Math worksheet focusing on LCM exercises with number lines, prime factorization, and long division methods.

Let's solve each problem step-by-step from the worksheet on LCM (Least Common Multiple).

---

Problem 1: Find the first 3 common multiples of the given using a number line.

#### (i) 3 and 4

Step 1: List multiples of each number
- Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, ...
- Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, ...

Common multiples: Numbers that appear in both lists.
- First 3 common multiples: 12, 24, 36

✔ Answer: 12, 24, 36

> 💡 On a number line, you would mark multiples of 3 and 4 and look for overlapping points.

---

#### (ii) 5 and 6

Multiples of 5: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, ...
Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, ...

Common multiples: 30, 60, 90, ...

✔ Answer: 30, 60, 90

---

✔ Problem 1 Answers:

- (i) 12, 24, 36
- (ii) 30, 60, 90

---

Problem 2: Find the LCM by Prime Factorization Method

We break down each number into its prime factors, then take the highest power of each prime.

---

#### (i) 12 and 20

Prime factorization:
- 12 = 2² × 3¹
- 20 = 2² × 5¹

Now take the highest powers:
- 2², 3¹, 5¹

So,
$$
\text{LCM} = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60
$$

✔ Answer: 60

---

#### (ii) 36, 27, and 18

Prime factorizations:
- 36 = 2² × 3²
- 27 = 3³
- 18 = 2¹ × 3²

Take highest powers:
- 2² (from 36)
- 3³ (from 27)

$$
\text{LCM} = 2^2 \times 3^3 = 4 \times 27 = 108
$$

✔ Answer: 108

---

#### (iii) 32 and 48

Prime factorizations:
- 32 = 2⁵
- 48 = 2⁴ × 3¹

Highest powers:
- 2⁵, 3¹

$$
\text{LCM} = 2^5 \times 3 = 32 \times 3 = 96
$$

✔ Answer: 96

---

✔ Problem 2 Answers:

- (i) 60
- (ii) 108
- (iii) 96

---

Problem 3: Find the LCM by Long Division Method

We divide all numbers simultaneously by prime numbers until all quotients are 1.

---

#### (i) 70, 110, 150

We'll use long division method:

```
| 70 110 150
--------------------
2 | 35 55 75
--------------------
5 | 7 11 15
--------------------
7 | 1 11 3
--------------------
11 | 1 1 3
--------------------
3 | 1 1 1
--------------------
| 1 1 1
```

Wait — this is not correct because we can't divide 7 and 11 at the same time. Let's do it properly.

We need to divide all three numbers together with common primes only.

Start over:

```
| 70 110 150
--------------------
2 | 35 55 75
--------------------
5 | 7 11 15
--------------------
7 | 1 11 3
--------------------
11 | 1 1 3
--------------------
3 | 1 1 1
--------------------
| 1 1 1
```

But now, we see that after dividing by 2 and 5, the remaining numbers are 7, 11, and 15.

We must divide only when all numbers are divisible by the same prime.

So let’s do it correctly:

Step 1: Divide by 2 (common divisor of 70, 110, 150)
- 70 ÷ 2 = 35
- 110 ÷ 2 = 55
- 150 ÷ 2 = 75

Now: 35, 55, 75

Step 2: Divide by 5 (common divisor)
- 35 ÷ 5 = 7
- 55 ÷ 5 = 11
- 75 ÷ 5 = 15

Now: 7, 11, 15

Now check if any common prime divides all three? No.

But we can still continue with individual divisions.

So we stop here and write:

$$
\text{LCM} = 2 \times 5 \times (\text{product of remaining numbers}) = 2 \times 5 \times 7 \times 11 \times 15
$$

But wait — this is incorrect because we must account for all prime factors, but we already divided out 2 and 5.

Actually, the LCM is:
$$
\text{LCM} = 2 \times 5 \times \text{LCM}(7, 11, 15)
$$

But since 7, 11, 15 have no common factors, their LCM is just their product.

So:
$$
\text{LCM} = 2 \times 5 \times 7 \times 11 \times 15
$$

But wait — we already divided by 2 and 5, so the full LCM is:

$$
\text{LCM} = 2 \times 5 \times 7 \times 11 \times 3 \times 5 = ?
$$

Wait — better to do it systematically.

Let’s do proper long division:

```
| 70 110 150
--------------------
2 | 35 55 75
--------------------
5 | 7 11 15
--------------------
7 | 1 11 15
--------------------
11 | 1 1 15
--------------------
3 | 1 1 5
--------------------
5 | 1 1 1
--------------------
| 1 1 1
```

No — this is not standard.

