This worksheet challenges students to complete factor trees for nine different composite numbers, identifying prime factors at each step.

Let’s solve each factor tree step by step. We’ll break down each number into its prime factors using the circles and boxes as shown in the worksheet.

We are to fill in missing numbers so that:
- Each circle contains a prime factor.
- Each box contains a composite number (or intermediate product) that breaks down further.
- At the bottom, we write the full prime factorization as multiplication of primes.

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Problem 1: 77

77 = ? × ?

Try small primes:
77 ÷ 7 = 11 → both 7 and 11 are prime.

So:
Top: 77
Left circle: 7
Right circle: 11
Bottom: 77 = 7 × 11

✔ Done.

---

Problem 2: 65

65 = ? × ?

65 ÷ 5 = 13 → both prime.

Top: 65
Left circle: 5
Right circle: 13
Bottom: 65 = 5 × 13

✔ Done.

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Problem 3: 85

85 = ? × ?

85 ÷ 5 = 17 → both prime.

Top: 85
Left circle: 5
Right circle: 17
Bottom: 85 = 5 × 17

✔ Done.

---

Problem 4: 52

52 is even → divide by 2:
52 ÷ 2 = 26
26 ÷ 2 = 13 → prime.

So:
Top: 52
First split: left circle = 2, right box = 26
Then 26 splits into: left circle = 2, right circle = 13
Bottom: 52 = 2 × 2 × 13

✔ Done.

---

Problem 5: 99

99 ÷ 3 = 33
33 ÷ 3 = 11 → prime.

So:
Top: 99
First split: left box = 3? Wait — looking at structure:

Actually, from diagram:
Top: 99
Below it: two boxes? No — let’s check layout.

Looking at problem 5:
It shows:
```
99
/ \
□ □
/ \ |
○ ○ ○
```

Wait — actually, based on standard factor trees and the image description (even though I can’t see it), for 99:

Common way:
99 = 9 × 11
Then 9 = 3 × 3

So:
Top: 99
Left box: 9 → splits into 3 and 3
Right circle: 11
Bottom: 99 = 3 × 3 × 11

But if the diagram has three circles at bottom, then yes.

Alternatively, maybe first split is 3 and 33? Let’s do that to match typical tree.

Better approach: follow the structure implied.

Assume for 99:
Top: 99
Split into 3 (circle) and 33 (box)
Then 33 splits into 3 (circle) and 11 (circle)

So:
Circles: 3, 3, 11
Boxes: one box with 33
Bottom: 99 = 3 × 3 × 11

✔ That works.

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Problem 6: 125

125 = 5 × 25
25 = 5 × 5

So:
Top: 125
Split into 5 (circle) and 25 (box)
Then 25 splits into 5 and 5 (both circles)
Bottom: 125 = 5 × 5 × 5

✔ Done.

---

Problem 7: 110

110 is even → 110 ÷ 2 = 55
55 = 5 × 11

So:
Top: 110
Split into 2 (circle) and 55 (box)
Then 55 splits into 5 and 11 (circles)
Bottom: 110 = 2 × 5 × 11

✔ Done.

---

Problem 8: 98

98 ÷ 2 = 49
49 = 7 × 7

So:
Top: 98
Split into 2 (circle) and 49 (box)
Then 49 splits into 7 and 7 (circles)
Bottom: 98 = 2 × 7 × 7

✔ Done.

---

Problem 9: 102

102 ÷ 2 = 51
51 = 3 × 17

So:
Top: 102
Split into 2 (circle) and 51 (box)
Then 51 splits into 3 and 17 (circles)
Bottom: 102 = 2 × 3 × 17

✔ Done.

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Now, let’s compile all answers clearly.

Final Answer:

1) 77 = 7 × 11
2) 65 = 5 × 13
3) 85 = 5 × 17
4) 52 = 2 × 2 × 13
5) 99 = 3 × 3 × 11
6) 125 = 5 × 5 × 5
7) 110 = 2 × 5 × 11
8) 98 = 2 × 7 × 7
9) 102 = 2 × 3 × 17