Place the numbers 1 through 9 in the available input boxes so that the results of performing the mathematical operations in any row and column (from left to right and top to bottom) always equals 13.
\begin{bmatrix}\square&+&\square&-&\square\\ -& &+& &\times\\ \square&\times&\square&+&\square\\ +&&-&&+\\ \square&\times&\square&-&\square\\ \end{bmatrix}
It takes me an hour to solve this. Do you have an easy way to solve this? Let's explore.
On
I tried it a bit, and my answer for this is $$8 \ \ 7 \ \ 2$$$$1 \ \ 9 \ \ 4$$ $$6 \ \ 3 \ \ 5$$ My approach was to notice that for $z$ in $x+y-z = 13$, $z$ can only be $1,2,3,4$, meaning the middle-left and bottom-middle ones could only be one of these four, and went on from there.
On
Label spaces A through I.
A+B, B+E,A+G are all at least 14. So A,B,E, and G are all greater than or equal to 5.
D x E must be less than 13 and E is 5 or more so D = 1 or 2.
If D=2 then E=5. If E = 5 then B+5-H=13 so B=9, H=1. So Gx1 -I =13 is impossible.
So D isn't 2. So D=1. So E+F = 13, so F is at least 4. C x F +I = 13 but F is at least 4 and I is at least 3 and C is at least 2.
If C is at least 3 then CxD+I is at least 14 so C is at most 2 and at least 2. So C is 2 and I is at least 3.
2x F + I =13 and F is at least 4 so we have either 2x4 + 5. Or 2x5 +3.
If F=4, I=5, E=9. GxH = 18 and neither G or H is 9 so G and H are 3 and 6. And G is at least 5 so G = 6, H =3 and B+9-3=13 so B=7. That leaves A =8. And that works as A+7-2=13 and A-1+6=13.
That's a solution.
any other solution would have F=5 and I =3 and E=8.
So A+B=11, GxH= 16, A+G=12, B-H=5. 1,2,3 are assigned so G and H are at least 4, so G xH =16 is impossible.
So only solution is
A=8,B=7,C=2,D=1,E=9,F=4,G=6,H=3,I=5.
My eye settled on the middle row and right column, which have the same pattern of operators. There aren't too many ways to make $13$ in that pattern. Looking down the first column, the first multiplicand of the middle row is subtracted, so it wants to be small, and I will assume that the first number in the second row is smaller than the second. This gives the second row as one of $149, 158, 167, 176,185,194,237,245,253,261,341$. These are also the possibilities for the third column, though the first two numbers could be switched. Starting down the list, each of $149,158,167,176$ fail because they require a $1$ in the top of the third column to match one we have used. If it is $185$ the third column needs to be $253$ and the other two numbers in the top row have to sum to $15$ and be $69$ as we have used $8$. They can't be $96$ because the middle bottom would have to be $1$ and $69$ fails because the bottom left needs to be $8$. We go on to $194$ where the right column is $245$. Again the top two need to sum to $15$ but now they must be $78$. The order $78$ doesn't work because of the center bottom, which would have to be $4$, but $87$ works. We are done with $$\begin {array} &8&7&2\\1&9&4\\6&3&5 \end {array}$$