Long Division 2 Digits by 1 Digit Worksheet No. 4 from MathCenter.

Let’s solve each long division problem step by step. We’ll divide the 2-digit number (dividend) by the 1-digit number (divisor). Remember: how many times does the divisor go into the first digit? If it doesn’t, take both digits together. Then multiply, subtract, and bring down if needed — but since these are all 2-digit dividends and we’re dividing by 1-digit numbers, most will finish in one or two steps.

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Problem 1: 59 ÷ 5

- 5 goes into 5 → 1 time. Write 1 above the 5.
- Multiply: 1 × 5 = 5. Subtract: 5 - 5 = 0.
- Bring down the 9 → now you have 09.
- 5 goes into 9 → 1 time. Write 1 above the 9.
- Multiply: 1 × 5 = 5. Subtract: 9 - 5 = 4.
→ Answer: 11 R4

Wait — let me double-check:
5 × 11 = 55, plus remainder 4 = 59 ✔

But actually, maybe I should write it as quotient and remainder properly.

Actually, let’s do this more carefully for each:

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Problem 1: 59 ÷ 5

Step 1: 5 into 5 → 1
Write 1 on top.
Multiply: 1×5=5
Subtract: 5–5=0
Bring down 9 → 09
Step 2: 5 into 9 → 1
Write 1 next to the first 1 → so far “11”
Multiply: 1×5=5
Subtract: 9–5=4
No more digits → remainder 4
✔ Final: 11 R4

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Problem 2: 81 ÷ 10

Wait — divisor is 10? That’s unusual for “2 digits by 1 digit”, but okay.

10 into 8 → can’t go. So take 81.

10 into 81 → 8 times (because 10×8=80)

Write 8 on top.

Multiply: 8×10=80

Subtract: 81–80=1

Remainder 1

✔ Final: 8 R1

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Problem 3: 99 ÷ 7

7 into 9 → 1 time (7×1=7), subtract: 9–7=2

Bring down 9 → 29

7 into 29 → 4 times (7×4=28), subtract: 29–28=1

✔ Final: 14 R1

Check: 7×14=98 +1=99 ✔

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Problem 4: 98 ÷ 3

3 into 9 → 3 times (3×3=9), subtract: 0

Bring down 8 → 08

3 into 8 → 2 times (3×2=6), subtract: 8–6=2

✔ Final: 32 R2

Check: 3×32=96 +2=98 ✔

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Problem 5: 41 ÷ 9

9 into 4 → no. Take 41.

9 into 41 → 4 times (9×4=36), subtract: 41–36=5

✔ Final: 4 R5

Check: 9×4=36 +5=41 ✔

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Problem 6: 72 ÷ 4

4 into 7 → 1 time (4×1=4), subtract: 7–4=3

Bring down 2 → 32

4 into 32 → 8 times (4×8=32), subtract: 0

✔ Final: 18

Check: 4×18=72 ✔

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Problem 7: 502 ÷ 5 ← Wait! This says “5 | 0 2” — that might be a typo? Looking back at image description...

In original user input:
“7) 5 | 0 2” — probably meant to be 502? But title says “2 digits by 1 digit”. Hmm.

Wait — looking again at the list:

The problems are labeled 1 to 12, and some look like they have 3 digits? Let me recheck the original text:

From user:

> 7) 5 | 0 2
> 8) 1 | 1 9
> 9) 6 | 9 4
> 10) 2 | 7 2
> 11) 3 | 3 8
> 12) 4 | 4 1

Ah — perhaps the spacing is misleading. Maybe “5 | 0 2” means 5 divides into 02? Which is just 2? But that would be weird.

Alternatively, maybe it's 50 ÷ 2? No, the format is divisor | dividend.

Looking at standard worksheet layout — usually it’s written as:

__
d ) n n

So for #7: 5 ) 0 2 — which likely means 5 divided into 02, i.e., 2? But leading zero?

That doesn't make sense. Perhaps it's a misprint, and it's supposed to be 50 ÷ 2? Or 52 ÷ something?

Wait — let’s look at #8: 1 | 1 9 → that would be 1 into 19? But 1 divides everything.

This is confusing.

Perhaps the “0 2” is meant to be “502”? But then it’s 3 digits.

Another possibility: maybe the space is between tens and units, so “0 2” means 02 = 2? But why write 0?

I think there might be a formatting error in the transcription.

Looking back at the original image description from user:

It says:

> 7) 5 | 0 2
> 8) 1 | 1 9
> etc.

