Practice sheet designed to help students understand and identify multiples of various numbers through listing and pattern recognition exercises.

Absolutely! Let’s solve each question step by step and explain the reasoning clearly.

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1. Write down the first ten multiples of 11.

Multiples of 11 are found by multiplying 11 by whole numbers: 1, 2, 3, ..., 10.

Answer:
> 11, 22, 33, 44, 55, 66, 77, 88, 99, 110

✔ *Explanation:*
11 × 1 = 11
11 × 2 = 22
11 × 3 = 33
...
11 × 10 = 110

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2. What do you notice?

Look at the pattern in the multiples of 11 from above:

> 11, 22, 33, 44, 55, 66, 77, 88, 99, 110

Answer:
> The first 9 multiples (up to 99) have repeating digits — like 11, 22, 33, etc. After that, it breaks the pattern with 110.

✔ *Explanation:*
This is a fun pattern! For single-digit multipliers (1–9), multiplying by 11 gives a two-digit number where both digits are the same. But at 10×11=110, the pattern changes because we now have a three-digit number.

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3. Which of these numbers are multiples of 7?

Given: 71, 47, 76, 49, 56

Check each one by dividing by 7 or using known multiplication facts.

- 71 ÷ 7 ≈ 10.14 → Not whole → ✘
- 47 ÷ 7 ≈ 6.71 → Not whole → ✘
- 76 ÷ 7 ≈ 10.85 → Not whole → ✘
- 49 ÷ 7 = 7 → ✔ (7×7)
- 56 ÷ 7 = 8 → ✔ (7×8)

Answer:
> 49, 56

✔ *Explanation:*
Only numbers that divide evenly by 7 (no remainder) are multiples of 7.

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4. Write down the first ten multiples of 20.

Multiply 20 by 1 through 10.

Answer:
> 20, 40, 60, 80, 100, 120, 140, 160, 180, 200

✔ *Explanation:*
20 × 1 = 20
20 × 2 = 40
...
20 × 10 = 200

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5. What do you notice? Look especially at the tens digit.

Look at the list:
> 20, 40, 60, 80, 100, 120, 140, 160, 180, 200

Answer:
> All multiples of 20 end in 0. The tens digit increases by 2 each time until 100, then continues increasing by 2 in the tens place (e.g., 120 has 2 in tens, 140 has 4, etc.). Also, every multiple of 20 is also a multiple of 10 and 2.

✔ *Explanation:*
Since 20 = 2 × 10, all its multiples must end in 0 (like multiples of 10). The tens digit alternates even numbers: 2, 4, 6, 8, 0 (in 100), then 2 again (in 120), etc.

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6. Which of these numbers are multiples of 9?

Given: 36, 47, 91, 99, 92

Check divisibility by 9 — a quick trick: add the digits; if sum is divisible by 9, so is the number.

- 36 → 3 + 6 = 9 → ✔ (9×4)
- 47 → 4 + 7 = 11 → ✘
- 91 → 9 + 1 = 10 → ✘
- 99 → 9 + 9 = 18 → ✔ (9×11)
- 92 → 9 + 2 = 11 → ✘

Answer:
> 36, 99

✔ *Explanation:*
The digit-sum rule for 9 is very useful here. Only 36 and 99 pass this test.

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7. Write down the first ten multiples of 12.

Multiply 12 by 1 through 10.

Answer:
> 12, 24, 36, 48, 60, 72, 84, 96, 108, 120

✔ *Explanation:*
12 × 1 = 12
12 × 2 = 24
...
12 × 10 = 120

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8. What do you notice? Is there a pattern?

Look at the list:
> 12, 24, 36, 48, 60, 72, 84, 96, 108, 120

Answer:
> All multiples of 12 are even. They increase by 12 each time. Also, they are all multiples of both 3 and 4 (since 12 = 3×4). You might also notice that the last digits cycle: 2, 4, 6, 8, 0, and repeat.

✔ *Explanation:*
Pattern: +12 each time. Since 12 is divisible by 2, 3, 4, 6 — all its multiples share those factors too. The last digits follow a repeating cycle: 2,4,6,8,0 — which is the same as multiples of 2 ending in even digits.

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9. Circle the numbers which are multiples of 6:

Given: 25, 46, 56, 66, 76, 86, 90

A number is a multiple of 6 if it is divisible by both 2 and 3.

Check each:

- 25 → odd → not divisible by 2 → ✘
- 46 → even → check sum: 4+6=10 → not divisible by 3 → ✘
- 56 → even → 5+6=11 → not divisible by 3 → ✘
- 66 → even → 6+6=12 → divisible by 3 → ✔
- 76 → even → 7+6=13 → not divisible by 3 → ✘
- 86 → even → 8+6=14 → not divisible by 3 → ✘
- 90 → even → 9+0=9 → divisible by 3 → ✔

Answer:
> Circle: 66, 90

✔ *Explanation:*
Multiples of 6 must be even AND have digit sums divisible by 3. Only 66 and 90 meet both conditions.

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10. Circle the numbers which are multiples of both 3 and 5:

Given: 45, 50, 25, 15, 60, 10, 11

A number divisible by both 3 and 5 must be divisible by 15 (LCM of 3 and 5).

Check each:

- 45 → ends in 5 → divisible by 5. Sum: 4+5=9 → divisible by 3 → ✔
- 50 → ends in 0 → divisible by 5. Sum: 5+0=5 → not divisible by 3 → ✘
- 25 → ends in 5 → divisible by 5. Sum: 2+5=7 → not divisible by 3 → ✘
- 15 → ends in 5 → divisible by 5. Sum: 1+5=6 → divisible by 3 → ✔
- 60 → ends in 0 → divisible by 5. Sum: 6+0=6 → divisible by 3 → ✔
- 10 → ends in 0 → divisible by 5. Sum: 1+0=1 → not divisible by 3 → ✘
- 11 → not divisible by 5 → ✘

Answer:
> Circle: 45, 15, 60

✔ *Explanation:*
To be divisible by both 3 and 5, a number must end in 0 or 5 (for 5) AND have digits adding to a multiple of 3 (for 3). Only 45, 15, and 60 satisfy both.

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## ✔ Final Summary of Answers:

1. 11, 22, 33, 44, 55, 66, 77, 88, 99, 110
2. First 9 multiples have repeating digits (e.g., 22, 33); 110 breaks the pattern.
3. 49, 56
4. 20, 40, 60, 80, 100, 120, 140, 160, 180, 200
5. All end in 0; tens digit increases by 2 each time.
6. 36, 99
7. 12, 24, 36, 48, 60, 72, 84, 96, 108, 120
8. Increase by 12; all even; last digits cycle 2,4,6,8,0; divisible by 2,3,4,6.
9. Circle: 66, 90
10. Circle: 45, 15, 60

Let me know if you’d like this formatted as a printable answer sheet! 😊