The answer is no. The product of all the products of adjacent pairs is a perfect square, but the product $\prod _{j=1}^{2024}j!$ is not. Edit: We can prove that the product of summations is not by considering p-adic of 1009 like @below

Yeah ok this is pretty easy for canada mo.. SFTSOC, it's possible. Just notice that the product of the product of the adjacent pairs of the numbers, each of $\{1, \dots, 2024\}$ are present twice, making it a perfect square. while if all of these products are in the set, $1! \cdot \cdots \cdot 2024!$ must be a perfect square as well. To prove that it's not, in $1! \cdot \cdots \cdot 2024!$ we'd have to take a prime less than (actually <$\frac{2024}{2}$, this method works, just slightly less motivated and more work). $\sqrt{2024}$, or it'll have an even power. (following numbers denote the power of 43 in their factorials). Notice that: $$1, 2, ..., 42 \implies 0$$$$43, 44, ..., 85 \implies 1$$$$\dots$$$$1978, \dots, 2020 \implies 46$$$$2021, 2022, 2023, 2024 \implies 47$$So the total power of 43 is $\frac{46\cdot47}{2} + 47\cdot4$. However, we still need to add $1$ (to)since $1849, \dots, 2024$ have a power of $43^2$ in them, so the total power is $\frac{46\cdot47}{2} + 4\cdot47 + 176$, which is odd.

I cited Nagura's result for this (that there is a prime between $n$ and $6n/5$ for $n \geq 25$).

lol i used 2-adic and paired up (2!, 3!) (4!, 5!) (6!, 7!) etc and we're left with 2024! and v2 (2024!) is odd by the formula \[n-s_2(n)=2017\]

Is this right? No. Suppose otherwise. If we take the product of all adjacent pairs, then the result is the product of the numbers squared, so \[ N = 1! \cdot 2! \cdots 2024!\]is a perfect square. Now we consider the $\nu_{1009}$ of this. For any $n < 1009$, we have $\nu_{1009} (n!) = 0$, if $1009 \le n < 2018$, we have $\nu_{1009}(n!) = 1$, and if $2018 \le n \le 2024$, we have $\nu_{1009}(n!) = 2$. Hence the parity of the $\nu_{1009}$ of $N$ is the number of positive integers $n$ with $1009 \le n < 2018$, which is just $1009$, hence $\nu_{1009}(N)$ is odd, so $N$ isn't a perfect square, contradiction.

Yes it's right ! Congrats on solving it (me who used 43 )

rrc orzz!!! We claim that it is no, because the product of all pairs must be a square as each term appears twice in the product. However, this is impossible as the product of $1!, 2!, \dots $ is not a square. To show this, consider powers of $2$. We can pair $(2k)!$ and $(2k+1)!$ which cancels (in terms of parity of the exponent of $2$), leaving us with $2024!$. We then calculate $\nu_2(2024) = 1012+506+253+126+63+31+15+7+3+1$ which is odd.

are Canada MO problems supposed to be increasing order or random

this was way too easy for the CMO why was this p2

Oops didnt realize $1009$ works. $v_2\left(\prod{i=1}^{2024} i!\right)=v_2\left(\prod_{j=1}^{2024} j^{2025-j}\right)$, which is congruent to $v_2(2024!)$ modulo $2$, which turns out to be $2017$, an odd number.

We claim the answer is no. Assume FTSOC it is possible. Note that $(a_1a_2 \cdots a_{2024})^2 = \prod_{i=1}^{2024} i!$. Every integer $i$ appears $2025-i$ times in the product, so odd integers appear an even number of times and even integers appear an odd number of times. Recall Nagura's Theorem: For $n \geq 25$, there is always a prime $p$ such that $n < p < \frac{6n}{5}$. By Nagura's Theorem, there exists a prime $p$ such that $840 < p < 1008$. In the product, this prime appears only in the terms $p$ and $2p$, since $1008 * 2 = 2016 < 2024$ and $840 * 3 = 2520 >2024$. The term $p$ appears an even number of times and the term $2p$ appears an odd number of times in the product, so $v_p(\prod_{i=1}^{2024} i!)$ is odd, which is a contradiction to $\prod_{i=1}^{2024} i!$ being a perfect square.

The answer is no. Suppose ftsoc that it's possible to write $2024$ natural numbers described. Multiplying together the product of every adjacent pair, we see that every number appears twice, so $P = 1!2! \cdots 2024!$ must be a perfect square. However, note that $1009$ is prime, so we can compute $$\nu_{1009}(P) = 1008 \cdot 0 + 1009 \cdot 1 + 7 \cdot 2 \equiv 1 \pmod 2,$$which implies that $P$ cannot be a perfect square, contradiction.

