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MIT PRIMES/Art of Problem Solving

CROWDMATH 2022: Factorizations in Additive Structures

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Solution to Problem 7.1
Stiffler   2
N Feb 3, 2023 by Stiffler
Hello!! In a recent post Julmath presented an example that solve part (a) of Problem 7.1. Then I will focus in solve part (b).

Solution of part (b)
2 replies
Stiffler
Jan 3, 2023
Stiffler
Feb 3, 2023
Solution to Problem 7.3
julmath   0
Jan 4, 2023
The answer is yes. I split the proof in a proposition, a lemma and a theorem.

Proposition: Let $\alpha $ be a positive algebraic number such that $\mathbb{N}_0[\alpha]$ is atomic. Suppose that $\mathbb{N}_0 [\alpha]$ is not finitely generated. Then $\mathbb{N}_0[\alpha]$ has infinitely many Betti elements.
Proof: We have that $\mathcal{A}(\mathbb{N}_0[\alpha])= \{ \alpha^n, n \in \mathbb{N}_0\}$. Let $(p(x),q(x))$ be the minimal pair of $\alpha$. Consider the set $$N=\{n\in \mathbb{N}\; |\; \exists x \in \mathbb{N}_0[\alpha], n \in \mathsf{L}(x) \land \vert \mathsf{Z}(x)\vert \geq 2\}$$We know that $p(1), q(1) \in N$, then $N$ is a nonempty set of positive integers. Let $m$ be its minimal element and let $x \in \mathbb{N}_0[\alpha]$ such that $m \in \mathsf{L}(x)$ and $\vert \mathsf{Z}(x)\vert \geq 2$. Consider $z \in \mathsf{Z}(x)$ such that $\vert z \vert=m$. We have that $z$ is not an isolated vertex in $\triangledown_x$ if and only in there exists $z' \in \mathsf{Z}(x)$, $z'\neq z$, such that $\gcd(z,z')>0$ . It is not hard to see that $z-\gcd(z,z'),z'-\gcd(z,z')\in \mathsf{Z}(y)$ for some $y \in \mathbb{N}_0[\alpha]$. Then $\vert z-\gcd(z,z')\vert \in N$ and $\vert z-\gcd(z,z')\vert<\vert z \vert$. This contradictis the minimality of $m$ in $N$. Hence $z$ is an isolated vertex in $\triangledown_x$. Then $x$ is a Betti element of $\mathbb{N}0[\alpha]$.
Now, for each $l \in \mathbb{N}$, consider $x_l=\alpha^lx$. It is clear that $m \in \mathsf{L}(x_l)$ and $\mathsf{Z}(x_l)\geq 2$, for all $l \in \mathbb{N}$. Following exactly the same reasoning as before, we obtain that there exists an isolated vertex in $\triangledown_{x_l}$. Therefore $x_l$ is a Betti element of $\mathbb{N}_0[\alpha]$ for each $l \in \mathbb{N}$. This concludes our proof.

Lemma: Let $V\subset \mathbb{N}^k$. We say that, for $v=(v_1,...,v_k)\in V$ and $v'=(v_1',...,v_k')\in V$, $v<v'$ if $v_i\leq v_i'$ for all $i=1,...,k$ and $v_j<v_j'$ for some $j=1,...,k$. Then, the set:
$$W=\{ v\in V: \nexists v' \in V , v'<v\}$$has only finitely many elements.

Proof: Let us proceed by induction on $k$. If $k=1$, we have that $V\subset \mathbb{N}$, and by the well-ordering theorem we have that $V$ has a minimal element $m$. It follows that $W=\{m\}$. Now, assume the Lemma is valid for $k=l$. Let $V\subset \mathbb{N}^{l+1}$. Consider the set:
$$ M=\{n \in \mathbb{N}: \exists v \in V, n= \max \{v_i\}\} $$Fix $n_0 \in M$, and $v \in V$ such that $n_0=\max \{v_i\}$. Then for every $v'\in V$ such that $\min\{v'_i\}>n_0$, we have that $v<v'$ and $v'\notin W$. Now, let $v'\in V$ such that $\min \{v_i'\} \leq n_0$. Then we have that for some $j\in [1,l+1]$ and some $s\in [1,n_0]$, $v'\in V_{j,s}$, where
$$V_{j,s}=\{ v \in V: v_j=\min\{v_i\}=s\}$$Let $f:V_{j,s}\rightarrow \mathbb{N}^l$, such that $f(v_1,...,v_{l+1})=(v_1,...v_{j-1},v_{j+1},...,v_{l+1})$. It is clear that for $x,y\in V_{j,s}$, $x<y\Leftrightarrow f(x)<f(y)$. Applying the lemma in $\mathsf{Im} f \subset \mathbb{N}^l$, we obtain that the set
$$W_{j,s}=\{v \in V_{j,s}: \nexists v' \in V_{j,s}, v'<v\}$$has only finitely many elements.
Since $W \subset \displaystyle\cup_{j,s} W_{j,s}$ we obtain that $W$ also has only finitely many elements. Our proof concludes

