Question

puzzle - only using 1-9 once

+ + = 15
+
+ * = 24
-
+ - = 14
=
3
=
12
=
4

Answer:

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<reasoning>

Define variables for the grid

Let the grid of 9 boxes be represented by variables \(a, b, c, d, e, f, g, h, i\) from top-left to bottom-right:
\[

$$\begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$$

\]
The equations from the image are:
Row 1: \(a + b + c = 15\)
Row 2: \(d + e \times f = 24\)
Row 3: \(g + h - i = 14\)
Column 1: \(a + d - g = 3\)
Column 2: \(b \times e - h = 12\)
Column 3: \(c \div f \div i = 4\)
The numbers used must be from \(1\) to \(9\) exactly once.

Solve the system by constraint satisfaction

Using the set of numbers \(\{1, 2, 3, 4, 5, 6, 7, 8, 9\}\):
From Column 3: \(c \div f \div i = 4 \implies c = 4 \times f \times i\).
Since \(c \le 9\), we must have \(f \times i \le 2.25\).
Thus, \(\{f, i\} = \{1, 2\}\).
If \(f = 1, i = 2\), then \(c = 8\).
If \(f = 2, i = 1\), then \(c = 8\).
Let's test \(c = 8\), \(f = 2\), \(i = 1\):
Remaining numbers: \(\{3, 4, 5, 6, 7, 9\}\).
From Row 1: \(a + b + 8 = 15 \implies a + b = 7\).
Since \(a, b \in \{3, 4, 5, 6, 7, 9\}\), the only pair that sums to \(7\) is \(\{3, 4\}\).
Case 1: \(a = 3, b = 4\).
Remaining numbers: \(\{5, 6, 7, 9\}\).
From Column 2: \(4 \times e - h = 12 \implies 4e - h = 12\).
If \(e = 5 \implies h = 8\) (already used).
If \(e = 6 \implies h = 12\) (too large).
Case 2: \(a = 4, b = 3\).
Remaining numbers: \(\{5, 6, 7, 9\}\).
From Column 2: \(3 \times e - h = 12 \implies 3e - h = 12\).
If \(e = 5 \implies h = 3\) (already used).
If \(e = 6 \implies h = 6\) (cannot repeat).
If \(e = 7 \implies h = 9\). This works!
So \(e = 7, h = 9\).
Remaining numbers: \(\{5, 6\}\).
Since \(a = 4\), from Column 1: \(4 + d - g = 3 \implies g - d = 1\).
With remaining numbers \(\{5, 6\}\), we must have \(g = 6, d = 5\).
Let's verify all equations with this assignment:
\(a=4, b=3, c=8, d=5, e=7, f=2, g=6, h=9, i=1\).

Verify all equations

Row 1: \(4 + 3 + 8 = 15\) (True)
Row 2: \(5 + 7 \times 2 = 5 + 14 = 19
eq 24\) (False, standard order of operations applies: \(d + (e \times f) = 24\)).
If Row 2 is evaluated left-to-right: \((5 + 7) \times 2 = 24\) (True).
Let's check if standard order of operations is intended or strictly left-to-right.
Usually, these puzzles are solved strictly left-to-right:
Row 1: \((a + b) + c = 15\)
Row 2: \((d + e) \times f = 24\)
Row 3: \((g + h) - i = 14\)
Column 1: \((a + d) - g = 3\)
Column 2: \((b \times e) - h = 12\)
Column 3: \((c \div f) \div i = 4\)

Let's re-verify with left-to-right:
Row 1: \((4 + 3) + 8 = 15\) (True)
Row 2: \((5 + 7) \times 2 = 24\) (True)
Row 3: \((6 + 9) - 1 = 14\) (True)
Column 1: \((4 + 5) - 6 = 3\) (True)
Column 2: \((3 \times 7) - 9 = 12\) (True)
Column 3: \((8 \div 2) \div 1 = 4\) (True)
All equations are perfectly satisfied using the numbers \(1\) to \(9\) exactly once.
</reasoning>

<answer>
The completed puzzle grid (using numbers 1 to 9 exactly once, evaluated left-to-right and top-to-bottom) is:

\[

$$\begin{array}{ccccc} \mathbf{4} & + & \mathbf{3} & + & \mathbf{8} & = & 15 \\ + & & \times & & \div & & \\ \mathbf{5} & + & \mathbf{7} & \times & \mathbf{2} & = & 24 \\ - & & - & & \div & & \\ \mathbf{6} & + & \mathbf{9} & - & \mathbf{1} & = & 14 \\ = & & = & & = & & \\ 3 & & 12 & & 4 & & \end{array}$$

\]
</answer>

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