Tricky Logic Puzzles for Adults - Steven Clontz - Free Printable

To solve this KenKen puzzle, we need to fill the $7 \times 7$ grid with numbers from 1 to 7. Each number must appear exactly once in every row and every column. The numbers in each "cage" (the bolded boxes) must combine using the specified operation ($+$, $-$, $\times$, $\div$) to equal the target number.

Here is the step-by-step logical deduction:

Step 1: Analyze Unique Combinations and Constraints
* Top-Left Cage ($22+$): This is a 4-cell cage in the top-left corner (R1C1, R1C2, R2C1, R2C2). The maximum sum of four distinct digits from 1-7 is $7+6+5+4=22$. Since the target is 22, these four cells must contain $\{4, 5, 6, 7\}$.
* This implies that any other cell in Row 1 or Row 2 cannot be 4, 5, 6, or 7 if it's outside this block? No, just that R1C1, R1C2, R2C1, R2C2 are 4,5,6,7 in some order. Consequently, the remaining cells in Rows 1 and 2 (R1C3-C7 and R2C3-C7) must contain $\{1, 2, 3\}$ and potentially others depending on columns, but specifically, R1C3 and R1C4 are part of a $5+$ cage.
* Row 1, Cells 3 & 4 ($5+$): This is a 2-cell horizontal cage. The possible pairs for sum 5 are $\{1,4\}$ or $\{2,3\}$. However, R1C1 and R1C2 are already occupied by $\{4,5,6,7\}$. So R1C3 cannot be 4, 5, 6, or 7. It must be 1, 2, or 3.
* If the pair is $\{1,4\}$, one cell is 4. But R1C1/R1C2 take up two of $\{4,5,6,7\}$. Wait, R1C1 and R1C2 are just two cells. They contain two numbers from $\{4,5,6,7\}$. The other two numbers from $\{4,5,6,7\}$ are in R2C1 and R2C2.
* Let's look at Column 1. R1C1 and R2C1 are in the $22+$ cage.
* Let's look at the $336\times$ cage in R1C5, R2C5, R3C5, R4C5. This is a vertical 4-cell cage. $336 = 7 \times 6 \times 4 \times 2$. So the cells $\{R1C5, R2C5, R3C5, R4C5\}$ contain $\{2, 4, 6, 7\}$.
* This means R1C5 and R2C5 are from $\{2, 4, 6, 7\}$.
* Back to Row 1: R1C1, R1C2 $\in \{4,5,6,7\}$. R1C5 $\in \{2,4,6,7\}$.
* The $5+$ cage at R1C3-R1C4 uses $\{1,4\}$ or $\{2,3\}$.
* The $4-$ cage at R1C6-R1C7. Possible pairs for difference 4: $\{1,5\}, \{2,6\}, \{3,7\}$.

Let's refine Row 1 possibilities.
Cells R1C1, R1C2 are from $\{4,5,6,7\}$.
Cell R1C5 is from $\{2,4,6,7\}$.
Remaining cells in Row 1: R1C3, R1C4, R1C6, R1C7.
Values available for Row 1: $\{1,2,3,4,5,6,7\}$.

Let's look at the $84\times$ cage in R3C7, R4C7, R5C7. Wait, looking at the grid, the $84\times$ cage is R3C7, R4C7, R5C7? No, let's trace the lines carefully.
The cage labeled $84\times$ covers R3C7, R4C7, R5C7. It looks like a vertical strip of 3 cells. $84 = 7 \times 6 \times 2$ or $7 \times 4 \times 3$? $7 \times 6 \times 2 = 84$. $7 \times 4 \times 3 = 84$. So the set is either $\{2,6,7\}$ or $\{3,4,7\}$.

Let's look at the $336\times$ cage again. It occupies R1C5, R2C5, R3C5, R4C5. Values: $\{2,4,6,7\}$.
This means R3C5 and R4C5 are from $\{2,4,6,7\}$.

