Question

66
=
10
-
:
-
-
11
+
×
×
12
+
+
+
13
×
:

Explanation:

🆕 New Concept Discovered: Backtracking Equations
Working backward from a known result to find missing values.

Step 1: Trace the path and identify the variables

Let's trace the snake-like path from the top-left empty box to the top-right box containing \( 66 \). We will represent the empty boxes as variables in order of their appearance along the path:

1. Start at the top-left box: \( x_1 \)
2. Go down: \( + \)
3. Next box: \( 13 \)
4. Go down: \( \times \)
5. Next box: \( x_2 \)
6. Go down: \( : \) (which represents division, \( \div \))
7. Next box (bottom-left): \( x_3 \)
8. Go right to the connector box: \( x_4 \)
9. Go right to the bottom of the second column: \( + \)
10. Go up: \( x_5 \)
11. Go up: \( + \)
12. Go up: \( 12 \)
13. Go up: \( \times \)
14. Next box (top of the second column): \( x_6 \)
15. Go right to the connector box: \( - \)
16. Go right to the top of the third column: \( x_7 \)
17. Go down: \( - \)
18. Next box: \( 11 \)
19. Go down: \( + \)
20. Next box: \( x_8 \)
21. Go down: \( \times \)
22. Next box (bottom of the third column): \( x_9 \)
23. Go right to the connector box: \( x_{10} \)
24. Go right to the bottom of the fourth column: \( : \) (division, \( \div \))
25. Go up: \( x_{11} \)
26. Go up: \( - \)
27. Next box: \( 10 \)
28. Go up: \( = \)
29. Top-right box: \( 66 \)

This is a classic snake puzzle where operations are performed sequentially from start to finish (left-to-right along the path, following standard left-to-right order of operations as a single continuous chain calculation).

Let's write down the continuous equation:
\[ ((((((((((x_1 + 13) \times x_2) \div x_3) + x_5) + 12) \times x_6) - x_7) - 11) + x_8) \times x_9) \div x_{11} - 10 = 66 \]

In these puzzles (often from Vietnamese primary school math challenges), the empty boxes are typically filled with the digits \( 1, 2, 3, 4, 5, 7, 8, 9 \) (since \( 10, 11, 12, 13, 66 \) are already placed on the board). Let's count the number of empty boxes:
There are 9 empty boxes to fill. The digits available to fill them are the single digits from \( 1 \) to \( 9 \):
\[ \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \]

Let the 9 empty boxes be represented by variables \( a, b, c, d, e, f, g, h, i \) in order of their positions along the path:
- Box 1 (top-left): \( a \)
- Box 2 (third down, first column): \( b \)
- Box 3 (bottom-left): \( c \)
- Box 4 (bottom connector 1): \( d \)
- Box 5 (bottom of second column): \( e \)
- Box 6 (top of second column): \( f \)
- Box 7 (top of third column): \( g \)
- Box 8 (fourth down, third column): \( h \)
- Box 9 (bottom of third column): \( i \)
- Box 10 (bottom connector 2): \( j \)
- Box 11 (bottom of fourth column): \( k \)

Looking closely at the grid layout:
- Column 1 has 5 vertical boxes: `[a]`, `[+]`, `[13]`, `[x]`, `[b]`, `[:]`, `[c]`. Wait, let's count the grid cells in Column 1:
- Row 1: `[a]` (empty)
- Row 2: `[+]`
- Row 3: `[13]`
- Row 4: `[x]`
- Row 5: `[b]` (empty)
- Row 6: `[:]`
- Row 7: `[c]` (empty)
- Row 7, Column 2: `[d]` (empty connector)
- Row 7, Column 3: `[+]`
- Row 6, Column 3: `[e]` (empty)
- Row 5, Column 3: `[+]`
- Row 4, Column 3: `[12]`
- Row 3, Column 3: `[x]`
- Row 2, Column 3: `[f]` (empty)
- Row 1, Column 3: `[-]` (Wait, Row 1 Column 3 is empty, Row 1 Column 4 is `[-]`, Row 1 Column 5 is empty? Let's look at the image grid structure).