No it is not possible. For it to be true we must have $\prod _{j=1}^{2024}j!$ as a perfect square. After checking, the p-adic value of $2$ turns out to be odd. We are done

TortilloSquad wrote: We claim the answer is no. Assume FTSOC it is possible. Note that $(a_1a_2 \cdots a_{2024})^2 = \prod_{i=1}^{2024} i!$. Every integer $i$ appears $2025-i$ times in the product, so odd integers appear an even number of times and even integers appear an odd number of times. Recall Nagura's Theorem: For $n \geq 25$, there is always a prime $p$ such that $n < p < \frac{6n}{5}$. By Nagura's Theorem, there exists a prime $p$ such that $840 < p < 1008$. In the product, this prime appears only in the terms $p$ and $2p$, since $1008 * 2 = 2016 < 2024$ and $840 * 3 = 2520 >2024$. The term $p$ appears an even number of times and the term $2p$ appears an odd number of times in the product, so $v_p(\prod_{i=1}^{2024} i!)$ is odd, which is a contradiction to $\prod_{i=1}^{2024} i!$ being a perfect square. I did the exact same thing in contest! The funny thing is, I learned of Nagura's Result while doing the PRIMES application pset. I found it pretty amusing in contest

Very silly. If $a_1,a_2, \ldots, a_{2024}$ are the numbers around the circle is \[ \left ( \prod_{k=1}^{2024} a_k \right )^2 = \prod_{k=1}^{2024} a_ka_{k+1} = \prod_{i=1}^{2024} k! \] But, \[ \nu_{47} \left ( \prod_{i=1}^{2024} k! \right ) = 47(1+2+ \cdots +42) + 4 \cdot 43 \]Which is odd. and thus the product can't be a square. $\blacksquare$

Huh The thing we want to prove is that $\prod_{i=1}^{2024} i!$ is not a perfect square. We want the prime to be as large as possible, so we start from the primes close to $\frac{2024}{2}=1012$. Note $1009$ is a prime. $1009\cdot 2=2018$. There are in total $2024-1009+1+2024-2018+1=1023$ $1009$s in the number. $v_{1009}(\prod_{i=1}^{2024}i!)=1023$, so there is no way the number is a square. Probably AMC level ngl

Let, the numbers on the circle are, $a_1,a_2,\cdots,a_{2024}$, Suppose, $b_i! $ are permutation of $1!, \cdots, 2024!$ for $i=1,2, \cdots,2024$, now suppose this is possible then, $a_ja_{j+1}=b_j! $ where, $a_1=a_{2025}$, Then $$b_1!b_3!\cdots b_{2023}!=b_2!b_3!\cdots b_{2024}!=a_1a_2...a_{2024}$$This will imply, $$b_1!b_2!b_3!b_4!\cdots b_{2024}!=(b_2!b_3!\cdots b_{2024}!)^2=K(say) $$Let, $\tau(n)$ be the total number of positive divisors of $n$ then we have, $$\tau(b_1!b_2!b_3!\cdots b_{2024}!)=\tau((b_2!b_4!\cdots b_{2024}!)^{2})$$We know that total number of divisors of perfect square is always odd. Then, $\tau(K)$ is always odd. But, $661$ is prime dividing $$b_1!b_2!\cdots b_{2024}!=1!2!\cdots 2024!=C$$But, $$v_{661}(C)=1\times(1321-661+1)+2\times (1982-1322+1)+3(2024-1983+1)=2109$$but, $\tau(C)$ is divisible by $\tau(661^{2109})=2110$ thus $\tau(C)$ must be even but since $K=C$ and $\tau(K)$ is odd, so contradiction!!! $\blacksquare$

Jane writes down $n$ natural numbers around the perimeter of a circle. She wants the $n$ products of adjacent pairs of numbers to be exactly the set $\{ 1!, 2!, \ldots, n! \}.$.From wich $n$ can she accomplish this? We are going to prove that for $n>=75$ she can not accomplish this. The product of all the products of adjacent pairs is a perfect square, but the product $\prod _{j=1}^{n}j!$ is not. If $n=odd$ we can just select an odd prime number bigger than $\frac{n}{2}$ and we are done. If $n=even$ we want to find an odd prime number belongs to $[\frac{n}{3},\frac{n}{2}]$ and we know that for $a>=25$ there exist a prime belongs to $[a,\frac{6a}{5}]$ so sinse $n>=75$ there exist a prime in $[\frac{n}{3},\frac{6n}{15}]=[\frac{n}{3},\frac{2n}{5}]$ done. Now consider the case $n<=74$ If $n$ is odd $>3$ as before. If $n=3$ we can't. If $n=enen$ then For $58<=n<=74$ we can take $p=29$ For$40<=n<=56$ we can take $p=19$ For $28<=n<=38$ we can get $p=13$ For$n=22,24,26$ we can get $p=11$ For $n=20$ we can get $p=7$ For $n=18,16,14,12,10$ we see $U_5$ For $n=8$ we see $U_2$ For $n=6$ we see $U_3$ For $n=4$ we see $U_2$ For $n=2$ we can get the numbers $1,2$

