$T$ -Tetrominoes KenKen Puzzle!

There is a very simple explanation why the solution must be unique with the equality as given in the correct answer. Here are the key points:

• $K$ must be an even number, since every Tetromino contains at least one even digit. If there is one even digit then x $K$ is even.
• It is impossible to have two times two or one times more than two repeated digits inside a Tetromino, like $\{1, 1, 2, 2\}$ , or $\{2, 2, 2, 4\}$ .
• The lowest possible product is $1*1*2*3=6$ . Nonetheless, you cannot achieve 6 with a summation, as it would require the previously mentioned forbidden (impossible) combination $\{1, 1, 2, 2\}$ .
• The 2nd lowest product is $1*2*2*3=12$ . On the other hand, this is the maximum possible summation $4+4+3+1=12$ . Hence, this is the equilibrium from the viewpoint of $+K$ and x $K$ . There is only one way to fill the central squares to get x $K=12$ and $+K=12$ . In that arrangement the equality $x+y+z+w=K$ holds.

---

First Step: Arithmetic

---

Since

• The product of all distinct digits is $4! = 24$ .
• The sum of all distinct digits is $10$ .
• The number of digits in every cage is the same as the one in each column and row.

we can safely make both product and sum equal to each other by the following equation $10 + p = \dfrac{24}{p}$ where $p = 2$ is the only positive solution that makes both sides equal. This shows that:

• For sum set, we can increase one of the digits by $2$ , which leads to the possible sum sets $\{4,3,3,1\}$ and $\{4,4,3,1\}$ .
• For product set, we can divide one of the digits by $2$ , which leads to the possible product sets $\{3,2,2,1\}$ and $\{4,3,1,1\}$ .

---

Final Step: Set Substitution

---

Since the given cages are not long column-shaped, we can check out each pair of product and sum sets. Let's test out sets with common repeated digits:

Case 1: If $\{4,3,3,1\}$ is the possible sum set, then (by intersection) this eliminates all choices of the product set. This can't happen since all four $3$ 's are used up by two separate sum cages.

Case 2: Likewise, if we choose either $\{3,2,2,1\}$ and $\{4,3,1,1\}$ , then we run out of possible sum sets with digits that are not repeated in each product set.

Thus, the sum set is $\{4,4,3,1\}$ and the product set is $\{3,2,2,1\}$ . So since the repeated digits $2$ and $4$ occur at the center 2-by-2 square,

$\boxed{x + y + z + w = 2(2 + 4) = 12 = K}$

So we have $2 \cdot 2 = 4$ different solutions of $12$ -valued cages, where:

• $2$ 's and $4$ 's are all fixed in these positions
• empty cells contain either $1$ or $3$ ; since the cage borders do not exist between them, we can substitute them in any way. This is similar to deadly pattern , where the cells with common candidates can't be specifically determined.