Solve the number pattern in the grid.

Let's analyze the pattern in this grid. We have a 4×3 table (4 rows, 3 columns), and we're to find the missing number in the bottom-right cell.

Here is the table:

```
| | |
6 | 5 | 12
----|---|----
12 |15 | 36
----|---|----
36 |60 |144
----|---|----
144 |300| ?
```

We are to find the value of the question mark in the last row, third column.

---

Step 1: Look at the pattern in each column.

#### Column 1:
- Row 1: 6
- Row 2: 12 → 6 × 2 = 12
- Row 3: 36 → 12 × 3 = 36
- Row 4: 144 → 36 × 4 = 144

So the pattern is:
Each term is multiplied by increasing integers:
6 → 12 (×2) → 36 (×3) → 144 (×4)

So it’s multiplying by 2, then 3, then 4.

This suggests that each row is generated by multiplying the previous row by the row number (starting from row 2).

But more importantly, notice:
- 6 × 2 = 12
- 12 × 3 = 36
- 36 × 4 = 144

So:
Row 1: 6
Row 2: 6 × 2 = 12
Row 3: 12 × 3 = 36
Row 4: 36 × 4 = 144

Yes, so each row is previous × (row number). But row numbers start at 1? Let's index them:

- Row 1: 6
- Row 2: 6 × 2 = 12
- Row 3: 12 × 3 = 36
- Row 4: 36 × 4 = 144

So the first column follows:
Each entry is the previous one multiplied by the row index.

Let’s check if the same applies to other columns.

---

#### Column 2:
- Row 1: 5
- Row 2: 15 → 5 × 3 = 15
- Row 3: 60 → 15 × 4 = 60
- Row 4: 300 → 60 × 5 = 300

Wait — not exactly multiplying by row number.

From row 1 to row 2: 5 → 15 → ×3
Row 2 to row 3: 15 → 60 → ×4
Row 3 to row 4: 60 → 300 → ×5

So multipliers: ×3, ×4, ×5 — that is, multiplier increases by 1 each time.

So the pattern is:
- Row 1: 5
- Row 2: 5 × 3 = 15
- Row 3: 15 × 4 = 60
- Row 4: 60 × 5 = 300

So it's like: multiply by (row number + 2)? Let's see:

- From row 1 to row 2: ×3 = (2+1)
- Row 2 to row 3: ×4 = (3+1)
- Row 3 to row 4: ×5 = (4+1)

So each step multiplies by (current_row_number + 1)

Alternatively, think of it as:
- Row n: value = previous × (n + 1) for n ≥ 2?

But let's see if there's a better way.

Wait — perhaps the entire pattern is based on multiplication across rows and columns.

Let’s look at each row.

---

Step 2: Analyze rows

#### Row 1: 6, 5, 12

Is there a relationship between these?

Try: 6 × 2 = 12 → but middle is 5.

6 + 5 = 11 ≠ 12

6 × 5 = 30 → too big

What about: 6 × 2 = 12, but where does 5 come in?

Not obvious.

#### Row 2: 12, 15, 36

12 × 3 = 36 → again, middle is 15

12 + 15 = 27 ≠ 36

15 × 2.4 = 36 → not helpful

Wait — what if we consider column-wise multiplication?

Let’s go back.

---

Step 3: Look at column patterns again

We already saw:

#### Column 1:
- 6 → 12 → 36 → 144
- 6 × 2 = 12
- 12 × 3 = 36
- 36 × 4 = 144

So: ×2, ×3, ×4 → multipliers increase by 1

#### Column 2:
- 5 → 15 → 60 → 300
- 5 × 3 = 15
- 15 × 4 = 60
- 60 × 5 = 300

So: ×3, ×4, ×5 → again increasing by 1

So the multiplier from row to row is increasing by 1, but starting at different values.

Now column 3:

#### Column 3:
- 12 → 36 → 144 → ?

Let’s compute ratios:

- 12 → 36 → 36 / 12 = 3
- 36 → 144 → 144 / 36 = 4

So: ×3, then ×4

So likely next is ×5 → 144 × 5 = 720

So the missing number might be 720

Let’s test this hypothesis.

