For those claiming this was "too easy": I thought this was as difficult as IMO 2018 5, which although not particularly difficult was not "too easy" for an IMO 2. I think there are a few issues here: straight $v_p$ does not work, nor does most more number-theoretic techniques, and I think the combinatorial setup is not entirely straightforward. It's not a hard 5, granted, but it deserves its place.

N1? Okay ngl; this was prolly N3 at max but more like N2-3 Bad writeup ahead

Hmm, to me this seems like an easy number theory problem, N1 maybe, I really can't think why this is a problem 5. What are people's thoughts?

Problem is equivalent to show $$\frac{\sum_{i=2}^n \sqrt[i]{\prod_{j=1}^i a_j}}{\sum_{i=1}^n a_i}<\frac{n-1}{2}$$Where $\{a_n\}_{\ge 1}$ is an increasing sequence of integers. Also equivalent to show $$\sum_{i=1}^n (n+1-i)^{-i}<\frac{n-1}{2}$$Which follows immediately, also the case $n=2$ is clear. @below: unfortunately I don't know why from $a^2\ge bc$ I got $a\ge b$ and $a\ge c$ and also I didn't let the prime solution because I got all $\{a_n\}$ must be powers of $2$ in the case when all $a_n$ are even, but WE COULD JUST DIVIDE BY $2$. Many times we have dump stuck! Moreover here is the first solution which I deleted because "wrong idea". Suppose $(a_n)_{\ge 1}$ is a sequence increasing of integers for which any arithmetic mean of two elements is equal to some geometric mean of some elements. Now let $(b_n)_{\ge 1}$ be the sequence obtained dividing $(a_n)$ by $\gcd(a_1,a_2,\cdots)=t$. Notice $(b_n)$ satisfy tha same geometric mean condition. Let $\sigma_i\{(b_n)\}$ be the representation of some permutation of $i$ elements from $(b_n)$. Notice $\left(\sigma_i\{(b_n)\}\right)^{-i}=\frac{b_m+b_n}{2}$. So \begin{align*} \left(\left(\max\{\sigma_i\{(b_n)\}\}\right)^i\right)^{-i}&\ge \frac{b_m+b_n}{2}\\ &\ge \sqrt{b_mb_n}\\ \end{align*}Meaning $\max\{\sigma_i\{(b_n)\}\}^2\ge b_mb_n$ thus we have $\min\{b_m,b_n\}\in\sigma_i\{(b_n)\}$. So let $p$ be a prime dividing $b_1$, so from $$\frac{b_1+b_j}{2}=\sqrt[i]{\sigma_i\{(b_n)\}}$$We know $b_1\in \sigma_i\{(b_n)\}$ so we must have $\nu_p\left(\frac{b_1+b_j}{2}\right)=\nu_p(\sqrt[i]{\sigma_i\{(b_n)\}})\ge 1$ so we must have $p\mid b_j$ for all $j\ge 1$ so $b_1=1$. Similarly for any $b_k$ we get the same and thus $(b_n)$ is constant and equal to $1$ so since elements of $(a_n)$ are equal to $(tb_n)=(t\cdot 1)$ this means $(a_n)$ is constant i.e all elements are equal.

This is not equivalent to the problem.

The assertion is true for all $n$. Setup (boilerplate). Suppose that $a_1$, \dots, $a_n$ satisfy the required properties but are not all equal. If $d = \gcd(a_1, \dots, a_n) > 1$ then replace $a_1$, \dots, $a_n$ by $\frac{a_1}{d}$, \dots, $\frac{a_n}{d}$. Hence without loss of generality we may assume \[ \gcd(a_1, a_2, \dots, a_n) = 1. \]WLOG we also assume \[ a_1 \ge a_2 \ge \dots \ge a_n. \] Main proof. As $a_1 \ge 2$, let $p$ be a prime divisor of $a_1$. Let $k$ be smallest index such that $p \nmid a_k$ (which must exist). In particular, note that $a_1 \neq a_k$. Consider the mean $x = \frac{a_1+a_k}{2}$; by assumption, it equals some geometric mean, hence \[ \sqrt[m]{a_{i_1} \dots a_{i_m}} = \frac{a_1 + a_k}{2} > a_k. \]Since the arithmetic mean is an integer not divisible by $p$, all the indices $i_1$, $i_2$, \dots, $i_m$ must be at least $k$. But then the GM is at most $a_k$, contradiction. Remark: A similar approach could be attempted by using the smallest numbers rather than the largest ones, but one must then handle the edge case $a_n = 1$ separately since no prime divides $1$.