Let's look at Row 3.
R3C5 $\in \{2,4,6,7\}$.
R3C7 is part of $84\times$.
R3C1, R3C2 are part of $4-$ and $11+$ cages?
Let's trace cages more precisely.
- Top Left 2x2: $22+$ ($\{4,5,6,7\}$).
- R1C3-R1C4: $5+$.
- R1C5-R4C5: $336\times$ ($\{2,4,6,7\}$).
- R1C6-R1C7: $4-$.
- R2C3-R2C4: $12+$. This is a 2-cell cage? Or L-shaped? Looking at the border, R2C3 and R2C4 form a horizontal pair. Sum 12. Pairs for 12 from 1-7: $\{5,7\}$. So $\{R2C3, R2C4\} = \{5,7\}$.
- R2C6-R3C6: $2\div$. Vertical pair.
- R3C1-R4C1: $4-$. Vertical pair.
- R3C2-R4C2-R4C3? No, R3C2 is part of $11+$. The cage $11+$ seems to cover R3C2, R4C2, R4C3? Let's check the shape. The line goes around R3C2, R4C2, and R4C3? Or R3C2, R3C3?
- Looking at R3C2, there is a vertical line to its right separating it from R3C3. There is a horizontal line below R3C2? No, the line continues down. So R3C2 and R4C2 are connected. Then from R4C2, it goes right to R4C3? The label $11+$ is in R3C2. Usually, the label is in the top-left cell of the cage. If the cage is R3C2, R4C2, R4C3, that's 3 cells. Sum 11.
- Alternatively, is it R3C2 and R3C3? No, thick line between them.
- Is it R3C2, R4C2, R5C2? No, thick line below R4C2? Let's look at R4C2. Below it is R5C2 which starts $18\times$. So R4C2 is the bottom of that column segment. To the right of R4C2 is R4C3. Is there a line between R4C2 and R4C3? No. So R4C2 and R4C3 are connected. Is R3C3 connected? There is a thick line between R3C2 and R3C3. So R3C3 is not in the $11+$ cage.
- So the $11+$ cage is likely an L-shape: R3C2, R4C2, R4C3. Sum = 11.
- R3C3-R4C3? No, R4C3 is in $11+$. R3C3 is part of $4-$ with R4C3? No, R3C3 has a $4-$ label. It connects to R4C3? No, R4C3 is in $11+$. It connects to R3C4? No, thick line. It connects to R4C3? Thick line. Wait.
- Let's re-examine R3C3. Label $4-$. It is a single cell label? No, it must have a partner. The box for $4-$ is R3C3 and... R4C3? If R4C3 is in $11+$, they can't share.
- Let's look at the grid lines again.
- R3C3 has a thick border on Left (from R3C2), Top (from R2C3), Right (from R3C4). Bottom? The line between R3C3 and R4C3 is thin. So R3C3 and R4C3 are in the same cage. BUT R4C3 is also connected to R4C2 (thin line) and R3C2 is connected to R4C2 (thin line). This would mean R3C2, R4C2, R4C3, R3C3 are all one big cage? That doesn't fit the labels $11+$ and $4-$.
- Correction: Look at R3C3. The label is $4-$. The cage likely includes R3C3 and R4C3? If so, R4C3 cannot be in the $11+$ cage.
- Let's look at the $11+$ label position. It is in R3C2. The cage includes R3C2. Does it include R3C3? Thick line. Does it include R4C2? Thin line. Does it include R4C3? From R4C2 to R4C3 is a thin line. So R3C2, R4C2, R4C3 are connected. If R3C3 is separate, then R4C3 is shared? Impossible.
- Alternative interpretation: The $11+$ cage is R3C2, R4C2, R3C3? No, thick line between R3C2/R3C3.
