I think it's 6, i might be wrong Becasue with 7 we can said that a triplete will have the same color. Still I have a doubt

JJ546655 wrote: I think it's 6, i might be wrong There's an example for $9$.

Here is an example for $9$. We represent each card by a vector $v\in\{0, 1, 2\}^3$, where each coordinate is one of the attributes (shape, color, number) and $0, 1, 2$ are the three different possibilities for each attribute. \begin{align*} v_1 &=(2, 2, 1) \\ v_2 &=(2, 1, 1) \\ v_3 &=(2, 0, 0) \\ v_4 &=(1, 2, 1) \\ v_5 &=(1, 1, 2) \\ v_6 &=(1, 0, 2) \\ v_7 &=(0, 2, 2) \\ v_8 &=(0, 1, 1) \\ v_9 &=(0, 0, 2) \end{align*}

Wrong construction. Here is a construction for $n=10$. \begin{align*} &(0,0,0) \\ &(0,0,1) \\ &(0,1,2) \\ &(0,2,2) \\ &(1,0,2) \\ &(2,0,2) \\ &(1,1,1) \\ &(2,2,0) \\ &(1,2,0) \\ &(2,1,1) \end{align*}Suppose that there is an example for $n\ge 11$. So there are $2$ triples $(x,y,z)$ which have the same coordinates $x,y$. WLOG $(x,y)=(0,0)$ and the $z$ coordinates are $0,1$. So these two triples are $(0,0,0),(0,0,1)$. Then there are at most $2$ triples with $(x,y)=(0,1)$ or $(0,2)$. Also, for $(x,y)=(1,0),(2,0)/(1,1),(2,2)/(1,2),(2,1)$. Thus there are at most $2+2+2+2+2=10$ triples. Contradicts to $n\ge 11$. So the answer is $10$.

DimensionX wrote: Here is a construction for $n=10$. \begin{align*} &(0,0,0) \\ &(0,0,1) \\ &(0,1,2) \\ &(0,2,2) \\ &(1,0,2) \\ &(2,0,2) \\ &(1,1,1) \\ &(2,2,0) \\ &(1,2,0) \\ &(2,1,1) \end{align*} $(1,0,2)+(1,1,1)+(1,2,0)=(0,0,0)$

bgn wrote: DimensionX wrote: Here is a construction for $n=10$. \begin{align*} &(0,0,0) \\ &(0,0,1) \\ &(0,1,2) \\ &(0,2,2) \\ &(1,0,2) \\ &(2,0,2) \\ &(1,1,1) \\ &(2,2,0) \\ &(1,2,0) \\ &(2,1,1) \end{align*} $(1,0,2)+(1,1,1)+(1,2,0)=(0,0,0)$ Oops!

bgn wrote: There are $27$ cards, each has some amount of ($1$ or $2$ or $3$) shapes (a circle, a square or a triangle) with some color (white, grey or black) on them. We call a triple of cards a match such that all of them have the same amount of shapes or distinct amount of shapes, have the same shape or distinct shapes and have the same color or distinct colors. For instance, three cards shown in the figure are a match be cause they have distinct amount of shapes, distinct shapes but the same color of shapes. What is the maximum number of cards that we can choose such that non of the triples make a match? Proposed by Amin Bahjati Am I the only one who thought this game was Set? Nice question BTW.

The answer is $n=9$. Interpret the cards as elements of $\mathbb F_3^3$, and note that the matching condition of three cards with corresponding elements $a,b,c$ is equivalent to $a+b+c=0$. A construction for $n=9$ is $(1,1,0),(-1,1,0),(1,-1,0),(-1,-1,0),(0,0,1),(0,1,-1),(1,0,-1),(0,-1,-1),(-1,0,-1)$. Now, we show that among any 10 elements of $\mathbb F_3^3$ there are three distinct ones summing to 0. Suppose we had a subset $A$ of $\mathbb F_3^3$ with $|A|=10$ and no three elements of $A$ summing to 0. First, we claim that there exist $w,x,y,z\in A$ with $w+x=y+z$. Otherwise, there cannot be any repeats if we take pairwise sums of the elements of $A$, a contradiction as $\binom{10}2>3^3$. Now, take a linear transformation that sends $w,x,y,z$ to $(1,1,0),(-1,1,0),(-1,-1,0),(1,-1,0)$. The condition is clearly preserved under this transformation. Now, note that there cannot be any other points $(p,q,0)$ in $A$ and that for any $(a,b,c)$ in $A$ with $c\ne 0$, none of the points $(\pm 1-a,\pm 1-b,-c)$ can be in $A$. In particular, this means that if we have at least two points with $z$-coordinate $1$, then the set of available points with $z$-coordinate $-1$ either has size less than 3 or consists of three collinear points. This means that we cannot have at least two points in both the planes $z=1$ and $z=-1$. Thus, one of those planes must have 5 points. It is not hard to see that the answer to the 2-dimensional version of this question is 4 by some Pigeonhole, so we have a contradiction and are done.