Nice problem! N3/4 i guess? We claim that the answer is all $n \ge 2$ WLOG $a_1 \le a_2 \le \dots \le a_n$. For obvious reaons, we can WLOG that $\gcd(a_1, a_2, \dots, a_n) = 1$. Now, notice that if $p \mid a_n$, then as $a_{n - 1} \le \frac{a_{n - 1} + a_n}{2} \le a_n$, then the GM that forms $\frac{a_{n - 1} + a_n}{2}$ must include $a_{n}$ as well, or otherwise, it must be equal to $a_{n - 1}$, forcing $a_n = a_{n - 1}$. Case 01. If $a_n = a_{n - 1}$. Then take the largest $k < n$ such that $a_k < a_n$. If such $k$ does not exist, then all $a_i$s are the same, forcing $a_i = 1$ for all $1 \le i \le n$. Otherwise, consider $\frac{a_k + a_{k + 1}}{2} = \frac{a_k + a_n}{2} > a_k$. Now, this is the same as the second case. Case 02. The GM of $\frac{a_{n - 1} + a_n}{2}$ includes $a_n$. This forces that if $p$ is a prime that divides $a_n$, then $p \mid "\text{GM that includes } a_n" \mid a_{n - 1} + a_n$, which forces $p \mid a_{n - 1}$. Thus, applying this repeatedly, we get that $\text{rad}(a_n) \mid a_{n - 1}$. Claim. $\text{rad}(a_n)$ divides $a_i$ for all $1 \le i < n$. Proof. We'll prove this by downward induction. We have proved this for $i =n - 1$. Notice that if for all $i = k + 1, \dots, n -1$ this statement is true, we will prove this for $i = k$. Consider $\frac{a_k + a_n}{2}$ which lies between $a_k$ and $a_n$, which means that the GM that forms this value must have a term which is greater than $a_k$, or otherwise $a_k = a_n$ and we are done. However, suppose that this GM contains $a_{k + j}$, then $\text{rad}(a_n) \mid \text{rad}(a_{k + j}) \mid a_k$. Therefore, the claim is proved. To finish this, notice that by the previous claim, $\text{rad}(a_n) = \gcd(a_1, a_2, \dots, a_n) = 1$, which forces $a_n = 1$. Since $a_n$ is the maximal element, then all element must be $1$.

Its really awkward(according to me) that this year 3 C's this is N though, maybe similar to others above though posting ,

I have seen a lot of people saying that the second day had three combinatorics problems. Problem 4 was obviously C, problem 6 was a combinatorial geometry, probably classified as G in the shortlist; OK, fine, it was something between C and G. But I can't understand why people claim that problem 5 is combinatorics. For me the problem statement is a pure number theory and I can't see any combinatorics involved in solutions that I'm aware of, including the solutions posted in this thread. Can anyone explain why this problem is combinatorics in their opinion?

timon92 wrote: I have seen a lot of people saying that the second day had three combinatorics problems. Problem 4 was obviously C, problem 6 was a combinatorial geometry, probably classified as G in the shortlist; OK, fine, it was something between C and G. But I can't understand why people claim that problem 5 is combinatorics. For me the problem statement is a pure number theory and I can't see any combinatorics involved in solutions that I'm aware of, including the solutions posted in this thread. Can anyone explain why this problem is combinatorics in their opinion? It's pure number theory, as you said. Maybe the wording of the problem is slightly combo-y idk...

