains are coins of value less than $\frac{1}{2i}.$
We perform a greedy algorithm and fit a remaining coin into any box that can fit it. I claim this fits all the coins.
FTSOC this is not true. Then each coin already fits more than $1-\frac{1}{2i+1}$ worth of coins. Over all $i$ boxes, this combines to a total value of $i-\frac{i}{2i+1}>i-\frac{1}{2}$ which is a contradiction. Q.E.D.
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math-olympiad-clown
30 posts
#76
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Can I ask how much point can I get ? ChatGPT says I can got 4~5 but I think I can’t get that much. :-D
we want to prove a stronger statement: for some coins it’s total value is less than n+1/2 .
Then they can be split into n+1 sets . So the problem is just a special case for n=99.

We will do this by induction ,for n=1,we need to prove the coins total value less than 1+1/2 can be split into two sets. And that’s easy because the deviation is less than 1/2 for all sets by greedy algorithm.
next we assume that for n=2~98 all follow the induction hypothesis.Now we need to go on to n=99.
we can deduce that we only need to consider the module of k for the numbers of all coins of Denomination 1/k.
This is because if for some k there are k*a+b for such 1/k coins, then we can merge the k*a coins into a coins of denomination of 1.
We can delete these sets of coins can that leads to a case for n=2~98,which follows the induction hypothesis.
So we only need to consider the module of k for the numbers of all coins of Denomination 1/k.
Now we want to prove we can split all of the coins into two parts for all finite collection of coins with total value at most 99.5 : one has total denomination less than 1 and the other one has total value less than 98.5 ,if so ,we can easily finish the question by induction.
But I failed to prove this :wacko:
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