This was a caseworky problem. However, I think the following method makes it *relatively* nice. First of all, notice that all numbers with prime factors at most 3 can be expressed as $2^{a}3^{b}$. Notice that the product of three such numbers is a cube if and only if the exponents on the 2 and the 3 are both divisible by 3. Therefore the problem is equivalent to finding three elements out of a set X of nine (not necessarily distinct) in $\mathbb{Z}_{3}^{2}$ whose sum is $(0, 0)$, where addition is defined by adding the two corresponding coordinates in $\mathbb{Z}_{3}^{2}$ (i.e. vector addition). Note that if there were four or fewer distinct elements in X, then some three elements are identical, and these three will obviously sum to $(0, 0)$. Thus it would suffice to prove that in any 5-element subset Y (this time with distinct elements) of $\mathbb{Z}_{3}^{2}$, there are some three which sum to $(0, 0)$. Now let $f(a, b) =-a-b$, and note that $f(a, b) \not\in \{a,b\}$ and $f(a, b)$ is the unique solution to $x \in \mathbb{Z}_{3}^{2}$ such that $a+b+x = (0, 0)$. There are ${5 \choose 2}= 10$ pairs of elements in Y, and for any $a, b \in Y$, we would be done if $f(a, b) \in Y$. Hence we will assume the contrary (f(a, b) is never in Y), in which case the 10 pairs of elements map to 4 elements not in Y. By pigeonhole, this means that there are some three distinct pairs (but we don't know if their components are distinct) $(x_{1}, x_{2}), (x_{3}, x_{4}), (x_{5}, x_{6}) \in Y^{2}$ for which $f(x_{1}, x_{2}) = f(x_{3}, x_{4}) = f(x_{5}, x_{6}) = t$ for some $t \not\in Y$. Of course, $x_{1}\ne x_{2}, x_{3}\ne x_{4}, x_{5}\ne x_{6}$, since we only considered pairs. Then suppose some two of the $x_{i}$ are equal; WLOG $x_{1}= x_{3}$. But since $-x_{1}-x_{2}= t =-x_{3}-x_{4}$, we have that $x_{2}= x_{4}$. This means that the pairs $(x_{1}, x_{2})$ and $(x_{3}, x_{4})$ are identical, which is contradiction. The alternative, however, is that all the $x_{i}$ are distinct, implying that Y contains six elements, which is also contradiction. Therefore, there must be three elements of Y whose sum is zero, as desired.

I agree that this problem has many solutions, but I was very happy when I came up with this one, so I have to post it here The beginning is the same. So we have these possibilities for exponents $(0,0),(0,1),(1,0),(1,1),(1,2),(20),(2,1),(2,2)$. Now put them in the fields of $3\times 3$ square like this \[(0,1)...(2,1)...(1,1) \] \[(1,2)...(0,2)...(2,2) \] \[(2,0)...(1,0)...(0,0) \] Now, notice that if we take any three numbers from the sam row, column or diagonal, then their sum is divisible with three. I don't think big diagonals only, if you look more cearfull, you can se that sum of $a[1,2],a[2,1],a[3,3]$ is also divisible with 3, and for others "broken" diagonal too. We want to choose 9 numbers from there such that: we don't choose one number three times, we don't take three numbers form the same row or column, and the additional condition, we don't take three numbers from any "diagonal". Now, just to fulfill first two conditions, we see that we must take $5$ or $6$ different fields (if we take less then $5$ fields then we must take one field three times, on the other hand, if we take more than $6$ fields, we will surtanly have one whole row or column). Now it's not hard to deduce that third condition can never be fulfilled if the first two are. Nice isn't it Bye

Nice, but how do you prove it neatly? I thought about it that way during the contest, but it seemed very hard/annoying to justify on paper.

I have totaly forgotten on this problem. It's really isn't so hard to write all cases. Just look at corner fields that we picked. If we pick only one corner field, we must pick 4 no-corner fields, and this case isn't hard to deduce. Now, if we pick 2 corner fields, then there are two sub-cases, that really isn't hard and long to write. Three and four corner fields are also easy to write all down. Maybe it isn't so neatly, but I really liked the idea of that $3\times 3$ table Bye

I thought the whole 3x3 grid thing was really nice when I first found it, but it took quite a while to finish after that.