The correct way is:

We divide only when the divisor divides ALL numbers.

So:

```
| 70 110 150
--------------------
2 | 35 55 75 ← 2 divides all?
→ 70/2=35, 110/2=55, 150/2=75 → yes
--------------------
5 | 7 11 15 ← 5 divides all? 7? no → stop
```

So only 2 and 5 are common divisors.

Now, after dividing by 2 and 5, we have:
- 7, 11, 15

Now, 15 = 3 × 5, so we need to include 3 and 5 again.

But since we already used 5 once, we need to consider maximum exponents.

Better to find LCM via prime factorization instead for clarity.

But let’s try long division properly.

Standard long division method for LCM:

We divide by primes one by one, even if only some numbers are divisible.

We keep going until all numbers become 1.

Let’s do it:

```
| 70 110 150
--------------------
2 | 35 55 75 ← 2 divides 70,110,150 → all go
--------------------
5 | 7 11 15 ← 5 divides 35,55,75 → all go
--------------------
7 | 1 11 15 ← 7 divides 7, but not 11 or 15 → skip
```

We can’t divide further unless we allow partial divisions.

Correct approach: Use continuous division.

We list the numbers and divide by smallest prime that divides at least one number.

But for LCM, we want to include all prime factors.

So:

```
| 70 110 150
--------------------
2 | 35 55 75 ← 2 divides all → yes
--------------------
5 | 7 11 15 ← 5 divides all → yes
--------------------
7 | 1 11 15 ← 7 divides 7 → 1, others remain
--------------------
11 | 1 1 15 ← 11 divides 11 → 1, others remain
--------------------
3 | 1 1 5 ← 3 divides 15 → 5
--------------------
5 | 1 1 1 ← 5 divides 5 → 1
--------------------
| 1 1 1
```

Now multiply all divisors used:
- 2, 5, 7, 11, 3, 5

So:
$$
\text{LCM} = 2 \times 5 \times 7 \times 11 \times 3 \times 5 = ?
$$

Calculate:
- 2 × 5 = 10
- 10 × 7 = 70
- 70 × 11 = 770
- 770 × 3 = 2310
- 2310 × 5 = 11,550

✔ Answer: 11,550

But let’s verify via prime factorization:

- 70 = 2 × 5 × 7
- 110 = 2 × 5 × 11
- 150 = 2 × 3 × 5²

Now take highest powers:
- 2¹, 3¹, 5², 7¹, 11¹

So:
$$
\text{LCM} = 2 \times 3 \times 5^2 \times 7 \times 11 = 2 × 3 × 25 × 7 × 11
$$

Calculate:
- 2 × 3 = 6
- 6 × 25 = 150
- 150 × 7 = 1050
- 1050 × 11 = 11,550

✔ Confirmed.

---

#### (ii) 25, 30, 150

Use long division method:

```
| 25 30 150
--------------------
2 | 25 15 75 ← 2 divides 30 and 150, but not 25 → no
```

So start with 5 (common divisor):

```
| 25 30 150
--------------------
5 | 5 6 30 ← 5 divides all → yes
--------------------
5 | 1 6 6 ← 5 divides 5 and 30, but not 6 → no
```

Now 5 divides 5 and 30 → so divide only those?

Wait — in long division method, we divide only when all numbers are divisible.

But 25, 30, 150:

- All divisible by 5? Yes:
- 25 ÷ 5 = 5
- 30 ÷ 5 = 6
- 150 ÷ 5 = 30

Now: 5, 6, 30

Again, divisible by 5? 5 and 30 → yes, but 6 → no → can't divide all.

So stop here.

Now, we need to continue with other primes.

But better to proceed as:

```
| 25 30 150
--------------------
5 | 5 6 30
--------------------
5 | 1 6 6 ← only 5 and 30 divisible by 5 → can't do
```

So instead, we use the method where we divide by primes that divide any number, and carry forward.

Standard way:

```
| 25 30 150
--------------------
2 | 25 15 75 ← 2 doesn't divide 25 → skip
3 | 25 10 75 ← 3 divides 30, 150 → 10, 25 → no
5 | 25 30 150 ← yes
--------------------
| 5 6 30
--------------------
5 | 1 6 6 ← 5 divides 5 and 30 → 1 and 6, 6 remains
--------------------
2 | 1 3 3 ← 2 divides 6 → 3, others unchanged
--------------------
3 | 1 1 1 ← 3 divides all → yes
--------------------
| 1 1 1
```

Wait — messy.

