Multiplication puzzle with trios of numbers

Question

I created a puzzle that I was curious what its properties are, and how it could be determined it is solvable or not. It consists of 6 rows of 3 of the same numbers, which in each row in order are 1, 2, 3, 4, 6, and 8.

1 1 1
2 2 2
3 3 3
4 4 4
6 6 6
8 8 8

The goal is to come up with 6 single digit multiplication equations (with only 3 numbers so not $2∗2∗2=8$) that allow no numbers to be left on the grid. An example equation could be $2∗3=6$ in which case you would cross off the 2, 3, and 6.

1 1 1
/ 2 2
/ 3 3
4 4 4
/ 6 6
8 8 8

Continuing making equations and crossing off numbers, is it possible to have all numbers crossed off? How would you figure that out and what about a grid with numbers that go higher than 8? I've been able to be left with only 3 numbers, and I have been able to solve a variant without the 3 8s at the end.

Answer 1

This puzzle is

not possible.

Here's why:

In order for there to be a solution, we must have six equations of the form $a * b = c$, where $a$, $b$, and $c$ are each one of the numbers in the grid. Analyzing one of these equations doesn't help us, so let's instead consider the product of all six equations - since both sides are equal, we can take $a * b = c$ and $d * e = f$ and safely combine them into $a * b * d * e = c * f$ and still have a valid equation.

So, we must be able to split these numbers into two groups which have the same product. (There are more constraints, but we can ignore these for now.) In order for that to be the case, the product of all of the numbers in the grid must be a square. In this case, we have $(1 * 2 * 3 * 4 * 6 * 8)^3 = 1528823808$, not a square number, so this puzzle doesn't work. The reason the puzzle works without the row of 8s is because $(1 * 2 * 3 * 4 * 6)^3 = 2985984$, which is a square number: you can divide those 15 numbers into two sides with products of 1728 (and with three 1s, you have some flexibility).

Regarding grids with higher numbers:

As I've shown, any puzzle where the product of all numbers is a square is theoretically possible. (To save you some time, you can just take one of each number; it's a lot easier to see if $144$ is a square than $144^3$.) For example, replacing your 8s with 9s makes a puzzle that has a solution:

1 * 9 = 9
3 * 3 = 9
1 * 6 = 6
2 * 3 = 6
1 * 4 = 4
2 * 2 = 4

Answer 2

You can solve the problem via integer linear programming as follows. Let $D=\{1,2,3,4,6,8\}$ be the set of digits. For $d\in D$, let integer decision variable $u_d$ be the number of times that $d$ is uncovered. For triple $(a,b,ab)\in D^3$, let integer decision variable $x_{a,b,ab}$ be the number of times the triple is crossed out. The problem is to minimize $\sum_{d\in D} u_d$ subject to \begin{align} 3x_{1,1,1} + x_{1,2,2} + x_{1,3,3} + x_{1,4,4} + x_{1,6,6} + x_{1,8,8} + u_1 &= 3 \\ 2x_{1,2,2} + 2x_{2,2,4} + x_{2,3,6} + x_{2,4,8} + u_2 &= 3 \\ 2x_{1,3,3} + x_{2,3,6} + u_3 = 3 2x_{1,4,4} + x_{2,2,4} + x_{2,4,8} + u_4 &= 3 \\ 2x_{1,6,6} + x_{2,3,6} + u_6 &= 3 \\ 2x_{1,8,8} + x_{2,4,8} + u_8 &= 3 \end{align} The optimal objective value turns out to be $3$, attained by $$x_{1,4,4}=x_{1,8,8}=1, x_{2,3,6}=3, u_1 = u_4 = u_8 = 1.$$

An alternative optimal solution is $$x_{1,3,3}=x_{1,4,4}=x_{1,6,6}=x_{2,2,4}=x_{2,3,6}=1,u_8=3,$$ which shows that omitting the three $8$s is feasible.