Multiplication worksheet for practicing 3-digit by 1-digit multiplication using the column method.

Let’s solve each section step by step.

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Section A: Multiply 3-digit numbers by 1-digit using column method

We multiply from right to left (ones, then tens, then hundreds).

1) 143 × 2
- 3 × 2 = 6 → write 6
- 4 × 2 = 8 → write 8
- 1 × 2 = 2 → write 2
→ Answer: 286

2) 231 × 3
- 1 × 3 = 3
- 3 × 3 = 9
- 2 × 3 = 6
→ Answer: 693

3) 524 × 2
- 4 × 2 = 8
- 2 × 2 = 4
- 5 × 2 = 10 → write 0, carry 1? Wait — no carrying needed here since we’re just multiplying digit by digit and writing below. Actually, 5×2=10, so we write “10” in front.
Wait — let’s do it properly:

Actually, for column multiplication:
Start from right:
- 4 × 2 = 8 → ones place
- 2 × 2 = 4 → tens place
- 5 × 2 = 10 → hundreds and thousands → write 10 in front → so 1048? Let me check: 524 × 2 = 1048 ✔️

But in column format, you write:

```
524
× 2
-----
1048
```

So yes, 1048

4) 221 × 5
- 1 × 5 = 5
- 2 × 5 = 10 → write 0, carry 1
- 2 × 5 = 10 + carry 1 = 11 → write 11

So:

```
221
× 5
-----
1105
```

Check: 221 × 5 = 1105 ✔️

---

Section B:

1) 153 × 3
- 3 × 3 = 9
- 5 × 3 = 15 → write 5, carry 1
- 1 × 3 = 3 + carry 1 = 4
→ 459

2) 263 × 4
- 3 × 4 = 12 → write 2, carry 1
- 6 × 4 = 24 + 1 = 25 → write 5, carry 2
- 2 × 4 = 8 + 2 = 10 → write 10
→ 1052

3) 378 × 5
- 8 × 5 = 40 → write 0, carry 4
- 7 × 5 = 35 + 4 = 39 → write 9, carry 3
- 3 × 5 = 15 + 3 = 18 → write 18
→ 1890

4) 167 × 2
- 7 × 2 = 14 → write 4, carry 1
- 6 × 2 = 12 + 1 = 13 → write 3, carry 1
- 1 × 2 = 2 + 1 = 3
→ 334

---

Section C:

1) 586 × 7
- 6 × 7 = 42 → write 2, carry 4
- 8 × 7 = 56 + 4 = 60 → write 0, carry 6
- 5 × 7 = 35 + 6 = 41 → write 41
→ 4102

2) 427 × 8
- 7 × 8 = 56 → write 6, carry 5
- 2 × 8 = 16 + 5 = 21 → write 1, carry 2
- 4 × 8 = 32 + 2 = 34 → write 34
→ 3416

3) 618 × 9
- 8 × 9 = 72 → write 2, carry 7
- 1 × 9 = 9 + 7 = 16 → write 6, carry 1
- 6 × 9 = 54 + 1 = 55 → write 55
→ 5562

4) 894 × 6
- 4 × 6 = 24 → write 4, carry 2
- 9 × 6 = 54 + 2 = 56 → write 6, carry 5
- 8 × 6 = 48 + 5 = 53 → write 53
→ 5364

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Section D: Fill in the boxes

These are puzzles where some digits are missing. We need to find what goes in the boxes.

1)
```
1 □ 2
× 4
-------
6 0 □
```

Let’s call the missing digit in top number “A”, and bottom result “B”.

So: (100 + 10*A + 2) × 4 = 600 + 0*10 + B = 600 + B

Try possible values for A (0–9):

If A = 5 → 152 × 4 = 608 → matches! Because 608 has 6, 0, 8 → so B = 8

✔ So first box is 5, second box is 8

2)
```
3 □ 6
× □
-------
1 □ 8 0
```

Let top be 3X6, multiplier Y, product is 1Z80.