Better to use:

Start over:

```
| 25 30 150
--------------------
5 | 5 6 30 ← divide all by 5 → 5,6,30
--------------------
5 | 1 6 6 ← divide 5 and 30 by 5 → 1 and 6; 6 remains
--------------------
2 | 1 3 3 ← divide 6 and 6 by 2 → 3 and 3
--------------------
3 | 1 1 1 ← divide 3 and 3 by 3 → 1
--------------------
| 1 1 1
```

Divisors used: 5, 5, 2, 3

So LCM = 5 × 5 × 2 × 3 = 150

But wait — 150 is one of the numbers. Is it divisible by 25 and 30?

- 150 ÷ 25 = 6 → yes
- 150 ÷ 30 = 5 → yes
- 150 ÷ 150 = 1 → yes

So LCM = 150?

Check via prime factorization:

- 25 = 5²
- 30 = 2 × 3 × 5
- 150 = 2 × 3 × 5²

Highest powers:
- 2¹, 3¹, 5²

So:
$$
\text{LCM} = 2 × 3 × 5^2 = 2 × 3 × 25 = 150
$$

✔ Answer: 150

---

#### (iii) 36, 60, 120

Use long division:

```
| 36 60 120
--------------------
2 | 18 30 60
--------------------
2 | 9 15 30
--------------------
3 | 3 5 10
--------------------
3 | 1 5 10
--------------------
5 | 1 1 2
--------------------
2 | 1 1 1
--------------------
| 1 1 1
```

Wait — 3 does not divide 5 or 10 → can't divide all.

So better:

Start:

```
| 36 60 120
--------------------
2 | 18 30 60 ← all divisible by 2
--------------------
2 | 9 15 30 ← 9 not divisible by 2 → stop
```

So divide by 2 twice? Only if all divisible.

After first 2: 18, 30, 60

Now divide by 2 again: 9, 15, 30 → 9 not divisible → no

So only one 2.

Now try 3:

- 18, 30, 60 → all divisible by 3
- 18÷3=6, 30÷3=10, 60÷3=20

Now: 6, 10, 20

Divide by 2: 3, 5, 10 → 3 not divisible → no

Divide by 2 again? Only 10 and 20 → no

Try 3: 3 and 10, 20 → only 3 → no

Try 5: 10 and 20 → yes, but 3 → no

So better to do:

Let’s do it properly:

```
| 36 60 120
--------------------
2 | 18 30 60 ← all divisible
--------------------
2 | 9 15 30 ← 9 not divisible → stop
```

Now try 3: 9, 15, 30 → all divisible by 3
- 9÷3=3, 15÷3=5, 30÷3=10

Now: 3, 5, 10

Now try 3: 3 divisible, 5 and 10 → no

Try 5: 5 and 10 → yes, but 3 → no

So stop.

But we missed.

Better to use continuous method:

```
| 36 60 120
--------------------
2 | 18 30 60
--------------------
2 | 9 15 30
--------------------
3 | 3 5 10
--------------------
3 | 1 5 10
--------------------
5 | 1 1 2
--------------------
2 | 1 1 1
--------------------
| 1 1 1
```

But 3 does not divide 5 or 10 → invalid.

So correct way:

Only divide when all numbers are divisible.

So:

- 36, 60, 120 → divisible by 2 → 18, 30, 60
- Again by 2 → 9, 15, 30 → 9 not divisible → stop
- Now try 3: 18, 30, 60 → divisible by 3 → 6, 10, 20
- Now 6, 10, 20 → divisible by 2 → 3, 5, 10
- Now 3, 5, 10 → no common divisor
- So we need to include remaining factors.

Wait — better to use:

```
| 36 60 120
--------------------
2 | 18 30 60
--------------------
2 | 9 15 30 ← can't divide 9 by 2 → stop
```

So only one 2.

Then divide by 3: 36÷3=12, 60÷3=20, 120÷3=40 → 12, 20, 40

Now divide by 2: 6, 10, 20

Again by 2: 3, 5, 10

Now 3, 5, 10 → no common divisor

So divisors used: 2, 3, 2, 2 → total 2³ × 3 × 5 × ?

Wait — this is messy.

Let’s do it correctly with proper steps.

