CrossNumber Puzzle

Name the cell on $x$ -th row and $y$ -th column as $RxCy$ .

From column 1, $R2C1 = 4 + R1C1 + R3C1 \ge 4+1+2 = 7$ . From column 3, $R2C3 = R1C3 + R3C3 - 11 \le 8+9-11 = 6$ . From row 2, $R2C3 > \frac{R2C1}{R2C2}$ . If $R2C2 = 1$ , then $6 \ge R2C3 > R2C1 \ge 7$ , contradiction. Thus $R2C2 \ge 2$ . Together with the fact that $R2C2$ divides $R2C1$ (so that row 2 results to an integer), there are only three possibilities for row 2: $(8,2,5), (8,4,3), (9,3,4)$ .

Case 1 : $(R2C1, R2C2, R2C3) = (9,3,4)$

Then $R1C1 + R3C1 = 5$ , and so $\{R1C1, R1C3\} = \{1,4\} \vee \{2,3\}$ . However, neither is possible, as they would repeat $4$ or $3$ used on $R2C3$ and $R2C2$ respectively. Thus this case is impossible.

Case 2 : $(R2C1, R2C2, R2C3) = (8,4,3)$

Likewise, this gives $R1C1 + R3C1 = 4$ , which forces $\{R1C1, R3C1\} = \{1,3\}$ (the other choice $\{2,2\}$ is impossible, as it would repeat the same digit). But now this repeats the $3$ on $R2C3$ .

Case 3 : $(R2C1, R2C2, R2C3) = (8,2,5)$

We have $\{R1C1, R3C1\} = \{1,3\}$ and $\{R1C3,R3C3\} = \{7,9\}$ . Thus by elimination, $\{R1C2, R3C2\} = \{4,6\}$ , and so we already have the answer: $R1C2+R2C1+R2C2+R2C3+R3C2 = (4+6) + 8 + 2 + 5 = \boxed{25}$ .

However, of course we want to keep going and see if the solution is unique.

From row 3, $R3C3 \ge 7$ , so $R3C1 \cdot R3C2 \le 6$ . But $R3C2 \ge 4$ , so $R3C1 \le \frac{6}{4}$ , forcing $R3C1 = 1$ . This means $R1C1 = 3$ .

Now, $R1C2 = R1C1 + R1C3 - 8 = R1C3 - 5 \le 9-5 = 4$ , but $R1C2 \ge 4$ . So equality is achieved, giving $R1C2 = 4$ , and the rest is easy.

The unique solution is:

* 4+8+2+5+6=25 *