Multiply 2-Digit by 1-Digit without Regrouping - Math Worksheets ... - Free Printable

Let's solve each of the 2-digit by 1-digit multiplication problems without regrouping. Since we're multiplying a two-digit number by a single digit, and there's no regrouping involved, we can multiply each digit of the two-digit number separately and then combine the results.

We'll go through each problem step-by-step:

---

Row 1

1. 21 × 1
- 1 × 1 = 1
- 20 × 1 = 20
→ 20 + 1 = 21

2. 24 × 2
- 4 × 2 = 8
- 20 × 2 = 40
→ 40 + 8 = 48

3. 33 × 3
- 3 × 3 = 9
- 30 × 3 = 90
→ 90 + 9 = 99

4. 32 × 2
- 2 × 2 = 4
- 30 × 2 = 60
→ 60 + 4 = 64

---

Row 2

5. 42 × 2
- 2 × 2 = 4
- 40 × 2 = 80
→ 80 + 4 = 84

6. 52 × 3
- 2 × 3 = 6
- 50 × 3 = 150
→ 150 + 6 = 156

7. 61 × 2
- 1 × 2 = 2
- 60 × 2 = 120
→ 120 + 2 = 122

8. 71 × 1
- 1 × 1 = 1
- 70 × 1 = 70
→ 70 + 1 = 71

---

Row 3

9. 31 × 1
- 1 × 1 = 1
- 30 × 1 = 30
→ 30 + 1 = 31

10. 74 × 2
- 4 × 2 = 8
- 70 × 2 = 140
→ 140 + 8 = 148

11. 34 × 2
- 4 × 2 = 8
- 30 × 2 = 60
→ 60 + 8 = 68

12. 62 × 4
- 2 × 4 = 8
- 60 × 4 = 240
→ 240 + 8 = 248

---

Row 4

13. 54 × 2
- 4 × 2 = 8
- 50 × 2 = 100
→ 100 + 8 = 108

14. 81 × 1
- 1 × 1 = 1
- 80 × 1 = 80
→ 80 + 1 = 81

15. 82 × 2
- 2 × 2 = 4
- 80 × 2 = 160
→ 160 + 4 = 164

16. 91 × 3
- 1 × 3 = 3
- 90 × 3 = 270
→ 270 + 3 = 273

---

Row 5

17. 94 × 2
- 4 × 2 = 8
- 90 × 2 = 180
→ 180 + 8 = 188

18. 43 × 3
- 3 × 3 = 9
- 40 × 3 = 120
→ 120 + 9 = 129

19. 53 × 2
- 3 × 2 = 6
- 50 × 2 = 100
→ 100 + 6 = 106

20. 52 × 2
- 2 × 2 = 4
- 50 × 2 = 100
→ 100 + 4 = 104

---

✔ Final Answers:

Here’s the completed grid:

```
21 × 1 = 21 24 × 2 = 48 33 × 3 = 99 32 × 2 = 64
42 × 2 = 84 52 × 3 = 156 61 × 2 = 122 71 × 1 = 71
31 × 1 = 31 74 × 2 = 148 34 × 2 = 68 62 × 4 = 248
54 × 2 = 108 81 × 1 = 81 82 × 2 = 164 91 × 3 = 273
94 × 2 = 188 43 × 3 = 129 53 × 2 = 106 52 × 2 = 104
```

---

🔍 Explanation:

- We break down the 2-digit number into tens and ones.
- Multiply each part by the single-digit multiplier.
- Add the two products together.
- Since no regrouping is needed (no carrying), this method works smoothly.

✔ All calculations are correct and follow the rule of multiplying without regrouping, meaning the product of each digit does not exceed 9 in any place value (but note: in some cases like 52 × 3 = 156, regrouping *would* be needed if done vertically — but since the problem says "without regrouping", it likely means these are designed to avoid carrying over, though some results do involve larger numbers. However, the instruction may refer to no carry-over during the process, so let’s double-check for any that might require regrouping.)

Wait — let's check: "Without Regrouping" implies that when multiplying, we don’t need to carry digits. But in practice, some of these multiplications do require regrouping when doing standard vertical multiplication.

For example:
- 24 × 2: 4×2=8, 2×2=4 → 48 → no carry
- 52 × 3: 2×3=6, 5×3=15 → 156 → here, 5×3=15 requires carrying (regrouping)

But the problem says "Without Regrouping", which suggests that all problems should be solvable without carrying digits.

So let’s re-express: perhaps the intention is to multiply digit by digit without carrying, i.e., only where the partial products don’t exceed 9 in the units place.

But actually, the term "without regrouping" usually refers to the absence of carrying during multiplication, meaning the product of the ones digit × multiplier must be less than 10.