Let's carefully map the grid coordinates:
- **Column 1**:
- Row 1: Empty box
- Row 2: `+`
- Row 3: `13`
- Row 4: `\times`
- Row 5: Empty box
- Row 6: `:`
- **Bottom Row (Row 6)**:
- Column 1: `:`
- Column 2: Empty box
- Column 3: `+`
- **Column 3**:
- Row 6: `+`
- Row 5: Empty box
- Row 4: `12`
- Row 3: `\times`
- Row 2: Empty box
- **Top Row (Row 1)**:
- Column 3: Empty box
- Column 4: `-`
- Column 5: Empty box
- **Column 5**:
- Row 1: Empty box
- Row 2: `-`
- Row 3: `11`
- Row 4: `+`
- Row 5: Empty box
- **Bottom Row (Row 6)**:
- Column 5: `\times`
- Column 6: Empty box
- Column 7: `:`
- **Column 7**:
- Row 6: `:`
- Row 5: Empty box
- Row 4: `-`
- Row 3: `10`
- Row 2: `=`
- Row 1: `66`

Let's list the sequence of elements along the snake path:
1. `[ ]` (let's call this \( v_1 \))
2. `+`
3. `13`
4. `\times`
5. `[ ]` (let's call this \( v_2 \))
6. `:`
7. `[ ]` (let's call this \( v_3 \))
8. `+`
9. `[ ]` (let's call this \( v_4 \))
10. `+`
11. `12`
12. `\times`
13. `[ ]` (let's call this \( v_5 \))
14. `-`
15. `[ ]` (let's call this \( v_6 \))
16. `-`
17. `11`
18. `+`
19. `[ ]` (let's call this \( v_7 \))
20. `\times`
21. `[ ]` (let's call this \( v_8 \))
22. `:`
23. `[ ]` (let's call this \( v_9 \))
24. `-`
25. `10`
26. `=`
27. `66`

There are exactly 9 empty boxes: \( v_1, v_2, v_3, v_4, v_5, v_6, v_7, v_8, v_9 \).
We must fill these 9 boxes using the digits \( 1, 2, 3, 4, 5, 6, 7, 8, 9 \) exactly once each.

Step 2: Simplify the equation using standard order of operations

Following standard mathematical order of operations (multiplication and division before addition and subtraction):
The expression is:
\[ v_1 + \frac{13 \times v_2}{v_3} + v_4 + 12 \times v_5 - v_6 - 11 + \frac{v_7 \times v_8}{v_9} - 10 = 66 \]

Let's simplify the constant terms by moving them to the right side:
\[ v_1 + \frac{13 \times v_2}{v_3} + v_4 + 12 \times v_5 - v_6 + \frac{v_7 \times v_8}{v_9} - 11 - 10 = 66 \]
\[ v_1 + \frac{13 \times v_2}{v_3} + v_4 + 12 \times v_5 - v_6 + \frac{v_7 \times v_8}{v_9} - 21 = 66 \]
\[ v_1 + \frac{13 \times v_2}{v_3} + v_4 + 12 \times v_5 - v_6 + \frac{v_7 \times v_8}{v_9} = 87 \]

Step 3: Determine values to make division terms integers

To make the terms \( \frac{13 \times v_2}{v_3} \) and \( \frac{v_7 \times v_8}{v_9} \) result in integers:
- Since 13 is a prime number and our digits are only \( 1 \) to \( 9 \), \( v_3 \) must divide either \( 13 \) (impossible unless \( v_3 = 1 \)) or \( v_2 \). Thus, \( \frac{v_2}{v_3} \) must be an integer.
- Let's try a common configuration that works well for this famous puzzle:
- Set \( v_2 = 2 \) and \( v_3 = 1 \). Then \( \frac{13 \times 2}{1} = 26 \).
- This leaves the digits \( \{3, 4, 5, 6, 7, 8, 9\} \) for the remaining variables.

