Day 158-161 - A Puzzle-Themed Week | Mr. Hussains 180 Blog - Free Printable

The image shows a person working on KenKen puzzles, specifically 5x5 grids. KenKen is a logic and arithmetic puzzle similar to Sudoku, where the goal is to fill a grid with numbers so that:

- Each row and column contains the digits 1 through 5 (for a 5x5 grid) without repetition.
- The numbers in each "cage" (a group of cells outlined by bold lines) must combine using the given operation (addition, subtraction, multiplication, or division) to produce the target number shown in the cage.

Let’s analyze one of the puzzles in the foreground (the leftmost one), which has some initial clues and partial work already done.

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Puzzle Layout (Left 5x5 KenKen Grid):

We’ll label the grid positions as (row, column), starting from top-left as (1,1).

Here's the layout of the cages and their targets:

```
(1,1) to (1,3): 10+ → sum of three cells = 10
(1,4) to (1,5): 3- → difference = 3
(2,1): 2| → division: one number divided by another = 2
(2,2) to (2,3): 6× → product = 6
(2,4): 2- → difference = 2
(3,1): 2- → difference = 2
(3,2) to (3,3): 5+ → sum = 5
(3,4) to (3,5): 6+ → sum = 6
(4,1): 5× → product = 5
(4,2): 2| → division = 2
(4,3): 6+ → sum = 6
(5,1): 6÷ → division = 6 (likely means 6/1 = 6, or 6/x = y)
(5,2) to (5,5): 15× → product = 15
```

Wait — let's clarify: In KenKen, the notation like `6÷` usually means the result of dividing two numbers in the cage is 6. But since we’re dealing with small integers (1–5), possible divisions are limited.

But looking at the actual handwritten notes:
- There is a “4” written in (4,1), and “5×” in (4,1), suggesting that cell (4,1) might be part of a 5× cage. But if it's a single cell, then the value must be 5. However, the "5×" is likely a cage spanning multiple cells.

Let’s re-analyze the left puzzle carefully.

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Step-by-step Solution Attempt

Let’s focus on the left 5x5 puzzle because it has more visible clues.

#### Cage Breakdown (Left Puzzle):

1. Top-left 3-cell cage: (1,1), (1,2), (1,3) → 10+
- Sum of three distinct numbers from 1–5 = 10
- Possible combinations: 1+4+5=10, 2+3+5=10
- So possibilities: {1,4,5} or {2,3,5}

2. (1,4)-(1,5): 3-
- Two cells, difference = 3
- Possible pairs: (1,4), (2,5), (4,1), (5,2)

3. (2,1): 2|
- Division: one number divided by the other = 2 → possible pairs: (1,2), (2,1), (2,4), (4,2), (3,6) invalid → only valid: (1,2), (2,1), (2,4), (4,2)
- Since values are 1–5, possible: (1,2), (2,1), (2,4), (4,2)

4. (2,2)-(2,3): 6×
- Product = 6 → possible pairs: (1,6) invalid, (2,3), (3,2), (1,6) no → so (2,3) or (3,2)

5. (2,4): 2-
- Difference = 2 → pairs: (1,3), (2,4), (3,1), (3,5), (4,2), (5,3)

6. (3,1): 2-
- Difference = 2 → same as above

7. (3,2)-(3,3): 5+
- Sum = 5 → possible pairs: (1,4), (2,3), (3,2), (4,1)

8. (3,4)-(3,5): 6+
- Sum = 6 → possible: (1,5), (2,4), (3,3), (4,2), (5,1)

9. (4,1): 5×
- Single cell? Then value must be 5. But “5×” suggests a product cage. Likely it's a single cell with value 5.

Wait — actually, looking at the image: (4,1) has a "5×" label, and there's a "4" written in it. That seems inconsistent unless it's a typo.

But wait — perhaps the "5×" cage includes (4,1) and (5,1)? Let’s check.

Actually, (4,1) has a "5×" label and the handwritten "4", but 4 ≠ 5, so contradiction.

Alternatively, maybe the "5×" cage is just (4,1) and (5,1), both together.

Looking at the image: the cage labeled "5×" appears to span (4,1) and (5,1). That makes sense.

So:
- (4,1)-(5,1): 5× → product = 5 → possible pairs: (1,5), (5,1)

10. (4,2): 2|
- Division = 2 → possible: (1,2), (2,1), (2,4), (4,2)

11. (4,3): 6+
- Sum = 6 → possible: (1,5), (2,4), (3,3), (4,2), (5,1)

12. (5,1): 6÷
- Division = 6 → only possible if 6/1 = 6, so one number is 6, but max is 5 → impossible?

Wait — this is confusing.

Actually, 6÷ likely means the quotient is 6. But 6 is greater than 5, so impossible.

Ah! But the notation may mean the result of the division is 6, but that can't happen with numbers ≤5.

So perhaps it's a typo, or I misread.

Wait — in the image, (5,1) has "6÷" written, but also a "4" written in it.

But if (4,1) and (5,1) form a 5× cage, then their product is 5 → possible only if (1,5) or (5,1)

But (5,1) has a "6÷" label? That doesn’t make sense.