Standard long division method:

We divide by smallest prime that divides all numbers.

Start:

- 36, 60, 120 → divisible by 2 → 18, 30, 60
- Again by 2 → 9, 15, 30 → 9 not divisible → stop
- So only two 2s? Wait, 36=2²×9, 60=2²×15, 120=2³×15 → so max 2³

So we need to include 2 three times.

So we can do:

```
| 36 60 120
--------------------
2 | 18 30 60
--------------------
2 | 9 15 30
--------------------
2 | 9 15 15 ← now 30÷2=15 → but 9 not divisible → invalid
```

So we can't do it that way.

Alternative: Use prime factorization.

- 36 = 2² × 3²
- 60 = 2² × 3 × 5
- 120 = 2³ × 3 × 5

Take highest powers:
- 2³, 3², 5¹

So:
$$
\text{LCM} = 2^3 \times 3^2 \times 5 = 8 × 9 × 5 = 72 × 5 = 360
$$

✔ Answer: 360

Now, let’s verify via long division:

We’ll do it step by step:

```
| 36 60 120
--------------------
2 | 18 30 60
--------------------
2 | 9 15 30
--------------------
3 | 3 5 10
--------------------
3 | 1 5 10
--------------------
5 | 1 1 2
--------------------
2 | 1 1 1
--------------------
| 1 1 1
```

But this is invalid because we didn't divide all numbers at each step.

Instead, we should:

1. Divide by 2: 36→18, 60→30, 120→60
2. Divide by 2: 18→9, 30→15, 60→30
3. Divide by 2: 30→15, but 9 not divisible → cannot divide all → stop

So only two 2s.

Now divide by 3: 9→3, 15→5, 30→10 → 3, 5, 10

Now divide by 3: 3→1, 5 and 10 → no → stop

Now divide by 5: 5→1, 10→2 → 1, 1, 2

Now divide by 2: 2→1 → 1,1,1

So divisors: 2, 2, 3, 5, 2

So LCM = 2 × 2 × 3 × 5 × 2 = 2³ × 3 × 5 = 8 × 3 × 5 = 120 → wrong!

Because 120 is not divisible by 36 → 120 ÷ 36 = 3.33...

So error.

The issue is that we must ensure all numbers are divisible at each step.

Correct long division method:

We can divide by a prime if it divides at least one number, but we must carry forward.

But standard method is to divide only when all numbers are divisible.

So:

- 36, 60, 120 → divisible by 2 → 18, 30, 60
- Again by 2 → 9, 15, 30 → 9 not divisible → stop
- So only one 2 used → but we need more.

This shows the limitation.

Better to accept that for multiple numbers, long division method requires care.

Alternatively, use:

We know LCM = 360.

Let’s do it right:

```
| 36 60 120
--------------------
2 | 18 30 60
--------------------
2 | 9 15 30
--------------------
3 | 3 5 10
--------------------
3 | 1 5 10
--------------------
5 | 1 1 2
--------------------
2 | 1 1 1
--------------------
| 1 1 1
```

But this assumes we can divide even if not all are divisible — which is not valid.

So the correct way is to use prime factorization for accuracy.

Thus, we conclude:

✔ Final Answers:

---

Problem 1: First 3 Common Multiples

(i) 3 and 4 → 12, 24, 36
(ii) 5 and 6 → 30, 60, 90

---

Problem 2: LCM by Prime Factorization

(i) 12 and 20
- 12 = 2² × 3
- 20 = 2² × 5
- LCM = 2² × 3 × 5 = 60

(ii) 36, 27, 18
- 36 = 2² × 3²
- 27 = 3³
- 18 = 2 × 3²
- LCM = 2² × 3³ = 4 × 27 = 108

(iii) 32 and 48
- 32 = 2⁵
- 48 = 2⁴ × 3
- LCM = 2⁵ × 3 = 32 × 3 = 96

---

Problem 3: LCM by Long Division Method

(i) 70, 110, 150
- LCM = 2 × 5 × 7 × 11 × 3 × 5 = 11,550

(ii) 25, 30, 150
- LCM = 2 × 3 × 5² = 150

(iii) 36, 60, 120
- LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360

---

✔ Final Answer Summary:

1.
(i) 12, 24, 36
(ii) 30, 60, 90

2.
(i) 60
(ii) 108
(iii) 96

3.
(i) 11,550
(ii) 150
(iii) 360

Let me know if you'd like visual number lines or diagrams!