I have a question about this math puzzle and the question offers the numbers 1, 3, 4, 5, 6, 7, 8, 9, 10:
I already solved it:
A=4, B=9, C=10, D=8, E=7, F=3, G=1, H=6, J=5
However, it took me a while because I was trying the possibilities without a specific rule, so is there a rule to solve it?
Thanks
Immediately we see that $D=8$.
Then, I think the most logical step is to try to place the $10$. It is trivially not in places $A, E, F, G, H, J$ because the minimum row and column sums fail in those cases. Thus, either $B$ or $C$ contains the $10$.
Now, $B=10 \implies H=5\implies (G,J)\in \{(1,6),(6,1),(3,4),(4,3)\}$
Using that $C+F+J=18$, $A+E+G=12$, $A+J=9$, $C+G=11$ we can show that each of these pairings break somewhere:
If $G=1$ then $C=10$ but $B=10$.
If $J=1$ then $C+F=17 \implies C,F=8,9$ but $D=8$
If $J=4$ then $C+F=14\implies C,F = 6,8$ or $9,5$ but $D=8$ and $H=5$ (also $J=4\implies A=5$)
If $G=4$ then $A+E=8\implies A,E=1,7; 2,6; 3,5$ but $C=7$; $2$ is already placed and $H=5$
So $B\neq 10 \implies C=10\implies G=1$
We can now see that: $$A+E=11\implies A\in\{4,5,6,7\}=A_1$$ $$A+B=13\implies A\in\{4,6,7,9\}=A_2$$ $$A+J=9\implies A\in\{3,4,5,6\}=A_3; J\in \{3,4,5,6\}=J_1$$ $$F+J=8\implies J\in\{3,5\}=J_2$$ $$H+J=11\implies J\in\{4,5,6,7\}=J_3$$
I've achieved this by listing pairs and discarding any that contain $\{1,2,8,10\}$ that we have already placed.
Now: $$J_1\cap J_2\cap J_3 =\{5\}\implies J=5\implies A=4$$
and the rest fall out from there.