Let’s verify which problems do not require regrouping:

Check the ones digit × multiplier:
- 21×1: 1×1=1 → OK
- 24×2: 4×2=8 → OK
- 33×3: 3×3=9 → OK
- 32×2: 2×2=4 → OK
- 42×2: 2×2=4 → OK
- 52×3: 2×3=6 → OK
- 61×2: 1×2=2 → OK
- 71×1: 1×1=1 → OK
- 31×1: 1×1=1 → OK
- 74×2: 4×2=8 → OK
- 34×2: 4×2=8 → OK
- 62×4: 2×4=8 → OK
- 54×2: 4×2=8 → OK
- 81×1: 1×1=1 → OK
- 82×2: 2×2=4 → OK
- 91×3: 1×3=3 → OK
- 94×2: 4×2=8 → OK
- 43×3: 3×3=9 → OK
- 53×2: 3×2=6 → OK
- 52×2: 2×2=4 → OK

✔ All ones digits × multiplier result in a single-digit product (<10), so no regrouping is required in the ones place.

Now check the tens digit × multiplier:
- For example, 52×3: tens digit 5×3=15 → this would require regrouping (carryover) in the tens place.

Ah! So even though the ones digit doesn't cause carry, the tens digit multiplication may.

But in standard multiplication without regrouping, we assume that no carrying occurs at all, so both parts must not produce a number ≥10.

Therefore, only those problems where both:
- ones digit × multiplier < 10
- tens digit × multiplier < 10

are truly "without regrouping".

Let’s check:

| Problem | Ones × mult | Tens × mult | Regroup? |
|--------|-------------|-------------|---------|
| 21×1 | 1×1=1 | 2×1=2 | No |
| 24×2 | 4×2=8 | 2×2=4 | No |
| 33×3 | 3×3=9 | 3×3=9 | No |
| 32×2 | 2×2=4 | 3×2=6 | No |
| 42×2 | 2×2=4 | 4×2=8 | No |
| 52×3 | 2×3=6 | 5×3=15 → 15 ≥10 → Yes, regrouping needed ✘ |
| 61×2 | 1×2=2 | 6×2=12 → 12 ≥10 → Regrouping needed ✘ |
| 71×1 | 1×1=1 | 7×1=7 | No |
| 31×1 | 1×1=1 | 3×1=3 | No |
| 74×2 | 4×2=8 | 7×2=14 → 14 ≥10 → Regrouping needed ✘ |
| 34×2 | 4×2=8 | 3×2=6 | No |
| 62×4 | 2×4=8 | 6×4=24 → 24 ≥10 → Regrouping needed ✘ |
| 54×2 | 4×2=8 | 5×2=10 → 10 ≥10 → Regrouping needed ✘ |
| 81×1 | 1×1=1 | 8×1=8 | No |
| 82×2 | 2×2=4 | 8×2=16 → 16 ≥10 → Regrouping needed ✘ |
| 91×3 | 1×3=3 | 9×3=27 → 27 ≥10 → Regrouping needed ✘ |
| 94×2 | 4×2=8 | 9×2=18 → 18 ≥10 → Regrouping needed ✘ |
| 43×3 | 3×3=9 | 4×3=12 → 12 ≥10 → Regrouping needed ✘ |
| 53×2 | 3×2=6 | 5×2=10 → 10 ≥10 → Regrouping needed ✘ |
| 52×2 | 2×2=4 | 5×2=10 → 10 ≥10 → Regrouping needed ✘ |

Only the following do not require regrouping:
- 21×1 → 21
- 24×2 → 48
- 33×3 → 99
- 32×2 → 64
- 42×2 → 84
- 71×1 → 71
- 31×1 → 31
- 34×2 → 68
- 81×1 → 81

Wait — 33×3 = 99 → 3×3=9, 30×3=90 → 90+9=99 → tens digit ×3 = 90, which is fine, but in vertical multiplication, you’d write:
```
33
× 3
----
99
```
→ 3×3=9, 3×3=9 → no carry → OK

Similarly, 32×2: 2×2=4, 3×2=6 → 64 → OK

But 42×2: 2×2=4, 4×2=8 → 84 → OK

So anywhere the tens digit × multiplier < 10, it's OK.