- Maybe the $11+$ cage is just R3C2 and R4C2? Sum 11. Pairs: $\{4,7\}, \{5,6\}$. Then what is R4C3? R4C3 is connected to R4C2? Yes, thin line. So R4C3 is in the same cage as R4C2. So the cage is R3C2, R4C2, R4C3. Sum 11.
- Then what about R3C3? It has label $4-$. It must connect to something. Top is R2C3 (thick). Left is R3C2 (thick). Right is R3C4 (thick). Bottom is R4C3 (thin). So R3C3 connects to R4C3. But R4C3 is already in the $11+$ cage. Contradiction.
- Let's re-read the grid lines.
- Maybe the line between R4C2 and R4C3 is THICK? If it is thick, then R4C3 is not in $11+$.
- If R4C2-R4C3 is thick, then $11+$ is just R3C2 and R4C2? Sum 11. $\{4,7\}$ or $\{5,6\}$.
- Then R3C3 ($4-$) connects to R4C3? If R3C3-R4C3 is thin, they form a pair. Difference 4.
- Then what is R4C3 connected to? Just R3C3? And R4C4? Line between R4C3 and R4C4 is thin. Label $3\div$ is in R4C4. So R4C3, R4C4, ...?
- Let's assume the standard KenKen layout where lines define cages clearly.
- Let's look at R4C3 and R4C4. Label $3\div$ is in R4C4. The cage likely includes R4C3 and R4C4? If so, R3C3 ($4-$) must connect to R3C4? No, thick line. R2C3? Thick. R3C2? Thick. R4C3? If R3C3 connects to R4C3, and R4C3 connects to R4C4, then R3C3, R4C3, R4C4 are one cage? Two labels? No.
- Okay, look at R3C3 label $4-$. Look at R4C4 label $3\div$.
- There is a vertical line between Col 3 and Col 4 in Row 3? Yes, thick.
- There is a vertical line between Col 3 and Col 4 in Row 4? No, thin. So R4C3 and R4C4 are connected.
- There is a horizontal line between Row 3 and Row 4 in Col 3? No, thin. So R3C3 and R4C3 are connected.
- This implies R3C3, R4C3, R4C4 are all in one cage. But there are two labels: $4-$ in R3C3 and $3\div$ in R4C4. This is impossible.
- Therefore, one of the lines I see as thin must be thick.
- Likely candidate: The line between R3C3 and R4C3 is THICK.
- If R3C3-R4C3 is thick, then R3C3 is isolated from below.
- R3C3 Left (R3C2) thick. Top (R2C3) thick. Right (R3C4) thick. Bottom (R4C3) thick. R3C3 would be a 1-cell cage? No, operations require at least 2 cells.
- So R3C3 MUST connect to someone.
- Maybe R3C3 connects to R3C4? If the line between R3C3 and R3C4 is thin, they form a pair. Label $4-$.
- Let's check the image again. Between R3C3 and R3C4, there is a vertical line. It looks thick.
- Between R3C3 and R2C3, horizontal line. Thick.
- Between R3C3 and R3C2, vertical line. Thick.
- Between R3C3 and R4C3, horizontal line. It looks THIN.
- So R3C3 connects to R4C3.
- Now, does R4C3 connect to R4C4? The line between R4C3 and R4C4 looks THIN.
- Does R4C3 connect to R5C3? Line between R4C3 and R5C3 is THICK (top of $28\times$ cage? No, $28\times$ is R6C3-R7C3? Or R5C3-R6C3? Let's check labels).
- Label $28\times$ is in R6C3? No, R6C3 has no label. R5C3 has no label. R6C3 is part of a cage with R6C2? No, R6C2 is $28\times$?
- Let's restart the cage identification based on standard visual cues.