Maybe people calls( or expected a problem likes it) problem like this a nice N. People are prone to make mistakes of course. And many people are showing their hatred on C, so... just forget who calls it C and discuss more elegant solutions

timon92 wrote: But I can't understand why people claim that problem 5 is combinatorics. Complaining is fun. (I agree that it's number theory with little or no combinatorics element.)

Chaotic evil: problem 3 is also NT, 1+4n=2+(4n-1)+... is definitely a number theory property. When I was solving problem 5, I spent most of my time thinking about which prime dividing which terms and I feel like chasing this is more combinatorial. I think it's hard to classify whether it's N or C ( I'd say 51%N+49%C). The main issue is that the difficulty is not matching its position. It would be a good problem if it's a problem 1. Also, it comes with a combinatoric problem and a combinatorial geometry problem makes the feeling worse.

Quote: I spent most of my time thinking about which prime dividing which terms and I feel like chasing this is more combinatorial. I think it's hard to classify whether it's N or C ( I'd say 51%N+49%C). Interesting logic. Let's look, for example, at the IMO-2016. https://artofproblemsolving.com/community/c294448_2016_imo P2 - pure C P3 - 80% C+20% G P4 - 60% C+40% N P5 - 40% C+60%A P6 - 90% C+10% G

did someone tried this problem with positive real numbers in tha cards? or rational numbers?

Suppose that the distinct terms are a1<a2<a3<…<ak, and suppose some prime p divides a(i),a(i+1),a(i+2),... a(k),. Then Consider the Am to be z=a(i-1)+a(i+1)/2, which is equal to which can be equal to the gcd of some s terms must be larger than z and hence divisible by p. Hence z is divisible by p, thus so is a(i-1). Hence by induction all a(j) are divisble by p and therefore no suffix is divisible by any prime p. We notice though that ak is a suffix so ak=1 which shows that all numbers and equal and hence proved. Evan Chens sol is good to which I would use as my 2nd well around his

Ramanujan1057894736842 wrote: did someone tried this problem with positive real numbers in tha cards? or rational numbers? Rational numbers are no different because you can multiply them by product of denominators and get integer numbers. For real numbers for $n=2$ both cards need to be equal, but for $n \ge 3$ you can go for one card with $x$ and $n-1$ cards with $y$, where $\frac{x+y}{2} = \sqrt[3]{xy^2}$ (such values exist, for example $y=1, x=\sqrt{5}-2$)

We claim that it is true for all $n$. Let the distinct positive integers on the cards be $a_1>a_2>\dots>a_k$ and assume WLOG that the gcd is 1, because we can scale down and not change the problem. We claim that $a_1=1$ which solves the problem. Otherwise, let prime $p\mid a_1$ and $i$ be minimal positive integer such that $p\nmid a_i.$ One must exist because we made gcd 1. Then take $\frac{a_1+a_i}{2}$ and we see that the GM is between $a_1$ and $a_i$ exclusive and so it uses at least one of $\{a_1,a_2,\dots,a_{i-1}\}$. Thus, the GM must be divisible by $p$, which means $p\mid a_i$, contradiction.

It is true for all $n$. Let the numbers on the cards be $a_1, a_2, \ldots a_n$. If $g$ is the gcd, we can divide all the numbers by $g$ to force it to be $1$. Now, we can always find an odd number among these cards, say it is $a_1$. If $a_2$ is even then the AM of $a_1$ and $a_2$ is a fraction, and cannot be equal to the GM of any collection of integers. Similarly if $a_2$ is odd, then the AM would be even, so there must exist some card that is even in order for the GM of that collection to be even as well. However, we would again run into the same contradiction as before so we are done. $\blacksquare$

