Prime factorization of 20123648 shown through a step-by-step tree diagram.

The image shows a factor tree for the number 20123648, breaking it down into its prime factors. Let's analyze and verify the solution step by step.

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🔍 Step 1: Understand the Factor Tree

We start with:
```
20123648
/ \
4352 4624
/ \ / \
64 68 68 68
/ \ / \ / \ / \
8 8 4 17 4 17 4 17
/ \ / \ / \ / \ / \ / \ / \ / \
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
```

But wait — let’s look carefully at the structure:

- 20123648 splits into 4352 × 4624
- Then:
- 4352 = 64 × 68
- 4624 = 68 × 68

So:
- 4352 = 64 × 68
- 4624 = 68 × 68

Let’s now break down each component.

---

🔍 Step 2: Break Down Each Number

#### ✔ Break down 64:
64 = 8 × 8
8 = 4 × 2 → 4 = 2 × 2
So:
- 8 = 2 × 2 × 2 = $2^3$
- So 64 = $8 × 8 = (2^3) × (2^3) = 2^6$

✔ So 64 = $2^6$

#### ✔ Break down 68:
68 = 4 × 17
4 = 2 × 2 = $2^2$
So: 68 = $2^2 × 17$

✔ So 68 = $2^2 × 17$

Now plug back in:

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🔍 Step 3: Compute Full Factorization

#### First branch: 4352 = 64 × 68
= $2^6 × (2^2 × 17)$ = $2^{8} × 17$

#### Second branch: 4624 = 68 × 68
= $(2^2 × 17) × (2^2 × 17)$ = $2^{4} × 17^2$

Now multiply both parts together:
- Total = 4352 × 4624 = $(2^8 × 17) × (2^4 × 17^2)$
- Combine powers:
- $2^{8+4} = 2^{12}$
- $17^{1+2} = 17^3$

✔ So, 20123648 = $2^{12} × 17^3$

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✔ Final Answer:

$$
\boxed{2^{12} \times 17^3}
$$

This matches what is written at the bottom of the image.

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📌 Verification (Optional Check):

Let’s compute $2^{12} × 17^3$ to verify:

- $2^{12} = 4096$
- $17^2 = 289$, so $17^3 = 289 × 17 = 4913$

Now:
- $4096 × 4913$

Let’s compute:
- $4096 × 4913 = ?$

Break it down:
- $4096 × 5000 = 20,480,000$
- $4096 × 87 = ?$ → Wait, better:
- Actually: $4913 = 5000 - 87$

So:
- $4096 × 4913 = 4096 × (5000 - 87) = 4096×5000 - 4096×87$

Compute:
- $4096 × 5000 = 20,480,000$
- $4096 × 87$:
$4096 × 80 = 327,680$
$4096 × 7 = 28,672$
Sum: $327,680 + 28,672 = 356,352$

Now subtract:
- $20,480,000 - 356,352 = 20,123,648$

✔ Matches: 20,123,648

---

✔ Conclusion:

The prime factorization of 20123648 is:
$$
\boxed{2^{12} \times 17^3}
$$

This is correct, and the factor tree correctly leads to this result. The highlighted numbers (2s and 17s) are all prime, and the exponents are properly counted.

✔️ Answer is verified and correct.