Re-evaluating Cages:

1. R1C1, R1C2, R2C1, R2C2: $22+$. Cells: 4. Values: $\{4,5,6,7\}$.
2. R1C3, R1C4: $5+$. Cells: 2. Values: $\{1,4\}$ or $\{2,3\}$.
3. R1C5, R2C5, R3C5, R4C5: $336\times$. Cells: 4. Values: $\{2,4,6,7\}$.
4. R1C6, R1C7: $4-$. Cells: 2. Values: $\{1,5\}, \{2,6\}, \{3,7\}$.
5. R2C3, R2C4: $12+$. Cells: 2. Values: $\{5,7\}$. (Only pair summing to 12 in 1-7).
* This fixes R2C3 and R2C4 as 5 and 7.
* Since R2C3, R2C4 are 5,7, they are NOT in $\{1,2,3,4,6\}$.
6. R2C6, R3C6: $2\div$. Cells: 2. Vertical.
7. R3C1, R4C1: $4-$. Cells: 2. Vertical.
8. R3C2, R4C2, R4C3? Or R3C2, R3C3?
* Let's look at R3C2 ($11+$) and R3C3 ($4-$).
* We established R2C3 is 5 or 7.
* R3C3 is below R2C3.
* If R3C3 is part of a cage with R3C4 ($4-$)? No, line looks thick.
* If R3C3 is part of a cage with R4C3 ($4-$)?
* Let's look at R4C3 and R4C4 ($3\div$).
* Hypothesis: The cage for $4-$ is R3C3 and R4C3. The cage for $3\div$ is R4C4 and R5C4? Or R4C4 and R4C5?
* R4C5 is in $336\times$. So R4C4 cannot connect to R4C5 (thick line between Col 4 and 5 in Row 4? Yes, R4C5 is end of $336\times$ vertical bar. So line to left is thick).
* So R4C4 connects to R4C3? If R4C3 is in $4-$ cage with R3C3, then R4C4 is separate?
* Label $3\div$ is in R4C4. It needs a partner.
* Partners for R4C4: R3C4 (above), R5C4 (below), R4C3 (left), R4C5 (right).
* R4C5 is taken.
* R3C4: Above R3C4 is R2C4 (5 or 7). Line between R3C4 and R2C4 is thick? Yes. Line between R3C4 and R3C3 is thick. Line between R3C4 and R3C5? R3C5 is in $336\times$. Line is thick. So R3C4 is isolated on Top, Left, Right. Must connect Down to R4C4.
* So R3C4 and R4C4 form the $3\div$ cage? But the label $3\div$ is in R4C4. And label $4-$ is in R3C3.
* If R3C4 and R4C4 are $3\div$, what is R3C3?
* R3C3 ($4-$) connects to R4C3? If so, R4C3 is in $4-$ cage.
* Does R4C3 connect to R4C4? Line between R4C3 and R4C4 is thin. If they are in different cages, this line should be thick.
* Contradiction again.
* Let's look really closely at the intersection of R3C3, R3C4, R4C3, R4C4.
* Maybe the $4-$ cage is R3C3 and R3C4? If the vertical line between them is thin.
* Maybe the $3\div$ cage is R4C3 and R4C4? If the vertical line between R3C3/R4C3 is thick and R3C4/R4C4 is thick?
* Let's assume:
* Cage A: R3C3, R3C4 ($4-$). Horizontal pair.
* Cage B: R4C3, R4C4 ($3\div$). Horizontal pair.
* This requires thick lines: R3C3-R4C3 (horizontal), R3C4-R4C4 (horizontal), R3C2-R3C3 (vertical), R3C4-R3C5 (vertical), R4C2-R4C3 (vertical), R4C4-R4C5 (vertical).
* Let's check the image for these thick lines.
* R3C3-R4C3: Looks thin.
* R3C4-R4C4: Looks thin.
* R4C2-R4C3: Looks thin.
* This hypothesis fails.

* Alternative Hypothesis:
* Cage $11+$: R3C2, R4C2, R4C3? (L-shape). Sum 11.
* Cage $4-$: R3C3, R4C3? No, overlap.
* Cage $4-$: R3C3, R3C4? No, line thick.
* Cage $4-$: R3C3, R2C3? No, line thick.
* Cage $4-$: R3C3, R4C3? (Vertical).
* Cage $3\div$: R4C4, R5C4? (Vertical).
* If R3C3-R4C3 is a cage ($4-$), then the line between R4C3 and R4C4 MUST be thick. In the image, it looks thin.
* If R4C4-R5C4 is a cage ($3\div$), then the line between R4C3 and R4C4 MUST be thick.
* So, if the line R4C3-R4C4 is actually THICK, then:
* R3C3-R4C3 is $4-$.
* R4C4-R5C4 is $3\div$.
* What about R3C4? It needs a cage. Neighbors: R2C4 (thick), R3C3 (thick), R3C5 (thick), R4C4 (thick). R3C4 is isolated! Impossible.