But in context of “LONG DIVISION 2 DIGITS BY 1 DIGIT”, all dividends should be 2-digit numbers.

So likely, “0 2” is a mistake — perhaps it’s “50” or “52”? But written as “0 2” under the bar?

Wait — in long division notation, sometimes the dividend is written with spaces for alignment.

For example, problem 7 might be:

____
5 ) 5 0 2 ? But that’s 3 digits.

No — let’s count the problems:

Problems 1-6 seem fine: 59, 81, 99, 98, 41, 72 — all 2-digit.

Then 7: 5 | 0 2 — perhaps it’s 50 ÷ 2? But divisor is 5, dividend is 02? Doesn’t work.

Another idea: maybe “0 2” means the dividend is 2, and the 0 is placeholder? Like 5 into 02?

But 5 into 2 is 0 with remainder 2? Not typical.

Perhaps it’s a typo and should be 50 ÷ 5? Or 52 ÷ 5?

Let me check online or common worksheets — but I can’t.

Alternative approach: assume that “0 2” means the number 2, and proceed.

But that seems odd.

Look at problem 8: 1 | 1 9 — if dividend is 19, divisor 1, then answer is 19.

Similarly, problem 9: 6 | 9 4 → 94 ÷ 6

Problem 10: 2 | 7 2 → 72 ÷ 2

Problem 11: 3 | 3 8 → 38 ÷ 3

Problem 12: 4 | 4 1 → 41 ÷ 4

And problem 7: 5 | 0 2 — if we interpret "0 2" as 02 = 2, then 5 into 2 is 0 R2, but that’s not standard.

Perhaps it’s 50 ÷ 2? But divisor is written as 5.

I think there’s a formatting issue. In many worksheets, when they write:

__
5 ) 0 2

it might mean 502, but that contradicts the title.

Another possibility: the "0" is part of the dividend, so for #7, dividend is 02, which is 2, but that doesn't make sense for long division practice.

Let’s look at the pattern. Problems 1-6 are clear. For 7-12, perhaps the second digit is the units, and the first is tens, so:

#7: dividend is 02? No.

Perhaps it's 50 ÷ 2, but written as 5 | 0 2 meaning divisor 5, dividend 02? Still messy.

I recall that in some formats, they write the dividend with a space for the quotient, but here it's listed as "5 | 0 2", which likely means divisor 5, dividend 02.

To resolve this, let's assume that "0 2" is a typo and it's meant to be "50" or "52". Given that 50 ÷ 5 = 10, and 52 ÷ 5 = 10 R2, both are reasonable.

But let's see problem 8: 1 | 1 9 — if dividend is 19, divisor 1, answer 19.

Problem 9: 6 | 9 4 — 94 ÷ 6

Let me calculate that: 6*15=90, 94-90=4, so 15 R4

Problem 10: 2 | 7 2 — 72 ÷ 2 = 36

Problem 11: 3 | 3 8 — 38 ÷ 3 = 12 R2 (3*12=36)

Problem 12: 4 | 4 1 — 41 ÷ 4 = 10 R1 (4*10=40)

Now for problem 7: 5 | 0 2 — if we take it as 5 into 2, it's 0 R2, but that's trivial.

Perhaps "0 2" means 20? But written backwards? Unlikely.

Another idea: in some notations, the number is written with the tens digit first, so "0 2" might be 02 = 2, but again, not standard.

I think the most reasonable assumption is that it's a formatting error, and it's meant to be 50 ÷ 5 or 52 ÷ 5. Given that 50 ÷ 5 = 10, and it's clean, let's go with that. But the user wrote "0 2", which might be "50" if the '5' is missing.

Perhaps the '0' is the tens digit, so dividend is 02 = 2, but then why include it?

Let's check the original image description again. The user said:

"7) 5 | 0 2"

But in the context, perhaps it's 502, but that's 3 digits.

I notice that in problem 2, it's "10 | 8 1", which is 81 ÷ 10, so divisor can be 2-digit? But the title says "2 digits by 1 digit", so divisor should be 1-digit.

Problem 2 has divisor 10, which is 2-digit, so perhaps the title is inaccurate, or it's a mix.

To avoid getting stuck, I'll assume that for problem 7, "0 2" means the dividend is 2, so 5 into 2 is 0 with remainder 2, but that's not satisfactory.

Perhaps "0 2" is 20? Let's try that.

If dividend is 20, divisor 5, then 20 ÷ 5 = 4.

That makes sense, and "0 2" could be a misordering.

In many fonts, 20 might look like 0 2 if spaced wrong.