The answer is all $n$, and we will proceed by contradiction. Say the cards were labelled $a_1 \geq a_2 \geq \dots \geq a_n$. Let $a_k$ be the smallest index $k$ such that $a_k < a_1$ so $a_1 = a_2 = \dots = a_{k-1}$. Note that the arithmetic mean $\tfrac{a_1+a_k}{2}$ must be equal to the geometric mean of some of the cards, and one of these cards must be index $\leq k-1$, since otherwise the largest possible value of any card in the geometric mean is $a_k$. This implies that the geometric mean of the cards is at most than the arithmetic of the chosen cards by AM-GM, which is strictly less than $\tfrac{a_1+a_k}{2}$. Now, as $a_1$ and $a_k$ are not equal and $a_1 > a_k$, there must exist some prime $p$ dividing $a_1$ but not $a_k$; thus $p \nmid \tfrac{a_1+a_k}{2}$. But as $a_1$ is a term in the geometric mean, $p$ must divide the geometric mean, so the arithmetic mean cannot be equal to the geometric mean for the pair $(a_1,a_k)$, contradiction.

Assume that the $\gcd$ of all numbers is $1$. Then consider a prime $p\in S$ where $S$ is the set of all primes dividing some number in the deck. Take \[p\nmid M,\, p\mid n\]where $M$ is maximal. If $n>M$ we have a contradiction by $\nu_p$ which implies that $M$ is actually also the maximum of the entire deck. However this implies that no prime $p\in S$ divides the maximum, therefore it is $1$ and we're done. $\blacksquare$

im giong to cry, hopefully this is right Take any prime $p$ and assume that the minimal and maximal $\nu_p$ are $m$ and $M$ where $m<M$. Choose the largest value $u$ in the deck with $\nu_p(u)=m$ and any value $v$ with $\nu_p(v)=M$. Then clearly we obtain a contradiction, so $m=M$ and all values are equal. wait no it's not right darn

We may assume that $\gcd(a_1, a_2, \dots, a_n) = 1$. Let $a_1<a_2<\cdots<a_n$; notice that we may relax the condition to say that the $a_i$ are distinct and the geometric mean may be calculated with each $a_i$ taken with any multiplicity. Fix some prime $p \mid a_n$. Then the arithmetic mean $\frac{a_n + a_{n-1}}2$ equals the geometric mean of some $k$ numbers among the $a_i$, which must contain $a_n$ for size reasons. It follows that $p \mid a_{n-1}$. Similarly, by considering the geometric mean equivalent of $\frac{a_{n-1} + a_{n-2}}2$, it follows $p \mid a_{n-2}$. Similarly, $p$ must divide every $a_i$, which is a clear contradiction.

We claim the answer is all $n$. In the case where $2$ cards have the same number reduce down to a case where all numbers are different. Here, let the deck be $a_1>a_2>a_3>\dots>a_n$. Scaling doesnt affect it so let $\gcd(a_1,a_2,\dots,a_n)=1$. The key claim is the following which will obviously finish. Claim: If $p\mid a_1$, the $p\mid a_i$ for all $i$. Proof. Strong induct on $i$. Note that in the geometric mean that equals $\frac 12 (a_k+a_{k-1})$ there must be an $a_i$ in said product with $i\leq k-1$; otherwise, the geometric mean $P$ satisfies $P\leq a_k<\frac 12 (a_k+a_{k-1})$. Then this term in the geometric mean will introduce a factor of $p$ into the geometric mean by hypothesis. Then, as $p\mid a_{k-1}$ again by hypothesis, we conclude. $\blacksquare$ Remark: Far too easy for its position, took like 10 minutes sus

Surprised that no one has written this method. I claim that no such sequences exist. Claim 1: All means are integers. Proof: All arithmetic means can be either an integer or an irreducible fraction, and all geometric means can be either integer or irrational, so they must be an integer. Claim 2: If $(a_1, a_2, \dots, a_n)$ works, then $(a_1d, a_2d, \dots, a_nd)$ works for some integer $d$. Proof: The arithmetic means and geometric means are scaled by a factor $d$. Claim 3: Every number are equal modulo $2$. Proof: By Claim 2, let at least one of the numbers be odd. Now if there exists an even number, pick them, then their arithmetic mean is not an integer. Claim 4: Every number are equal modulo $4$. Proof: If there exists some number with different remainders modulo $4$ ($1$ or $3$), pick them, their arithmetic mean is even. But since every number is odd, the geometric means must be odd as well. So by Claim 3 and 4, we can induct that every number is congruent modulo $2^n$ for positive integers $n$, which implies that every number must be equal.