Theorem: Let $\alpha $ be a positive algebraic number such that $\mathbb{N}_0[\alpha]$ is atomic and finitely generated. Then $\mathbb{N}_0[\alpha]$ has only finitely many Betti elements.
Proof: Let $m_\alpha$ be the minimal polynomial of $\alpha$, $n+1=\vert \mathcal{A}(\mathbb{N}_0[\alpha])\vert$. Suppose that $b$ is a Betti element of $\mathbb{N}_0[\alpha]$. Then, there exist $z,z'\in \mathsf{Z}(b)$ such that $z$ and $z'$ are not in the same connected component in $\bigtriangledown_b$. We can say that $z=a(\alpha)$, $z'=b(\alpha)$, where $a(x), b(x) \in \mathbb{N}[x]$. It is known that $a(x)-b(x)=m_\alpha(x)r(x)$, for some $r\in \mathbb{Z}[x]$. In this case we will say that $m_\alpha(x)r(x)$ produces a Betti element.
Consider the set $P=\{ t(x)\in \mathbb{Z}[x]: m_\alpha(x) \vert t(x), \mathsf{gr}(t)\leq n\}$. For $g(x)=g_nx^n+...+g_0 \in P$ and $h(x)=h_nx^n+...+h_0 \in P$ we say that $gRh$ if and only if for each $i\in [0,n]$, $p_ih_i>0$ or $p_i=h_i=0$.
It is not hard to see that $R$ is an equivalence relation that has finitely many equivalence classes. Let $C$ be an equivalence class of $R$. Because of the definition of $R$ it is not hard to see that for some $k \in \mathbb{N}$ and $i_1,...,i_k \in [0,n]$, there exists a map $f:C\rightarrow \mathbb{N}^k$ such that for $t(x)=t_nx^n+...+t_0 \in C$, $f(t)=(t_{i_1},...,t_{i_k})$ where $\{t_{i_1},...,t_{i_k}\}$ is the set of positive coeficients of $t$. Now, note that if we have that for $t(x),s(x)\in C$, $f(s)<f(t)$ then
$$t^+(x)-s(x)=t^+(x)-s^+(x)+s^-(x)=T(x)\in \mathbb{N}[x] $$It follows that there exists $x\in \mathbb{N}_0[\alpha]$ such that $$z=t^+(\alpha)\in \mathsf{Z}(x)$$, $$z'=t^+(\alpha)-s^+(\alpha)+s^-(\alpha) \in \mathsf{Z}(x)$$, $$z''=t^-(\alpha)\in \mathsf{Z}(x)$$. Note that $z'$ connects $z$ and $z''$ in $\bigtriangledown_x$. Then $t(x)$ does not produce a Betti element. Using Lemma 3 on $\mathsf{Im}f$ it follows that only finitely many polynomials in $C$ produce a Betti element. Since there are only finitely many equivalence classes in $R$, we conclude that $\mathbb{N}_0[\alpha]$ must have only finitely many Betti elements.
0 replies
julmath
Jan 4, 2023
0 replies
Till when is This years CrowdMath active ?
Rolo123   1
N Jan 2, 2023 by felixgotti
I just joined CrowdMath today, Can I still participate or should I start from next year ?
1 reply
Rolo123
Nov 20, 2022
felixgotti
Jan 2, 2023
Problem 4
julmath   1
N Jan 2, 2023 by felixgotti
Hello! Here I share an example where I am able to find the catenary degree of $\mathbb{N}_0[\alpha]$.
Suppose that the minimal pair of $\alpha>1$ is $(p(x),n)$, where $p(1)<n$ and $p$ is not monic. It is not hard to see that $\mathbb{N}0[\alpha]$ is atomic and $\mathcal{A}(\mathbb{N}0[\alpha])= \{\alpha^n: n \in \mathbb{N}0\}$. Let $x \in \mathbb{N}0[\alpha]$, and $z \in \mathsf{Z}(x)$, $z=\sum_{i=0}^l a_i\alpha^i$. If $a_j\geq n$ for some $j \in [ 0,l] $ then there exists $z_1 \in \mathsf{Z}(x)$ given by
$$z_1=\sum_{i=0}^{j-1}a_i\alpha^i+(a_j-n)\alpha^j+p(\alpha)\alpha^j+\sum_{i=j+1}^la_i\alpha^i $$and $\vert z_1\vert < \vert z \vert$. Also $d(z,z_1)=n$. Following this procedure we can create a finite sequence $z=z_0, z_1,z_2,...,z_k$ such that $d(z_i,z_{i+1})=n$ and $z_k=\sum_{i=0}^mb_i\alpha^i$ satisfies that $b_i<n$ for all $i \in [ 0,m ]$.
Now, assume that $z', z'' \in \mathsf{Z}(x)$, $z'\neq z''$ such that $z'=\sum_{i=0}^sc_i\alpha^i$, $z''=\sum_{i=0}^{s}d_i\alpha^i$ and $c_i,d_i\in [ 0,n-1 ] $ for all $i \in [b 0,s]$. Then
$$\sum_{i=0}^s(c_i-d_i)x^i=[p(x)-n]r(x)$$
where $r\in \mathbb{Z}[x]$. It is clear that $\vert c_i-d_i \vert <n $ for all $i \in [ 0,s]$, but the coefficient of the monomial of least degree in the right side of (1) is a multiple of $n$. This is a contradiction. Then there exists a unique factorization $z^0 \in \mathsf{Z}(x)$ such that every atom appears in $z^0$ less than $n$ times. Also note that $z^0$ is the factorization of minimal length of $x$. We have proved that for every $x \in \mathbb{N}0[\alpha]$ and $z\in \mathsf{Z}(x)$ there exists a n-chain of factorizations that connects $z$ and the unique factorization of minimal length of $x$. It follows that $c(\mathbb{N}0[\alpha])\leq n$. Following the same reasoning as before, we can prove that there is no $z\in \mathsf{Z}(p(\alpha))$ such that $d(z,n\cdot 1)<n$, so $c(p(\alpha))=n$. Then $c(\mathbb{N}_0[\alpha])=n$.
1 reply
julmath
Dec 2, 2022
felixgotti
Jan 2, 2023
Proposition Problem 4
Stiffler   1
N Jan 2, 2023 by felixgotti
In the solution of open problem $5$ Banghenz found and $\alpha > 1$ such that $\mathbb{N}_0[\alpha]$ is a non-FGM monoid with $\mathsf{c}(\mathbb{N}_0[\alpha]) = \infty$. In a recent post Julmath presents an example, again with $\alpha > 1$, where $\mathsf{c}(\mathbb{N}_0[\alpha]) < \infty$. Here I address a result that I found that contrasts with their examples since it focuses in the case $\alpha < 1$.