So pattern in column 3:
- Row 1: 12
- Row 2: 12 × 3 = 36
- Row 3: 36 × 4 = 144
- Row 4: 144 × 5 = 720

Yes — consistent with the others.

Now compare to other columns:

- Col 1: ×2, ×3, ×4 → so multiplier starts at 2
- Col 2: ×3, ×4, ×5 → starts at 3
- Col 3: ×3, ×4, ×5 → wait! Row 1 to row 2: ×3, row 2 to row 3: ×4 → so also starts at 3?

Wait — col 1: multiplier from row1→row2 is ×2

Col 2: row1→row2 is ×3

Col 3: row1→row2 is ×3

So col 1 has multiplier ×2, while others start at ×3.

But col 1:
- 6 → 12 → ×2
- 12 → 36 → ×3
- 36 → 144 → ×4

So multipliers: ×2, ×3, ×4

Col 2:
- 5 → 15 → ×3
- 15 → 60 → ×4
- 60 → 300 → ×5

Multipliers: ×3, ×4, ×5

Col 3:
- 12 → 36 → ×3
- 36 → 144 → ×4
- 144 → ? → ×5 → 720

So yes — all columns follow the pattern:
From row n to row n+1, multiply by (n+1) for n=1,2,3

Wait:

- From row 1 to row 2: multiply by 2 (for col 1), by 3 (for col 2 and 3)
- From row 2 to row 3: multiply by 3 (col 1), by 4 (col 2 and 3)
- From row 3 to row 4: multiply by 4 (col 1), by 5 (col 2 and 3)

But col 1 is different: it uses ×2, ×3, ×4

Others use ×3, ×4, ×5

So maybe not uniform.

But let's check if the multiplication factor depends on the column.

Alternatively, perhaps the value in each cell is related to the product of row and column indices?

Let’s try indexing rows and columns.

Let’s define:

- Row 1: R1
- Row 2: R2
- Row 3: R3
- Row 4: R4

- Col 1: C1
- Col 2: C2
- Col 3: C3

Let’s suppose the value at (i,j) follows some formula.

Look at C1:

- R1C1 = 6
- R2C1 = 12 = 6 × 2
- R3C1 = 36 = 12 × 3
- R4C1 = 144 = 36 × 4

So: R1C1 = 6
R2C1 = 6 × 2
R3C1 = 6 × 2 × 3
R4C1 = 6 × 2 × 3 × 4 = 6 × 24 = 144 ✔

So RnC1 = 6 × (2 × 3 × ... × n) for n ≥ 2?

Wait:

- R1C1 = 6
- R2C1 = 6 × 2 = 12
- R3C1 = 12 × 3 = 36 = 6 × 2 × 3
- R4C1 = 36 × 4 = 144 = 6 × 2 × 3 × 4

So general: RnC1 = 6 × (product from k=2 to n of k) = 6 × (n! / 1!) = 6 × n! / 1? Wait:

Actually:
For n=1: 6
n=2: 6 × 2 = 12
n=3: 6 × 2 × 3 = 36
n=4: 6 × 2 × 3 × 4 = 144

So: RnC1 = 6 × (n!) / 1! ? For n=4: 6 × 24 = 144 → yes
But for n=1: 6 × 1! = 6 → ok
But 6 × 1! = 6 → yes
n=2: 6 × 2! = 6 × 2 = 12 → yes
n=3: 6 × 6 = 36 → yes
n=4: 6 × 24 = 144 → yes

Wait — so RnC1 = 6 × n!

But 6 × 1! = 6 × 1 = 6
6 × 2! = 6 × 2 = 12
6 × 3! = 6 × 6 = 36
6 × 4! = 6 × 24 = 144 → yes!

So Column 1: Value = 6 × (row number)!

Let’s check Column 2:

- R1C2 = 5
- R2C2 = 15
- R3C2 = 60
- R4C2 = 300

Check if it's 5 × n!:

- n=1: 5 × 1! = 5 → matches
- n=2: 5 × 2! = 5 × 2 = 10 → but actual is 15 → no

No.