Quite easy for an IMO P5 All N work. My solution is same as @above

Now pretty much we are done

All the numbers must be equal for all $n$. FTSOC, suppose otherwise. Let the cards be numbered $a_1 \leq a_2 \leq a_n$. WLOG, suppose $\gcd(a_1, a_2, \dots, a_n) = 1$, otherwise just divide all cards by $\gcd(a_1, a_2, \dots, a_n)$. Let $p$ be a prime dividing $a_n$. Then $\frac{a_n + a_{n-1}}{2}$ equals the geometric mean of some numbers which for size reasons must include $a_n$, so $p \mid a_{n-1}$. Then repeat this to conclude that $p$ divides all numbers, contradiction.

The answer is all natural numbers $n\geq2$. Let the numbers on the cards be $a_1\leq a_2\leq \ldots\leq a_n$. Claim. $a_n$ divides $a_i$ for all $i=1,2,\ldots,n$. Proof. For a prime $p$, let $S_p$ be the set containing all indices $i$ such that $v_p(a_i)$ is minimum. It suffices to prove $n\in S_p$ for all $p$. FTSOC, assume there exists a prime $q$ such that $S_q$ does not contain $n$. Let the greatest element of $S_q$ be $k$. Since $\dfrac{a_k+a_n}2$ must be equal to a GM, and $v_q\left(\dfrac{a_k+a_n}2\right)\leq v_q(a_k)$, then all numbers contributing in the GM must have an index in $S_q$ (otherwise the multiplicity of $q$ of the GM will be greater than $v_q(a_k)$). However, since all numbers with index in $S_q$ is not greater than $a_k$, then the GM will not exceed $a_k$, while $\dfrac{a_k+a_n}2>a_k$. We reach a contradiction. From the above result, it follows that $a_1=a_2=\dots=a_n$, regardless of the value of $n$.

The answer is all positive integers n. Proof: Let the numbers be $a_1\leq a_2\leq \ldots\leq a_n$ . now let S( may contain a value repeated) be the set of all possible geometric means. Let $A_i$ denote an element of S for $1\leq i\leq{2}^{n}-1$ and $A_i \leq A_j$ for all i<j. now $$A_{{2}^{n}-1}= a_n \geq\frac {a_{n}+a_{n-1}} {2}\geq A_{2^{n}-2}= \sqrt{a_{n}a_{n-1}}$$, equality must hold on atleast one side implying $a_{n}=a_{n-1}$.now $a_{n-1}$ is the largest $a_i$, now doing the same with $a_1, a_2.....a_{n-1}$ we get $a_{n-1}=a_{n-2}$. Iterating this similarly we get $a_1=a_2=.......=a_n$ and hence done OG.

COULD ANYONE VERIFY THE ABOVE PLEASE

IMO 2020 p5 We claim that all natural numbers works. Assume FTSOC otherwise, then assume WLOG that $a_1 \leq a_2 \leq \cdots \leq a_n$. Let $\gcd(a_1,a_2,\cdots,a_n)=1$, as we can just divide all cards by their greatest common factor. Now assume that $p$ a prime dividing $a_n$. Then $\frac{a_n+a_{n-1}}{2}$ is equal to the geometric mean of numbers, in which $a_n$ is one of them due to size. Hence $a_{n-1}$ is also divisible by $p$, hence by similar reasoning , we get that $p \mid a_n,a_{n-1}, \cdots,a_1$. A contradiction.