Proposition: Let $n > 2$ be a positive integer. For every $d > 1$ there exists a non-FGM rank-d monoid $\mathbb{N}_0[\alpha]$ with $\alpha < 1$ and $\mathsf{c}(x) = n$ for every $x \in M$.

Proof
1 reply
Stiffler
Dec 2, 2022
felixgotti
Jan 2, 2023
Problem 7.2 solution
gourmet_salad   0
Jan 2, 2023
This can be viewed as an extension of existing results in the paper "Factorization invariants of Puiseux monoids generated by geometric sequences" (arxiv: https://arxiv.org/abs/1904.00219). In that paper, three results are shown about $M = \mathbb{N}_0[q]$, for rational $q = a/b$:
Lemma 1.
- $M$ is atomic iff $q$ is not the reciprocal of any integer
- If $M$ is atomic, the Betti elements of $M$ are $$\left\{ \frac{\mathsf{n}(q)^{m+1}}{\mathsf{d}(q)^m}\ \Big|\ m \in \mathbb{N}_0\right\}$$- If $M$ is atomic, the set of lengths of any element is an arithmetic progression with difference $a - b$


Lemma 2. Let $q$ be a positive rational, and let $n \geq 2$ such that $\sqrt[n]{q}$ is an irreducible n-th root of $q$. Then the polynomial $x^n - q$ is irreducible in $\mathbb{Q}$.
Proof: This is seen in the textbook Algebra by S. Lang, in page 297.