Try: 5 × something

R1: 5
R2: 15 = 5 × 3
R3: 60 = 15 × 4 = 5 × 3 × 4
R4: 300 = 60 × 5 = 5 × 3 × 4 × 5

So: R1C2 = 5
R2C2 = 5 × 3
R3C2 = 5 × 3 × 4
R4C2 = 5 × 3 × 4 × 5

So: RnC2 = 5 × (product from k=3 to n of k) for n ≥ 3?

Wait:

- R1: 5
- R2: 5 × 3
- R3: 5 × 3 × 4
- R4: 5 × 3 × 4 × 5

So it's: 5 × 3 × 4 × 5 × ... up to row number?

But R1: just 5
R2: 5 × 3
R3: 5 × 3 × 4
R4: 5 × 3 × 4 × 5

So it's like: 5 × (3 × 4 × 5 × ... × n) for n ≥ 2?

But for R2: 5 × 3 = 15 → yes
R3: 5 × 3 × 4 = 60 → yes
R4: 5 × 3 × 4 × 5 = 300 → yes

But R1: 5 → which would be 5 × (nothing?) → so base case.

So RnC2 = 5 × (product from k=3 to n of k) for n ≥ 3, and R1C2 = 5, R2C2 = 5×3

But note: 3 to n includes 3,4,...,n

So for n=2: only k=3? But 3 > 2 → so empty product → 1 → 5×1=5 → but actual is 15 → doesn't work.

Alternative idea: RnC2 = 5 × (n+1)! / 2!? Let's try.

n=1: 5 × 2! / 2! = 5 → ok
n=2: 5 × 3! / 2! = 5 × 6 / 2 = 15 → ok
n=3: 5 × 4! / 2! = 5 × 24 / 2 = 5 × 12 = 60 → ok
n=4: 5 × 5! / 2! = 5 × 120 / 2 = 5 × 60 = 300 → ok!

Perfect!

So RnC2 = 5 × (n+1)! / 2!

But 2! = 2, so RnC2 = 5 × (n+1)! / 2

Similarly, check column 3:

R1C3 = 12
R2C3 = 36
R3C3 = 144
R4C3 = ?

Try: RnC3 = ? × (n+1)! / something

Try: RnC3 = 12 × (n+1)! / 3!? Let's see.

n=1: 12 × 2! / 6 = 12 × 2 / 6 = 4 → no

Try: RnC3 = 12 × (n+1)! / 2! ?
n=1: 12 × 2! / 2 = 12 × 2 / 2 = 12 → ok
n=2: 12 × 3! / 2 = 12 × 6 / 2 = 36 → ok
n=3: 12 × 4! / 2 = 12 × 24 / 2 = 144 → ok
n=4: 12 × 5! / 2 = 12 × 120 / 2 = 12 × 60 = 720 → so answer is 720

Yes!

So the pattern is:

For column j, the value in row n is:

- C1: 6 × n!
- C2: 5 × (n+1)! / 2
- C3: 12 × (n+1)! / 2

Wait, but that seems arbitrary.

Alternatively, notice:

In column 3:
- R1: 12
- R2: 36 = 12 × 3
- R3: 144 = 36 × 4
- R4: ? = 144 × 5 = 720

Same as before.

And since the ratio pattern is consistent: ×3, ×4, ×5

And earlier columns also follow:

- C1: ×2, ×3, ×4 → so next would be ×5 → 144 × 5 = 720
- C2: ×3, ×4, ×5 → next ×6 → 300 × 6 = 1800 (but not needed)

So for column 3, the pattern is:

- Row 1: 12
- Row 2: 12 × 3 = 36
- Row 3: 36 × 4 = 144
- Row 4: 144 × 5 = 720

So the multiplier increases by 1 each time, starting from ×3.

Thus, the missing number is 720

---

Final Answer:

$$
\boxed{720}
$$

✔ The pattern in each column is multiplicative, with the multiplier increasing by 1 each row:

- Column 3: 12 → 36 (×3) → 144 (×4) → ? (×5) = 144 × 5 = 720