* Let's try one more configuration which is common in these puzzles:
* $11+$ Cage: R3C2, R4C2, R3C3? No, line thick.
* $11+$ Cage: R3C2, R4C2, R5C2? No, R5C2 is start of $18\times$?
* Let's look at R5C2. Label $18\times$ is in R5C1? No, R5C1 is $18\times$.
* Cage $18\times$: R5C1, R6C1? Or R5C1, R5C2?
* Label $18\times$ is in R5C1. Box extends to R5C2? Line between R5C1-R5C2 is thin. Box extends to R6C1? Line between R5C1-R6C1 is thin.
* So $18\times$ is likely an L-shape: R5C1, R5C2, R6C1? Or R5C1, R6C1, R6C2?
* Let's look at R6C1. Label $28\times$ is in R6C1? No, R6C1 has no label. R7C1? No.
* Wait, the label $28\times$ is in R6C1? No, it's in R6C1's row?
* Let's read the labels in Column 1:
* R1C1: $22+$
* R3C1: $4-$
* R5C1: $18\times$
* R6C1: $28\times$ ?? No, the label $28\times$ is in the cell R6C1? Or R7C1?
* Actually, looking at the bottom left:
* R5C1 has label $18\times$.
* R6C1 has label $28\times$.
* This implies R5C1 and R6C1 are in DIFFERENT cages.
* So the line between R5C1 and R6C1 is THICK.
* Then R5C1 ($18\times$) must connect to R5C2? Line is thin. So R5C1-R5C2 is a cage? $18 = 2 \times 9$ (no), $3 \times 6$. Pair $\{3,6\}$.
* R6C1 ($28\times$) must connect to R6C2? Line is thin. And R7C1? Line is thin.
* So R6C1-R6C2-R7C1? $28 = 4 \times 7 \times 1$? Or $2 \times 2 \times 7$ (duplicate). $4 \times 7 \times 1$. Set $\{1,4,7\}$.

* Let's re-verify the middle section with this "Thin/Thick" logic.
* If R5C1-R6C1 is thick, then R5C1 is separate from R6C1.
* R5C1 ($18\times$) connects to R5C2. Does it connect to R4C1? R4C1 is in $4-$ cage with R3C1. Line R4C1-R5C1 is thick.
* So R5C1, R5C2 is a cage? $18\times$. Pair $\{3,6\}$.
* R6C1, R6C2, R7C1 is a cage? $28\times$. Values $\{1,4,7\}$.
* Now back to R3/R4 middle.
* R3C1-R4C1 is $4-$. Pair $\{1,5\}, \{2,6\}, \{3,7\}$.
* R3C2 ($11+$). Connects to R4C2? Thin line. Connects to R3C3? Thick line. Connects to R2C2? Thick line (R2C2 is in $22+$).
* So R3C2-R4C2 is a vertical pair? Sum 11. $\{4,7\}$ or $\{5,6\}$.
* If R3C2-R4C2 is the cage, what is R4C3?
* R4C3 connects to R4C2? Thin line. So R4C3 is in the same cage as R4C2?
* If R3C2, R4C2, R4C3 is the cage ($11+$), then R3C3 is separate.
* R3C3 ($4-$). Connects to R4C3? Thin line. BUT R4C3 is in $11+$. Conflict.
* Therefore, the line between R4C2 and R4C3 MUST be THICK.
* If R4C2-R4C3 is thick, then R3C2, R4C2 is the $11+$ cage. Sum 11. $\{4,7\}$ or $\{5,6\}$.
* Then R4C3 is free. Connects to R4C4? Thin line.
* R4C4 ($3\div$). Connects to R3C4? Thin line.
* So R3C4, R4C4, R4C3? No, R3C3 is still orphaned.
* R3C3 ($4-$). Connects to R3C4? Thick line. Connects to R4C3? Thin line.
* So R3C3, R4C3 is the $4-$ cage.
* Then R4C3 is in $4-$ cage.
* Then R4C3 cannot connect to R4C4. So line R4C3-R4C4 must be THICK.
* Then R4C4 ($3\div$) connects to R3C4? Thin line. So R3C4, R4C4 is the $3\div$ cage.
* This works!
* Summary of Middle-Left Cages:
* $11+$: R3C2, R4C2. (Sum 11).
* $4-$: R3C3, R4C3. (Diff 4).
* $3\div$: R3C4, R4C4. (Div 3).
* Lines Check:
* R3C2-R4C2: Thin (Same cage).
* R4C2-R4C3: Thick (Different cages).
* R3C3-R4C3: Thin (Same cage).
* R4C3-R4C4: Thick (Different cages).
* R3C4-R4C4: Thin (Same cage).
* R3C2-R3C3: Thick.
* R3C3-R3C4: Thick.
* R4C2-R3C2: Thin.
* R4C3-R3C3: Thin.
* R4C4-R3C4: Thin.
* This seems consistent with the visual weight of the lines.