Corollary. Let $q$ be a positive rational, and let $\alpha \coloneqq \sqrt[n]{q}$ be a positive irreducible n-th root of $q$. Then $1, \alpha, \alpha^2, \dots, \alpha ^{n-1}$ are linearly independent in $\mathbb{Q}$.
Proof: The proof follows from Lemma 2; If these powers of $\alpha$ were not linearly independent, the polynomial $x^n - q$ (whose unique positive real root is $\alpha$) would be reducible in $\mathbb{Q}$.

Claim. Let $q$ be a positive rational that is not the reciprocal of any integer, and let $\alpha \coloneqq \sqrt[n]{q}$ be a positive irreducible n-th root of $q$. Let $f: \mathsf{Z}(\mathbb{N}_0[q])^n \rightarrow \mathsf{Z}(\mathbb{N}_0[\alpha])$ be the map defined by $f(z_1, z_2, \dots, z_n) = z_1 + \alpha z_2 + \alpha^2z_3 + \dots + \alpha^{n-1}z_n$. Then the following statements hold:
$(i)$ $f$ is an isomorphism.
$(ii)$ If $\pi(f(z_1, z_2, \dots, z_n)) = \pi(f(z_1', z_2', \dots, z_n'))$, then $\pi(z_i) = \pi(z_i')$ for all $i \in [1, n ]$. ($\pi$ is the isomorphism mapping a collection of atoms to the element they form.)
Proof. $(i)$ It's easy to check that $f$ is additive and maps the identity to itself, and thus it is a homomorphism. To show that it is an isomorphism, we will show that every factorization in $\mathsf{Z}(\mathbb{N}_0[\alpha])$ corresponds to a unique $n$-tuple of factorizations in $\mathsf{Z}(\mathbb{N}_0[q])^n$, showing both surjectivity and injectivity. Since $\mathbb{N}_0[\alpha]$ is generated by $\{\alpha^k \mid k \in \mathbb{N}_0\}$, its set of atoms must be a subset of this generating set. Moreover, as $\alpha^n =  q$, any atom $\alpha^{k'}$ can be written in the form $\alpha^rq^k$, where $k' = kn + r$, $k \geq 0$, and $r \in [0, n-1]$. Thus, any factorization in $z \in \mathsf{Z}(\mathbb{N}_0[\alpha])$ can be written as $$z = \sum_{r = 0}^{n-1}\sum_{k=0}^{\infty}c_{r,k}\alpha^rq^k$$For some coefficients $c_{r,k}\in\mathbb{N}_0$. This can be rewritten further as
$$z=\sum_{r = 0}^{n-1}\alpha^r(\sum_{k=0}^{\infty}c_{r,k}q^k)$$Lemma 1 shows that the set of atoms of $\mathbb{N}_0[q]$ is its generating set. Combining this with the previous equation, it's clear to see that the unique $n$-tuple of factorizations in $\mathsf{Z}(\mathbb{N}_0[q])^n$ that maps to $z$ is $(\sum_{k=0}^{\infty}c_{0,k}q^k, \sum_{k=0}^{\infty}c_{1,k}q^k, \dots, \sum_{k=0}^{\infty}c_{n-1,k}q^k)$. Thus, $f$ is an isomorphism.
$(ii)$ $\pi(f(z_1, z_2, \dots, z_n)) = \pi(f(z_1', z_2', \dots, z_n'))$ implies that $\pi(z_1) + \alpha \pi(z_2) + \dots + \alpha^{n-1}\pi(z_n) = \pi(z_1') + \alpha \pi(z_2') + \dots + \alpha^{n-1}\pi(z_n')$, and the linear independence shown in Corollary 1 shows that $\pi(z_i) = \pi(z_i')$ for all $i \in [1, n]$.