Let’s recheck:
- 21×1: 2×1=2 → OK
- 24×2: 2×2=4 → OK
- 33×3: 3×3=9 → OK
- 32×2: 3×2=6 → OK
- 42×2: 4×2=8 → OK
- 52×3: 5×3=15 → 15 ≥10 → Not allowed
- 61×2: 6×2=12 → 12 ≥10 → Not allowed
- 71×1: 7×1=7 → OK
- 31×1: 3×1=3 → OK
- 74×2: 7×2=14 → Not allowed
- 34×2: 3×2=6 → OK
- 62×4: 6×4=24 → Not allowed
- 54×2: 5×2=10 → Not allowed
- 81×1: 8×1=8 → OK
- 82×2: 8×2=16 → Not allowed
- 91×3: 9×3=27 → Not allowed
- 94×2: 9×2=18 → Not allowed
- 43×3: 4×3=12 → Not allowed
- 53×2: 5×2=10 → Not allowed
- 52×2: 5×2=10 → Not allowed

So only the following can be solved without regrouping:
- 21×1 = 21
- 24×2 = 48
- 33×3 = 99
- 32×2 = 64
- 42×2 = 84
- 71×1 = 71
- 31×1 = 31
- 34×2 = 68
- 81×1 = 81

The rest require regrouping (carrying), so they are not valid for "without regrouping".

But the worksheet says “2-Digit by 1-Digit Multiplication Without Regrouping”, so likely the problems are selected such that no carrying is needed, but clearly many do.

Wait — maybe "without regrouping" here means you don't need to carry during the multiplication process, but the final answer may have more digits.

But in reality, if the tens digit × multiplier ≥10, you must regroup.

So perhaps the worksheet has an error, or "without regrouping" is being used loosely.

Alternatively, maybe "without regrouping" means you multiply each digit independently and add, as long as you don't carry.

But in that case, the method is still valid, even if you get a large number.

Given that, and since the problem asks to solve them, and the instructions say “Multiply,” I will assume that the goal is to compute the correct answers, even if some require regrouping, but the label “Without Regrouping” might be misleading or apply only to the ones digit.

However, looking back at the original image, it says “2-Digit by 1-Digit Multiplication Without Regrouping”, so likely the problems were chosen so that no carrying is needed.

But as shown, several problems do require regrouping.

Let’s check which ones are truly without regrouping:

We need:
- ones digit × multiplier < 10
- tens digit × multiplier < 10

So:
- 21×1: 1×1=1, 2×1=2 → OK
- 24×2: 4×2=8, 2×2=4 → OK
- 33×3: 3×3=9, 3×3=9 → OK
- 32×2: 2×2=4, 3×2=6 → OK
- 42×2: 2×2=4, 4×2=8 → OK
- 52×3: 2×3=6, 5×3=15 → 15≥10 → NO
- 61×2: 1×2=2, 6×2=12 → 12≥10 → NO
- 71×1: 1×1=1, 7×1=7 → OK
- 31×1: 1×1=1, 3×1=3 → OK
- 74×2: 4×2=8, 7×2=14 → NO
- 34×2: 4×2=8, 3×2=6 → OK
- 62×4: 2×4=8, 6×4=24 → NO
- 54×2: 4×2=8, 5×2=10 → NO
- 81×1: 1×1=1, 8×1=8 → OK
- 82×2: 2×2=4, 8×2=16 → NO
- 91×3: 1×3=3, 9×3=27 → NO
- 94×2: 4×2=8, 9×2=18 → NO
- 43×3: 3×3=9, 4×3=12 → NO
- 53×2: 3×2=6, 5×2=10 → NO
- 52×2: 2×2=4, 5×2=10 → NO

So only these 8 problems are truly "without regrouping":
- 21×1 = 21
- 24×2 = 48
- 33×3 = 99
- 32×2 = 64
- 42×2 = 84
- 71×1 = 71
- 31×1 = 31
- 34×2 = 68
- 81×1 = 81

But the worksheet has 20 problems.

So either:
- The label is incorrect, or
- "Without regrouping" means something else.

Alternatively, perhaps "without regrouping" refers to the fact that you are not adding numbers that require carrying, but in multiplication, it's about partial products.

In any case, since the task is to solve the problems, and the image shows 20 problems, I will provide the correct answers for all, as per standard multiplication.

---

✔ Final Answer Grid (All Correct):

```
21 × 1 = 21 24 × 2 = 48 33 × 3 = 99 32 × 2 = 64
42 × 2 = 84 52 × 3 = 156 61 × 2 = 122 71 × 1 = 71
31 × 1 = 31 74 × 2 = 148 34 × 2 = 68 62 × 4 = 248
54 × 2 = 108 81 × 1 = 81 82 × 2 = 164 91 × 3 = 273
94 × 2 = 188 43 × 3 = 129 53 × 2 = 106 52 × 2 = 104
```

These are the correct solutions.

If the intent was without regrouping, only the first few are valid, but since the worksheet includes all, it's likely a general exercise.

Let me know if you'd like only the "without regrouping" ones solved.