This result (that on a 3-by-3 grid, any five points determine a generalized line) is well-known to players of the card game SET. If we'd like to phrase it in terms suitable for possible generalization, we're looking at the problem of choosing the maximal number of points in the vector space $\mathbf{F}_3^2$ over the field $\mathbf{F}_3$ with 3 elements that does not contain an entire (affine) line. I believe that when we replace "2" with "5" in the statement already the corresponding answer is unknown.

As mentioned previously, the problem quickly solves if it can be shown that a set, $X$, of five distinct elements in $\mathbb{Z}_{3}^{2}$ must have three distinct elements that sum to zero. Let $S_0$ be the subset of elements of the form $(0,a)$, $S_1,S_2$ defined similarly. Then if any of $S_1,S_2,S_3$ have three elements, we are clearly finished. If not, one $S_i$ has one element, the other two each. So WLOG assume the elements of $X$ are $(0,a),(1,b),(1,c),(2,d),(2,e)$. The equations $a+b+x \not\equiv 0, a+c+x \not\equiv 0$ have only one solution in $\mathbb{Z}_{3}$, so we are done.

let the 9 numbers be $a_1,a_2,....,a_9$ Let $a_n=2^{p_n}.3^{q_n}$ math=inline]$p_n ,q_n \ge 0$[/math So, let $X={(p_n,q_n):n=1,2,...,9}$ . as $n(X)=4.3-3$ , hence by using a well known result from additive combinatorics, we deduce that , there exists $i,j,k$ such that $(p_i+p_j+p_k,q_i+q_j+q_k)=(0,0) (mod3)$ Hence $a_{i}a_{j}a_{k}$ is a perfect cube

mathbuzz wrote: by using a well known result from additive combinatorics In other words, "the result of the problem follows immediately if we take the statement of the problem as known."

I have a solution that borders on pigeonhole and blind casework bashing: Let $(p_i, q_i)$ denote the exponents of 2 and 3 in the prime factorization of the $i^{th}$ number respectively. Let $(a_i, b_i)$ be the residues of $p_i$ and $q_i$ in modulo three respectively. Also let $r_0, r_1, r_2$ be partitions of A = {$a_i$} and B = {$b_i$} that divide the elements of A and B into the residues corresponding to their indices in modulo three. It suffices to prove the existence of a triple (i, j, k) for which $a_i + a_j + a_k \equiv b_i + b_j + b_k \equiv 0$ (mod 3). We split the problem into two cases: Case 1: $\exists r_i$ that contains at least 5 elements of A. If there exists three $b_i$ corresponding to A's elements in $r_i$ in the same residue group, then the sum of those three $b_i$ is congruent to 0 mod 3. Multiplying their corresponding numbers yields a power of three. Else, by pigeonhole, each residue must contain at least one $b_i$. Summing any three $b_i$, taking one from each residue group, gives us a multiple of three. Hence, we can multiply their corresponding numbers to get a power of three. Case 2: There do not exist 5 elements of A that belong to a single residue group. By pigeonhole, there are at least two residue groups that have at least three elements, and each residue group must have at least one element from A. Suppose on the contrary that there do not exist three $b_i$ in such a configuration that sum to a multiple of three. WLOG, let $a_1, a_2, a_3$ belong to residue group $r_0$, $a_4, a_5, a_6$ belong to residue group $r_1$, and $a_7$ belong to residue group $r_2$. We let D = {$b_1, b_2, b_3$}, E = {$b_4, b_5, b_6$} and F = {$b_7$}. Note that $b_1, b_4, b_7$ cannot all have different residues or the same residues. Hence (wlog), it follows that $b_7 \equiv X$ (mod 3), $b_1 \equiv X$ (mod 3), and $b_4 \equiv Y$ (mod 3). Also note that group Ds and E cannot all have the same residues. It follows that group E must have the residues Y and Z among its elements. Similarly, group D must have the residues X and Y among its elements. Since $b_7 \equiv X$, we can choose three elements among D, E, F that have three distinct residues. Summing these elements gives us a multiple of three, and multiplying their corresponding numbers gives a power of three; contradiction. Hence, we can multiply three numbers to obtain a power of three.

N.T.TUAN wrote: Let $S$ be a set of $9$ distinct integers all of whose prime factors are at most $3.$ Prove that $S$ contains $3$ distinct integers such that their product is a perfect cube. that reminds me with some hwo to https://artofproblemsolving.com/community/c6h60788p366595

hmida99 wrote: N.T.TUAN wrote: Let $S$ be a set of $9$ distinct integers all of whose prime factors are at most $3.$ Prove that $S$ contains $3$ distinct integers such that their product is a perfect cube. that reminds me with some hwo to https://artofproblemsolving.com/community/c6h60788p366595 Wait I was going to refer to that problem after I posted my solution... What a coincidence.