Corollary 1. Let $q$ be a positive rational that is not the reciprocal of any integer, and let $\alpha \coloneqq \sqrt[n]{q}$ be a positive irreducible n-th root of $q$. The set of atoms of $\mathbb{N}_0[\alpha]$ is its generating set, $\{\alpha^k \mid k \in \mathbb{N}_0\}$.
Proof: Any element that is generated by at least two generating elements can not be an atom, so $\mathcal{A}(N_0[\alpha]) \subseteq \{\alpha^k \mid k \in \mathbb{N}_0\}$. On the other hand, by the isomorphism $f$, the factorizations of $\alpha^{k'} = \alpha^rq^k$ (where $k' = kn+r$) are isomorphic to the factorizations of the element $(0, 0, \dots, 0, q^k, 0, \dots, 0)$ in $\mathbb{N}_0[q]^n$, where $q^k$ is at the $r$-th index. As shown in Lemma 1, $q^k$ is an atom of $N_0[q]$, thus $\alpha^{k'}$ must be an atom, hence $\mathcal{A}(N_0[\alpha]) = \{\alpha^k \mid k \in \mathbb{N}_0\}$.

Corollary 2.Let $q$ be a positive rational that is not the reciprocal of any integer, and let $\alpha \coloneqq \sqrt[n]{q}$ be a positive irreducible n-th root of $q$. The set of Betti elements of $\mathbb{N}_0[\alpha]$ is $$\left\{ \frac{\mathsf{n}(q)^{m+1}}{\mathsf{d}(q)^m}\alpha^r\ \Big|\ m \in \mathbb{N}_0\ \text{and}\ r \in [ 0, n-1 ]\right\}.$$Proof: For a fixed $r \in [0, n-1]$, we identify all Betti elements of the form $t\alpha^r$, where $t \in \mathbb{Q}_{>0}$. By the mapping $f$, the factorizations of $t\alpha^r$ are isomorphic to the factorizations of the element $(0, 0, \dots, 0, t, 0, \dots, 0)$ in $\mathbb{N}_0[q]^n$, where $t$ is at the $r$-th index. [Previous result] shows that the only Betti elements of $\mathbb{N}_0[q]$ are $\left\{ \frac{\mathsf{n}(q)^{m+1}}{\mathsf{d}(q)^m}\ \Big|\ m \in \mathbb{N}_0 \right\}$. Thus, by isomorphism, the only Betti elements of this form are $$\left\{ \frac{\mathsf{n}(q)^{m+1}}{\mathsf{d}(q)^m}\alpha^r\ \Big|\ m \in \mathbb{N}_0\right\}$$
Lastly, we prove that all elements of $\mathbb{N}_0[\alpha]$ that involve at least two different (linearly independent) powers of $\alpha$ can not be Betti elements. Let $a = t_0 + t_1\alpha + t_2\alpha^2 + \dots + t_{n-1}\alpha^{n-1}$ be such an element, where $t_i \in \mathbb{Q}_{\geq0}$, and at least two of the $t_i$'s are nonzero. By the mapping $f$, the factorizations of $a$ are isomorphic to the factorizations of $(t_0, t_1, \dots, t_{n-1})$ in $\mathbb{N}_0[q]^n$. Let $z$ and $z'$ be two arbitrary factorizations of $a$, and let $(z_0, z_1, \dots, z_{n-1})$ and $(z_0', z_1', \dots, z_{n-1}')$ be the $n$-tuples of factorizations corresponding to them by $f$. $\pi(z) = \pi(z') = a$, and so by property $(ii)$ in Claim 1, $\pi(z_i) = \pi(z_i')$ for all $i$. As a result, $(z_0, z_1, z_2 \dots, z_{n-1}) \rightarrow (z'_0, z_1, z_2 \dots, z_{n-1}) \rightarrow (z'_0, z'_1, z_2 \dots, z_{n-1}) \rightarrow \dots \rightarrow (z'_0, z'_1, z'_2, \dots, z'_{n_1})$ is a sequence consisting of factorizations of $(t_0, t_1, \dots, t_{n-1})$. Moreover, since at least two of these factorizations are non-zero, each two consecutive factorizations have non-zero greatest common divisor, and so this is a path within the factorization graph $\mathcal{G}((t_0, t_1, \dots, t_{n-1}))$. Therefore, $(t_0, t_1, \dots, t_{n-1})$ is not a Betti element, and by isomorphism, $a$ is not a Betti element.