Step 2: Solving with the Identified Cages

1. Row 2 Analysis:
* R2C3, R2C4 are $\{5,7\}$ ($12+$).
* R2C1, R2C2 are from $\{4,5,6,7\}$ ($22+$ cage).
* R2C5 is from $\{2,4,6,7\}$ ($336\times$ cage).
* R2C6, R3C6 are $2\div$.
* R2C7? Part of $4-$ cage with R1C7? No, R1C6-R1C7 is $4-$.
* R2C7 is part of $2\div$ cage? No, R2C6-R3C6 is $2\div$.
* R2C7 is part of $4-$ cage with R1C7? No, R1C6-R1C7 is horizontal.
* R2C7 is part of a cage with R3C7? No, R3C7 is $84\times$.
* R2C7 is part of a cage with R2C6? No, R2C6 is in $2\div$.
* Wait, R2C7 is isolated?
* R1C7 is in $4-$ (with R1C6).
* R2C7 neighbors: R1C7 (thick), R2C6 (thick? R2C6 is in $2\div$ with R3C6. Line R2C6-R2C7 is thick?), R3C7 (thick? R3C7 is $84\times$).
* If R2C7 is a 1-cell cage, it's invalid.
* Let's check R2C6-R2C7. Is it a cage? No label.
* Let's check R1C7-R2C7. Vertical pair? Label $4-$ is in R1C6. So R1C6-R1C7 is the cage.
* So R2C7 must connect to R3C7? If R2C7-R3C7 is thin, they are in $84\times$.
* $84\times$ cage: R2C7, R3C7, R4C7, R5C7?
* Let's check the height of $84\times$.
* Label $84\times$ is in R3C7.
* Cells: R3C7, R4C7, R5C7?
* If R2C7 is included, it's 4 cells. $84 = 2 \times 6 \times 7 \times 1$? Or $3 \times 4 \times 7 \times 1$?
* If R2C7 is NOT included, it's 3 cells (R3,4,5). $84 = 2 \times 6 \times 7$ or $3 \times 4 \times 7$.
* Let's look at R2C7's connections.
* R2C6 is in $2\div$ with R3C6. So R2C6-R3C6 is a vertical pair.
* R2C7 is to the right of R2C6. Line R2C6-R2C7 is likely THICK.
* R1C7 is above R2C7. Line R1C7-R2C7 is likely THICK (since R1C7 is in $4-$).
* R3C7 is below R2C7. Line R2C7-R3C7? If THICK, R2C7 is orphaned.
* So R2C7-R3C7 MUST be thin.
* Thus, R2C7 is part of the $84\times$ cage.
* So $84\times$ cage is R2C7, R3C7, R4C7, R5C7. (4 cells).
* Product 84. Factors from 1-7: $84 = 7 \times 6 \times 2 \times 1$ or $7 \times 4 \times 3 \times 1$ or $7 \times 3 \times 2 \times 2$ (dup) or $6 \times ? $.
* $84 = 7 \times 6 \times 2 \times 1$. Set $\{1,2,6,7\}$.
* $84 = 7 \times 4 \times 3 \times 1$. Set $\{1,3,4,7\}$.
* $84 = 7 \times 3 \times 2 \times 2$ (Invalid).
* $84 = 6 \times ? $. $84/6 = 14$. $14 = 7 \times 2 \times 1$. Set $\{1,2,6,7\}$.
* So the set is either $\{1,2,6,7\}$ or $\{1,3,4,7\}$.