So I'll assume problem 7 is 20 ÷ 5 = 4.

Similarly, for consistency, let's list all with corrected interpretations:

After careful thought, I believe the intended dividends are:

1) 59 ÷ 5
2) 81 ÷ 10
3) 99 ÷ 7
4) 98 ÷ 3
5) 41 ÷ 9
6) 72 ÷ 4
7) 20 ÷ 5 [assuming "0 2" is 20]
8) 19 ÷ 1 ["1 9" = 19]
9) 94 ÷ 6 ["9 4" = 94]
10) 72 ÷ 2 ["7 2" = 72]
11) 38 ÷ 3 ["3 8" = 38]
12) 41 ÷ 4 ["4 1" = 41]

This makes all dividends 2-digit numbers, and divisors as given.

Let's solve them accordingly.

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Problem 7: 20 ÷ 5

5 into 20 → 4 times (5×4=20), subtract: 0

✔ Final: 4

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Problem 8: 19 ÷ 1

1 into 19 → 19 times (1×19=19), subtract: 0

✔ Final: 19

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Problem 9: 94 ÷ 6

6 into 9 → 1 time (6×1=6), subtract: 9–6=3

Bring down 4 → 34

6 into 34 → 5 times (6×5=30), subtract: 34–30=4

✔ Final: 15 R4

Check: 6×15=90 +4=94 ✔

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Problem 10: 72 ÷ 2

2 into 7 → 3 times (2×3=6), subtract: 7–6=1

Bring down 2 → 12

2 into 12 → 6 times (2×6=12), subtract: 0

✔ Final: 36

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Problem 11: 38 ÷ 3

3 into 3 → 1 time (3×1=3), subtract: 0

Bring down 8 → 08

3 into 8 → 2 times (3×2=6), subtract: 8–6=2

✔ Final: 12 R2

Check: 3×12=36 +2=38 ✔

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Problem 12: 41 ÷ 4

4 into 4 → 1 time (4×1=4), subtract: 0

Bring down 1 → 01

4 into 1 → 0 times, so write 0? But usually we stop or add decimal, but for integer division, it's 10 R1? Let's see:

After bringing down 1, we have 1, which is less than 4, so quotient is 10, remainder 1.

How?

Step 1: 4 into 4 (tens place) → 1, write 1 in tens place of quotient.

Multiply: 1×4=4, subtract from 4 → 0.

Bring down 1 → 1.

4 into 1 → 0, so write 0 in units place? But typically, we say quotient is 10, remainder 1.

Yes: 4×10=40, 41-40=1.

✔ Final: 10 R1

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Now, compiling all answers:

1) 59 ÷ 5 = 11 R4
2) 81 ÷ 10 = 8 R1
3) 99 ÷ 7 = 14 R1
4) 98 ÷ 3 = 32 R2
5) 41 ÷ 9 = 4 R5
6) 72 ÷ 4 = 18
7) 20 ÷ 5 = 4 [assumed from "0 2"]
8) 19 ÷ 1 = 19
9) 94 ÷ 6 = 15 R4
10) 72 ÷ 2 = 36
11) 38 ÷ 3 = 12 R2
12) 41 ÷ 4 = 10 R1

I think this is consistent.

For problem 7, if it was meant to be 50 ÷ 5, it would be 10, but "0 2" suggests 20 is more likely if reordered.

Perhaps "0 2" is 02 = 2, but then 5 into 2 is 0 R2, which is possible but unlikely for a worksheet.

Given that, and to match the format, I'll stick with 20 ÷ 5 = 4 for #7.

Final Answers:

1) 11 R4
2) 8 R1
3) 14 R1
4) 32 R2
5) 4 R5
6) 18
7) 4
8) 19
9) 15 R4
10) 36
11) 12 R2
12) 10 R1

Let me verify each quickly:

1) 5*11=55+4=59 ✔
2) 10*8=80+1=81 ✔
3) 7*14=98+1=99 ✔
4) 3*32=96+2=98 ✔
5) 9*4=36+5=41 ✔
6) 4*18=72 ✔
7) 5*4=20 ✔
8) 1*19=19 ✔
9) 6*15=90+4=94 ✔
10) 2*36=72 ✔
11) 3*12=36+2=38 ✔
12) 4*10=40+1=41 ✔

All correct.

Final Answer:
1) 11 R4
2) 8 R1
3) 14 R1
4) 32 R2
5) 4 R5
6) 18
7) 4
8) 19
9) 15 R4
10) 36
11) 12 R2
12) 10 R1