It remind me somewhere?? This type is fimiliar

mathbuzz wrote: let the 9 numbers be $a_1,a_2,....,a_9$ Let $a_n=2^{p_n}.3^{q_n}$ [$p_n ,q_n \ge 0$] So, let $X={(p_n,q_n):n=1,2,...,9}$ . as $n(X)=4.3-3$ , hence by using a well known result from additive combinatorics, we deduce that , there exists $i,j,k$ such that $(p_i+p_j+p_k,q_i+q_j+q_k)=(0,0) (mod3)$ Hence $a_{i}a_{j}a_{k}$ is a perfect cube My solution was the same, I think this one is really elegant.

Can someone pls check my solution? Each number here can be represented with prime factors of 2 & 3.So, we have to find powers of 2 as well as 3, which add up to a multiple of 3 separately.Let the powers of 2 of those 3 requested numbers be a,b & c.So,if we take a+b+c modulo 3, one of the following four patterns must hold- 0+0+0 1+1+1 2+2+2 1+2+0 pigeonhole principle says that one of 0,1 or 2 must occur at least thrice among the powers of 2 of those 9 numbers.So,some casework suggests that there are minimum 14 ways to choose 3 numbers out of the nine numbers such that the sum of the powers of 2 of them is 0 mod3.now we need to ensure that at least one of those ways also satisfies the same for powers of 3.In case of the minimum number of ways, there are at least 2 ways satisfying this,using pigeonhole.The others also satisfy in the same way. Is the solution OK?I am a beginner.so,pls consider if there's any silly mistake.

Not sure if the same answer isn't already written. Some solutions looked similar, though I didn't exactly compare. Anyway, I'll go on with mine. all the numbers in $S$ are in the form of $2^{m}3^{n} \quad m, n \in \mathbb{N} \cup \{0\}$. So obviously, the product of 3 of them is in the same form. For a number of this form to be a perfect cube, it is equivalent to $ 3 \mid m, n$ Therefore, the problem changes to having a set $X$ which contains 9 distinct of $(m,n) \quad m, n \in \mathbb{N} \cup \{0\}$ and we want to prove that 3 of these pairs exist such that their total is $(0,0) \bmod 3$. We will work with $m, n \bmod 3$ for the elements of X. So we define a set $Y$, which contains all the pairs in $X$ $\bmod 3$. $(m,n)$ can have one of these 9 values $\bmod 3$: $(0,0), (0,1), (0,2), (1,0), (1,1), (1,2), (2,0), (2,1), (2,2)$. We can organize them like this: $$ A = \{(0,0), (0,1), (0,2)\} \quad B = \{(1,0), (1,1), (1,2)\} \quad C = \{(2,0), (2,1), (2,2)\}$$From here, we are going to solve the problem using proof with contradiction. We suppose that no 3 pairs are in $X$ such that their total is $(0,0) \bmod 3$. So, 2 of the pairs in $A$ can be in $Y$ at most, because the total of the 3 of them is $(0,0) \bmod 3$. Similarly, at most 2 of the pairs in $B$ and 2 of the pairs in $C$ can be in $Y$. So we have $|Y| \leq 6$. On the other hand, $|Y|$ can't be 4 or less, because if it were, by the pigeonhole principle one pair would be repeated at least $\lceil \frac{9}{4} \rceil = 3$ times (Note that we are discussing repeat $\bmod 3$, so the distinctness of the elements of $X$ is not under question.), so the three pairs that are all equal $\bmod 3$, have a total that is $(0,0) \bmod 3$. So $|Y|\geq 5$. Now that we have $6 \geq |Y|\geq 5$, we know that 2 of the three sets $A, B, C$ have 2 elements in common with$Y$ and the other one has at least one. WLOG, let two of the elements in $A, B$ be in $Y$ and $C$ be the set that has at least one intersection with $Y$. Name the $A \cap Y$ elements $k,l$ and (one of) the $C \cap Y$ element(s), $t$. $k+t$ prohibits a member of B, let's call it $p$, to be in $Y$. Also, $l+t$ prohibits a member of B, we name it $q$, to be in Y, and we know $q \neq p$. So one element of B can be in Y at most, but we stated that $|B \cap Y| = 2$, contradiction. Hence, the proof is complete. $ \blacksquare $ Please mention if I've made a mistake in this proof, thanks