Wait — perhaps the "6÷" is for a different cage.

Looking again: the bottom-left corner has a cage labeled "6÷" and "15×".

Wait — in the image, the bottom-left cage (cells (5,1)) is labeled "6÷", but the handwritten note says "4" in (4,1), and (5,1) has no number yet.

But the cage labeled "5×" appears to be (4,1) and (5,1), so their product is 5 → only possible if one is 1 and the other is 5.

But (5,1) is also in a "6÷" cage? That can't be.

Wait — perhaps the "6÷" is a separate cage.

Let me try to map the cages properly.

From standard KenKen layout:

In the left puzzle:

- Top-left: (1,1), (1,2), (1,3) → 10+
- (1,4), (1,5) → 3-
- (2,1) → 2|
- (2,2), (2,3) → 6×
- (2,4) → 2-
- (3,1) → 2-
- (3,2), (3,3) → 5+
- (3,4), (3,5) → 6+
- (4,1) → 5× ← possibly single cell?
- (4,2) → 2|
- (4,3) → 6+
- (5,1) → 6÷
- (5,2), (5,3), (5,4), (5,5) → 15×

But that would make (5,1) a single cell with "6÷" — impossible.

Alternatively, perhaps the "6÷" is a cage involving (5,1) and (5,2)? But (5,2) is part of the 15× cage.

Wait — look at the image: the bottom-right cage spans (5,2) to (5,5) → 15× → four cells → product = 15.

15 = 1×3×5 or 1×1×3×5, but all numbers 1–5, no repeats.

Possible: 1,1,3,5 → but duplicates not allowed → so must be 1,3,5 and one more number → 15 = 1×3×5×1 → duplicate 1 → invalid.

Wait: 15 = 1×3×5×1 → no.

Or 1×1×3×5 → still duplicate.

But 15 = 3×5×1×1 → same.

No way to get 15 from four distinct numbers 1–5?

Wait: 1×3×5 = 15 → need four numbers → so one of them must be 1, and others multiply to 15.

So: 1,1,3,5 → duplicate 1 → invalid.

Or 1,3,5,1 → same.

So no solution? Impossible.

But wait — 15× could be achieved with 1,1,3,5 → but duplicates not allowed.

Unless the cage has fewer cells.

Looking at the image: the "15×" cage appears to cover (5,2), (5,3), (5,4), (5,5) — four cells.

But 15 cannot be made from four distinct numbers from 1–5 without repeating.

Wait: 1×3×5 = 15 → needs three numbers.

But four cells → must have a repeated number? Not allowed.

So contradiction.

Unless one of the numbers is 1, and others are 3,5,1 — still duplicate.

Wait — 15 = 1×3×5×1 → duplicate 1 → invalid.

Or 15 = 3×5×1×1 → same.

Is there any combination of four distinct numbers from 1–5 whose product is 15?

Try: 1×2×3×5 = 30 → too big

1×2×3×4 = 24

1×2×3×1 = 6

1×3×5×1 = 15 → but duplicate 1 → invalid

So impossible.

Therefore, the "15×" cage must have only three cells.

Looking back at the image: the "15×" cage is in the bottom row, but it may only include (5,2), (5,3), (5,4) — not (5,5)?

But in the image, the bold line seems to go from (5,2) to (5,5), so four cells.

Alternatively, maybe the puzzle has a typo.

But wait — perhaps the "15×" is a 3-cell cage: (5,2), (5,3), (5,4), and (5,5) is separate.

But in the image, the cage is drawn as a single block from (5,2) to (5,5).

Alternatively, maybe the numbers are not all distinct? But KenKen rules require no repeats in rows/columns.

Another possibility: the "15×" is a 2-cell cage? But 15 = 3×5 → possible.

But visually, it's a large cage.

Perhaps it's a 3-cell cage: 1×3×5 = 15 → yes!

So if the "15×" cage has three cells, say (5,2), (5,3), (5,4), and (5,5) is separate.

But in the image, it looks like it spans all four.

This suggests the puzzle may have an error, or my interpretation is wrong.

Alternatively, maybe the "15×" is meant to be 15 in a 3-cell cage.

But let’s assume it’s a 3-cell cage: (5,2), (5,3), (5,4) → product = 15 → possible: 1,3,5 in some order.

Then (5,5) is separate.

But in the image, the cage outline goes to (5,5), so likely it's included.

Wait — perhaps the "15×" is for (5,2), (5,3), (5,4), and (5,5) is part of a different cage.

But the "6÷" is in (5,1), and "5×" in (4,1) and (5,1)? No.

Let’s try to resolve this.

After careful analysis of the image, here’s what I observe:

In the left puzzle, the bottom-left area:

- (4,1) has a "5×" label and a handwritten "4"
- (5,1) has a "6÷" label and no number
- (5,2) to (5,5) has "15×"

But if (4,1) is "5×", and it's a single cell, then it must be 5, but handwritten is "4" — contradiction.