So: (300 + 10*X + 6) × Y = 1000 + 100*Z + 80 + 0 = 1080 + 100*Z

Try small Y values.

Try Y = 5:

Then 3X6 × 5 ends with 0 → good (since 6×5=30)

Now, 3X6 × 5 = ?

Let’s try X = 1 → 316 × 5 = 1580 → that’s 1,5,8,0 → matches pattern 1□80 → so Z = 5

✔ So top box is 1, multiplier box is 5, bottom box is 5

Check: 316 × 5 = 1580 ✔️

3)
```
□ □ 8
× 9
-------
5 □ 0 □
```

Let top number be AB8, multiply by 9 → get 5C0D

So: (100*A + 10*B + 8) × 9 = 5000 + 100*C + 0*10 + D = 5000 + 100*C + D

Since last digit of product is D, and we’re multiplying by 9, and last digit of top is 8 → 8×9=72 → so D = 2, carry 7

Now, middle digit: B×9 + carry 7 must end with 0 (because product has 0 in tens place)

So: (B×9 + 7) mod 10 = 0 → B×9 ends with 3 → because 3+7=10 → ends with 0

What digit B makes B×9 end with 3?

Try B=7 → 7×9=63 → yes! Ends with 3 → plus carry 7 → 63+7=70 → write 0, carry 7

Now, hundreds place: A×9 + carry 7 → should give us something that leads to 5 in thousands and C in hundreds.

Product starts with 5, so total is between 5000 and 5999.

So: (A×9 + 7) must be such that when written, it gives two digits: first digit is 5 or part of 5...

Actually, let’s compute full value.

We have:

Top number: A78 (since B=7)

Multiply by 9:

A78 × 9 = ?

We know:

8×9=72 → write 2, carry 7

7×9=63 +7=70 → write 0, carry 7

A×9 +7 = ? → this will be the thousands and hundreds.

Total product is 5C02

So A×9 +7 must equal 50 + C? Not exactly — actually, since we’re at hundreds place, A×9 +7 gives the thousands and hundreds digits.

For example, if A=6 → 6×9=54 +7=61 → so write 61 → meaning 6 in thousands, 1 in hundreds → so product would be 6102 → but we need 5C02 → too big.

If A=5 → 5×9=45 +7=52 → so 5 in thousands, 2 in hundreds → product = 5202 → which fits 5□0□ → so C=2, D=2

Check: 578 × 9 = ?

578 × 9:

8×9=72 → 2, carry 7

7×9=63+7=70 → 0, carry 7

5×9=45+7=52 → 52

So 5202 ✔️

And 5202 matches 5□0□ → so boxes: first two are 5 and 7, then 2 and 2

Wait — the problem shows:

```
□ □ 8
× 9
-------
5 □ 0 □
```

So top: first box = 5, second box = 7

Bottom: third box (hundreds) = 2, fourth box (ones) = 2

✔ So answers: 5, 7, 2, 2

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Section E: Work out the products (multiply all together)

1) 5 × 8 × 4 × 7

Do step by step:

5 × 8 = 40

40 × 4 = 160

160 × 7 = 1120

✔ 1120

2) 9 × 3 × 4 × 6

9 × 3 = 27

27 × 4 = 108

108 × 6 = 648

✔ 648

---

Final Answer:

Section A:
143 × 2 = 286
231 × 3 = 693
524 × 2 = 1048
221 × 5 = 1105

Section B:
153 × 3 = 459
263 × 4 = 1052
378 × 5 = 1890
167 × 2 = 334

Section C:
586 × 7 = 4102
427 × 8 = 3416
618 × 9 = 5562
894 × 6 = 5364

Section D:
First puzzle: 152 × 4 = 608
Second puzzle: 316 × 5 = 1580
Third puzzle: 578 × 9 = 5202

Section E:
1) 5 × 8 × 4 × 7 = 1120
2) 9 × 3 × 4 × 6 = 648