Thanks fedex (@manteca) for the help Lets say Jane wrote down the numbers $a_1,a_2 \dots a_{2024}$. Let $N$ be such that $N=(a_1\times a_2)(a_2\times a_3)\dots (a_{2024}\times a_1)=a_1^2 a_2^2\dots a_{2024}^2$, then $N$ is a perfect square. Suppose the affirmation of the problem is possible, then $A=1!2!\dots 2024!$ is a perfect square, then all the prime factors of $A$ are elevated to an even number. Lets chose the prime $1009$, there are $(2017-1009+1)+2(2024-2018+1)=1023$ $1009$ factors in $A$, but $1023 \equiv 1 \mod 2$ wich is odd therefore Jane cant accomplish what she wants.

No, she cannot. Let the numbers written around the perimeter of the circle be $x_1, x_2,\dots, x_{2024}$. Consider the product of the $2024$ products: $\prod_{i=1}^{2024} x_ix_{i+1}=\prod_{i=1}^{2024} x_i^2=\prod_{i=1}^{2024} i!$, where $x_{2024}=x_1$. This suggests that $P=\prod_{i=1}^{2024} i!$ must be a perfect square; in other words, in the prime factorization of $P$, all the exponents must be even. But $\prod_{i=1}^{2024} i! = 2024!\prod_{i=1}^{1011} (2i+1)((2i)!)^2$, so clearly, $v_2\left(\frac{P}{2024!}\right)\equiv 0\pmod{2}$ and by Legendre's, $v_2(2024!)=2017$. Therefore, $v_2(P)=v_2\left(\frac{P}{2024!}\right)+v_2(2024!)\equiv 1 \pmod{2}$, meaning that the exponent of $2$ is odd, a contradiction. $\blacksquare$

Since there is $1!$, there are 2 "$1$" on the circle, and they are next to each other. Consider $1013$, which is a prime number. For all $1\leq i\leq 1012,v_{1013}(i!)=0$, and for all $1013\leq i \leq 2024, v_{1013}(i!)=1$. They indicate that there are 1012 numbers which can all be divided by $1013$ and they cannot be next to each other. But it is contradictory to the 2 "$1$" next to each other. Thus Jane cannot accomplish this.$\square$

We need to prove that, $P = \prod_{i=1}^{2024} i! = K^2$ is not possible for any integer $K$ Now $P = \prod_{i=1}^{2024} i! = 2^{1012} \cdot 1012! \cdot C^2$ where $C = \prod_{i=0}^{1011}(2i+1)!$ $\implies 1012! = A^2$ for some integer $A$, which is absurd due to Bertrands postulate i.e. there exists a prime between $ n \geq 2$ and $2n$ Also $v_2(1012!) = 506+253+126+63+31+15+7+3+1 = 1005$ a contradiction

Alright when we take product of all pairs we get (2024)! As a perfect square But when we find power of 2 in (2024)! We will get a odd number

Pretty easy. Suppose (s)he can do it for some natural numbers. Label the numbers as $a_1,a_2,a_3,...,a_{2024}$ $(a_0=a_{2024},a_{2025}=a_1)$ such $a_i$ and $a_{i+1}$ are adjacent $a_ia_{i+1}=i!$ Now $a_1a_2=1!$,$a_2a_3=2!$....$a_{2024}a_1=2024!$ Now multiplying all gives us $(a_1a_2a_3...a_{2024})^2 =1!2!3!...2034!=K$ Now $K$ must be a square number. That means for all $primes$ $p$, $v_p(K)$ is $even$. Now take prime $911$ (Don't get me wrong for choosing $911$ that's what came to my mind) which is a prime. Now $v_{911}(911!),v_{911}(912!),...,v_{911}(1821!)$ are all $1$ where there are odd number of $1$. Now $v_{911}(1822!),v_{911}(1823!),...,v_{911}(2024!)$ are all 2 So $v_{911}(K)$ is odd which is contradiction $\square$

Let the numbers be $a_1, a_2, .... a_{2024}$. they situated on circle in this order . after that $(a_1*a_2*a_3*...*a_{2023}*a_{2024})^2=1!*2!*3!*....*2023!*2024!$ from there we get $1!*2!*3!*....*2023!*2024!$ must be a square. $v_2$ $works$ We can match some numbers like; $2!, 3!$ $==>$ $v_2(2!)=v_2(3!)$ For a generelization $v_2(2k)!=v_2(2k+1)!$ so, $v_2$ will be even ($d*2$ for some d ) only $2024!$ lefts and from legendres formula we find $v_2(2024!)=odd$. $even+odd=odd$ . So a contradiction for being perfect square. Answer is no