Unless "5×" is a cage including (4,1) and (5,1), so their product is 5.

So (4,1) × (5,1) = 5 → possible pairs: (1,5), (5,1)

But (4,1) has "4" written — not 1 or 5 — so contradiction.

Unless the "4" is a mistake.

Perhaps the person is trying to solve it, and the "4" is a guess.

But 4 is not a factor of 5, so impossible.

So either:
- The "5×" is not a two-cell cage, or
- The "4" is incorrect.

But the "5×" label is near (4,1), and the cage extends down to (5,1), so likely it's a two-cell cage.

So (4,1) × (5,1) = 5 → only possible if one is 1, the other is 5.

So (4,1) must be 1 or 5, and (5,1) the other.

But handwritten "4" in (4,1) is wrong.

Similarly, (5,1) has "6÷" — division = 6.

But if (5,1) is 1 or 5, then division by what?

If (5,1) is 5, and (5,2) is 1, then 5÷1=5, not 6.

6÷ requires quotient 6, which is impossible with numbers ≤5.

So "6÷" must be a cage with two cells: e.g., (5,1) and (5,2), with (5,1)/(5,2)=6 or vice versa.

But 6 is >5, so impossible.

Unless it's (5,2)/(5,1)=6 → then (5,2)=6, impossible.

So "6÷" is impossible.

Unless it's a typo, and it's "6÷" meaning the result is 6, but that can't happen.

Wait — in some KenKen variants, "6÷" means the quotient is 6, but it's impossible.

Alternatively, it might be "6÷" as in the result is 6, but with subtraction or something else.

No — the symbol is "÷", so division.

So this suggests the puzzle might have a typo.

Alternatively, perhaps the "6÷" is for a cage with (5,1) and (5,2), and the intended answer is 6, but with numbers 3 and 2: 6/1=6, but 6 not allowed.

Wait — perhaps the "6÷" is a mistake, and it's "1÷" or "2÷".

But in the image, it's clearly "6÷".

Given the confusion, let’s switch to the right puzzle, which appears to have more progress.

On the right puzzle, there is a "15×" cage in the bottom-left, and handwritten "3" and "5" in (4,1) and (5,1), respectively.

And the "15×" cage is likely (4,1) and (5,1) → 3×5=15 → possible.

So perhaps in the left puzzle, the "5×" is a typo, and it should be "15×" or something.

But in the left puzzle, the "5×" is near (4,1), and "6÷" near (5,1), which is conflicting.

Given the complexity and the fact that the image shows a person actively solving, and the puzzle may have errors or be partially filled incorrectly, I will provide a general approach to solving KenKen puzzles.

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General Strategy for Solving KenKen:

1. Understand the rules:
- Fill each row and column with numbers 1 to 5 (for 5x5).
- No repeats in any row or column.
- Each cage has a target number and operation.
- Numbers in the cage must combine (using the operation) to give the target.

2. Start with cages that have few possibilities:
- For example, a "2-" cage in two cells: possible pairs are (1,3), (2,4), (3,1), (4,2), (3,5), (5,3) — but only if numbers are within range.
- A "2|" (division) cage: possible pairs: (1,2), (2,1), (2,4), (4,2)

3. Use elimination:
- If a cage must contain certain numbers, eliminate those from other cells in the same row/column.

4. Look for unique combinations:
- For example, a "15×" in three cells: only possible with 1,3,5 in some order.

5. Check for conflicts:
- If a cell must be 5 due to a cage, ensure it doesn't violate row/column uniqueness.

6. Work step by step:
- Start with cages that have only one possible combination.

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Example: Solve a Simple Cage

Take the "3-" cage in the top-right of the left puzzle: (1,4) and (1,5) → difference = 3.

Possible pairs: (1,4), (2,5), (4,1), (5,2)

But in row 1, we already have a 10+ cage in (1,1)-(1,3), which must sum to 10.

Possible sums: 1+4+5=10 or 2+3+5=10

So row 1 must contain 1,4,5 or 2,3,5 in the first three cells.

For the last two cells (1,4) and (1,5), they must be the remaining numbers.

If first three are 1,4,5, then (1,4) and (1,5) must be 2 and 3 → but |2-3|=1 ≠ 3 → not possible.

If first three are 2,3,5, then remaining are 1 and 4 → |1-4|=3 → works.

So (1,4) and (1,5) must be 1 and 4 in some order.

So the "3-" cage is satisfied.

Now, in row 1, cells (1,1)-(1,3) are 2,3,5 in some order.

So we can eliminate 1 and 4 from those cells.

This is how you solve it.

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Conclusion

The image shows a person solving KenKen puzzles, which are logic-based arithmetic challenges. The key is to use the operations and constraints to deduce possible values, ensuring no row or column repeats.

While the specific puzzle has some unclear or potentially conflicting labels (like "6÷" and "5×" in adjacent cells), the general method involves:

- Identifying cages and their requirements
- Listing possible number combinations
- Using row/column uniqueness to eliminate options
- Filling in cells step by step

If you'd like, you can send a clearer image or describe the exact puzzle, and I can solve it completely.