Corollary 3. Let $q$ be a positive rational that is not the reciprocal of any integer, and let $\alpha \coloneqq \sqrt[n]{q}$ be a positive irreducible n-th root of $q$. For any $a \in \mathbb{N}_0[\alpha]$, $\mathsf{L}(a)$ is an arithmetic progression with difference $|\mathsf{n}(q) - \mathsf{d}(q)|$.
Proof: Let $a = t_0 + t_1\alpha + t_2\alpha^2 + \dots + t_{n-1}\alpha^{n-1}$, where $t_i \in \mathbb{Q}_{\geq0}$ for all $i$. By the mapping $f$, the factorizations of $a$ are isomorphic to the factorizations of $T = (t_0, t_1, \dots, t_{n-1})$ in $\mathbb{N}_0[q]^n$. The length of a certain factorization $(z_0, z_1, \dots, z_{n-1})$ of $T$ is equal to $\sum|z_i|$. Let $d = |\mathsf{n}(q) - \mathsf{d}(q)|$. From Lemma 1, the length of the factorization $z_i$ of $t_i$ can take any value in $\{\min\mathsf{L}(t_i) + dk | k \in [0, \frac{\max\mathsf{L}(t_i) - \min\mathsf{L}(t_i)}{d}]\}$. Moreover, since the terms of the $n$-tuple are independent, $\min\mathsf{L}(T) = \sum\min\mathsf{L}(t_i)$. Thus, $|(z_0, z_1, \dots, z_{n-1})|$ can take any value in

$$\{\min\mathsf{L}(T) + dk | k \in [0, \frac{\max\mathsf{L}(T) - \min\mathsf{L}(T)}{d}]\}$$.

Hence, $\mathsf{L}(T)$ is an arithmetic progression with difference $d$, and by isomorphism, the same applies to $\mathsf{L}(a)$.

Please let me know if anything is unclear!
0 replies
gourmet_salad
Jan 2, 2023
0 replies
The last problem of CrowdMath 2022 has been posted!
felixgotti   0
Jan 1, 2023
Hi everyone! Check out the tab of problems. The last problem of CrowdMath 2022 (Problems 7), which has several parts, has just been posted. The problem is aimed to better understand the sets of lengths, catenary degrees, and the sets of Betti elements of the monoids we are studying in our research project. There is no resource corresponding to this last problem, but I have stated the new and needed definitions as part of the problem statement. Feel free of course to post any question you may have. Enjoy this last problem, and Happy New Year!!!
0 replies
felixgotti
Jan 1, 2023
0 replies
Primes in polynomials
FelJac   2
N Dec 25, 2022 by FelJac
Are there any polynomial of degree higher than one (e.g., n²+1), that contains infinitely many prime numbers?
Do any of you have any idea how we might prove this for a certain polynomial. This is an open problem, and a seemingly very difficult one. All thoughts about partial progres or so, would be very interesting to discuss. (Sorry for bad English).
2 replies
FelJac
Aug 20, 2022
FelJac
Dec 25, 2022
Problem 6 Solution
bangzheng   3
N Dec 9, 2022 by Stiffler
Fix $\alpha \in \mathbb{R}_{> 0}$. Prove (or maybe disprove) that $\mathbb{N}_0[\alpha]$ is not $k$-furcus for any $k \in \mathbb{N}$.

Solution
3 replies
bangzheng
May 9, 2022
Stiffler
Dec 9, 2022
Can I join Crowdmath?
Red_Fox_1293   5
N Oct 6, 2022 by BabaLama
Hello all.
I am currently in eighth grade doing calculus. I have participated in many aops classes in the past, and I feel like I could contribute nicely to crowdmath. However, I see that people under highschool need pre-approval to participate in crowdmath. How do I get that, and can I participate in crowdmath? Any help is welcome.
Thanks,
-Red_Fox_1293
5 replies
Red_Fox_1293
Feb 24, 2022
BabaLama
Oct 6, 2022
Till when is This years CrowdMath active ?
Rolo123   1
N Jan 2, 2023 by felixgotti
I just joined CrowdMath today, Can I still participate or should I start from next year ?
1 reply
Rolo123
Nov 20, 2022
felixgotti
Jan 2, 2023
Till when is This years CrowdMath active ?
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Rolo123
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