2. Determining R1C6, R1C7 ($4-$)
* R1C6, R1C7 difference is 4. Pairs: $\{1,5\}, \{2,6\}, \{3,7\}$.
* R1C7 is in Col 7. Col 7 contains R2C7..R5C7 ($84\times$) and R6C7, R7C7.
* R1C7 cannot be in the $84\times$ cage.

3. Solving Row 1
* R1C1, R1C2 $\in \{4,5,6,7\}$.
* R1C3, R1C4 $\in \{1,4\}$ or $\{2,3\}$ ($5+$).
* R1C5 $\in \{2,4,6,7\}$ ($336\times$).
* R1C6, R1C7 $\in \{1,5\}, \{2,6\}, \{3,7\}$ ($4-$).

Let's test R1C3,R1C4.
Case A: $\{1,4\}$.
Then R1C3, R1C4 are 1,4.
R1C1, R1C2 are from $\{5,6,7\}$ (since 4 is used in R1C3/4? No, R1C1/2 are $\{4,5,6,7\}$ subset. If 4 is in R1C3, R1C1/2 cannot be 4? No, R1C1/2 are distinct from R1C3/4. So if R1C3=4, R1C1/2 $\neq 4$. So R1C1/2 $\in \{5,6,7\}$ plus one more from $\{4,5,6,7\}$? No, R1C1/2 are TWO cells. They take TWO values from $\{4,5,6,7\}$. R1C3/4 take TWO values from $\{1,2,3,4,5,6,7\}$.
If R1C3/4 is $\{1,4\}$, then 4 is used in Row 1.
So R1C1, R1C2 cannot be 4. They must be $\{5,6,7\}$ minus one? No, they are 2 cells. They must be $\{5,6\}, \{5,7\}, \{6,7\}$.
But R1C1/2 are part of $\{4,5,6,7\}$ group. If 4 is not in R1C1/2, it must be in R2C1/2.
R1C5 $\in \{2,6,7\}$ (4 used).
R1C6/7 pair from $\{1,5\}, \{2,6\}, \{3,7\}$.
If R1C3/4 is $\{1,4\}$, 1 is used. So R1C6/7 cannot be $\{1,5\}$.
So R1C6/7 is $\{2,6\}$ or $\{3,7\}$.

Case B: $\{2,3\}$.
R1C3/4 is $\{2,3\}$. 2,3 used.
R1C1/2 $\in \{4,5,6,7\}$.
R1C5 $\in \{4,6,7\}$ (2 used? No, R1C5 from $\{2,4,6,7\}$. If 2 is in R1C3/4, R1C5 $\neq 2$. So $\{4,6,7\}$).
R1C6/7 pair from $\{1,5\}, \{2,6\}, \{3,7\}$.
2,3 used. So $\{2,6\}$ invalid (2 used). $\{3,7\}$ invalid (3 used).
So R1C6/7 MUST be $\{1,5\}$.
This forces R1C6, R1C7 to be 1,5.
Then R1C5 must be from $\{4,6,7\}$.
R1C1, R1C2 from $\{4,6,7\}$? No, R1C1/2 from $\{4,5,6,7\}$. 5 used in R1C6/7. So R1C1/2 from $\{4,6,7\}$.
Remaining value for Row 1?
Values used: 1,5 (C6/7), 2,3 (C3/4).
Remaining for C1,2,5: $\{4,6,7\}$.
R1C1, R1C2 take two of $\{4,6,7\}$.
R1C5 takes the last one.
So R1C5 $\in \{4,6,7\}$.

Let's check Case A again ($\{1,4\}$).
R1C3/4 $\{1,4\}$.
R1C6/7 $\{2,6\}$ or $\{3,7\}$.
If $\{2,6\}$, 2,6 used.
R1C5 $\in \{7\}$? (From $\{2,4,6,7\}$, 2,4,6 used? 4 in C3/4, 2,6 in C6/7. So R1C5=7).
Then R1C1/2 from $\{5\}$? No, R1C1/2 from $\{4,5,6,7\}$. 4,6,7 used. So R1C1/2=$\{5, ?\}$. Only 5 left? Need 2 numbers.
Available for R1: $\{1,2,3,4,5,6,7\}$.
Used: 1,4 (C3/4), 2,6 (C6/7), 7 (C5).
Remaining