Let's substitute these into our simplified equation:
\[ v_1 + 26 + v_4 + 12 \times v_5 - v_6 + \frac{v_7 \times v_8}{v_9} = 87 \]
\[ v_1 + v_4 + 12 \times v_5 - v_6 + \frac{v_7 \times v_8}{v_9} = 61 \]

Step 4: Test combinations for the remaining variables

We want to choose \( v_5 \) such that \( 12 \times v_5 \) is large enough but doesn't exceed the target.
- If \( v_5 = 3 \):
\[ v_1 + v_4 + 36 - v_6 + \frac{v_7 \times v_8}{v_9} = 61 \]
\[ v_1 + v_4 - v_6 + \frac{v_7 \times v_8}{v_9} = 25 \]

Now we need to choose \( v_7, v_8, v_9 \) from the remaining digits \( \{4, 5, 6, 7, 8, 9\} \) such that \( \frac{v_7 \times v_8}{v_9} \) is an integer:
- Let's try \( v_7 = 7 \), \( v_8 = 8 \), and \( v_9 = 4 \):
\[ \frac{7 \times 8}{4} = 14 \]
- Substituting this back in:
\[ v_1 + v_4 - v_6 + 14 = 25 \]
\[ v_1 + v_4 - v_6 = 11 \]

The remaining unused digits are \( \{5, 6, 9\} \).
Let's assign them to \( v_1, v_4, v_6 \):
- Let \( v_1 = 5 \)
- Let \( v_4 = 9 \)
- Let \( v_6 = 3 \) (Wait, \( 3 \) is already used for \( v_5 \). The unused digits are indeed \( 5, 6, 9 \)).
- Let's try:
\[ v_1 + v_4 - v_6 = 9 + 5 - 6 = 8 eq 11 \]
\[ v_1 + v_4 - v_6 = 9 + 6 - 5 = 10 eq 11 \]

Let's adjust our choices:
- Try \( v_5 = 4 \):
\[ v_1 + v_4 + 48 - v_6 + \frac{v_7 \times v_8}{v_9} = 61 \]
\[ v_1 + v_4 - v_6 + \frac{v_7 \times v_8}{v_9} = 13 \]
- Remaining digits: \( \{3, 5, 6, 7, 8, 9\} \)
- Let's choose \( v_7 = 9 \), \( v_8 = 8 \), and \( v_9 = 6 \):
\[ \frac{9 \times 8}{6} = 12 \]
- Substituting this back in:
\[ v_1 + v_4 - v_6 + 12 = 13 \]
\[ v_1 + v_4 - v_6 = 1 \]
- The remaining unused digits are \( \{3, 5, 7\} \).
- Let's assign them to satisfy \( v_1 + v_4 - v_6 = 1 \):
\[ 3 + 5 - 7 = 1 \]
This works perfectly!
- \( v_1 = 3 \) or \( 5 \)
- \( v_4 = 5 \) or \( 3 \)
- \( v_6 = 7 \)

Step 5: Verify the complete set of values

Let's write down the complete assignment of digits \( 1 \) through \( 9 \):
- \( v_1 = 5 \)
- \( v_2 = 2 \)
- \( v_3 = 1 \)
- \( v_4 = 3 \)
- \( v_5 = 4 \)
- \( v_6 = 7 \)
- \( v_7 = 9 \)
- \( v_8 = 8 \)
- \( v_9 = 6 \)

Let's plug these values back into the original equation to verify:
\[ 5 + \frac{13 \times 2}{1} + 3 + 12 \times 4 - 7 - 11 + \frac{9 \times 8}{6} - 10 \]
\[ = 5 + 26 + 3 + 48 - 7 - 11 + 12 - 10 \]
\[ = 31 + 3 + 48 - 7 - 11 + 12 - 10 \]
\[ = 34 + 48 - 7 - 11 + 12 - 10 \]
\[ = 82 - 7 - 11 + 12 - 10 \]
\[ = 75 - 11 + 12 - 10 \]
\[ = 64 + 12 - 10 \]
\[ = 76 - 10 \]
\[ = 66 \]

The equation is perfectly balanced and all digits from \( 1 \) to \( 9 \) are used exactly once.

Answer:

One valid solution to fill the empty boxes from start (top-left) to finish (bottom-right) using the digits \( 1 \) to \( 9 \) exactly once is:

• Box 1 (top-left): \( 5 \)
• Box 2 (second empty box down): \( 2 \)
• Box 3 (bottom-left): \( 1 \)
• Box 4 (bottom connector): \( 3 \)
• Box 5 (bottom of second column): \( 4 \)
• Box 6 (top of second column): \( 7 \)
• Box 7 (top of third column): \( 9 \)
• Box 8 (middle of third column): \( 8 \)
• Box 9 (bottom of third column): \( 6 \)