Missing digits subtraction problems for 3rd-grade math practice.

Let's solve each subtraction problem in the worksheet step by step. The task is to fill in the missing digits so that the subtraction is correct.

We'll go through each problem one by one, using logic and subtraction rules (including borrowing when needed).

---

Problem 1:

```
1 _ 6
- 4 1 5
-------
2 8 1 1
```

Wait — this seems off. The top number is a 3-digit number (1_6), subtracting another 3-digit number (415), but the result is 2811, which is a 4-digit number.

That’s impossible:
- The largest 3-digit number is 999.
- 999 - 100 = 899 → still only 3 digits.

So a 3-digit minus 3-digit cannot give a 4-digit result.

But here:
```
1 _ 6
- 4 1 5
-------
2 8 1 1
```

This can't be right unless there's a typo. But wait — look at the result: 2811. That’s larger than the minuend (1_6) — which is less than 200. So this subtraction is invalid as written.

Wait — perhaps the minuend is actually a 4-digit number, but it's shown as:

```
[ ] 1 [ ] 6
- 4 1 5
---------
2 8 1 1
```

Looking again — the first row has four boxes: `[ ] 1 [ ] 6` — so it's a 4-digit number, with the second digit being 1, and the fourth digit being 6.

Ah! So the format is:

```
A 1 B 6
- 4 1 5
--------
2 8 1 1
```

So we are subtracting a 3-digit number from a 4-digit number, resulting in a 4-digit number.

Let’s write it properly:

```
A 1 B 6
- 4 1 5
-------
2 8 1 1
```

We need to find digits A and B such that:

```
A1B6 - 415 = 2811
```

So:
A1B6 = 2811 + 415 = 3226

Now check:
3226 - 415 = ?

Calculate:
3226 - 400 = 2826
2826 - 15 = 2811 ✔

So the original number is 3226

Thus:
- First digit: A = 3
- Third digit: B = 2

✔ So missing digits: A=3, B=2

Answer for Problem 1:
```
3 1 2 6
- 4 1 5
-------
2 8 1 1
```

---

Problem 2:

```
6 _ 3 _
- _ 2 _ 7
--------
5 1 8 1
```

Let’s denote:

```
6 C 3 D
- E 2 F 7
--------
5 1 8 1
```

We know:
6C3D - E2F7 = 5181
So:
6C3D = 5181 + E2F7

We can try to work from the right (units place):

#### Units column:
D - 7 = 1 → So D = 8? But if D < 7, we borrow.

Let’s do column-wise subtraction.

Start from the right:

Units place:
D - 7 = 1 → D = 8? But if no borrow, then D = 8.
But if D < 7, then D + 10 - 7 = 1 → D = -2 → not possible.
So D must be ≥ 7 → D - 7 = 1 → D = 8

So D = 8

→ No borrow needed.

Tens place:
3 - F = 8 → Not possible unless borrow.

So we borrowed from hundreds? Wait — tens digit: 3 - F = 8?

But 3 < F → must borrow from hundreds.

But hundreds digit is 3 → so we can borrow.

So tens place:
(3 + 10) - F = 8 → 13 - F = 8 → F = 5

And we borrowed 1 from hundreds → so hundreds digit becomes 2 (from 3)

Now hundreds place:
After borrowing, we have 2 (from 3) - 2 = ? But result is 1 → so 2 - 2 = 0 ≠ 1

But result is 1 → so maybe we need to borrow from thousands?

Wait: current hundreds digit after borrow: 2
Subtracting 2 → 2 - 2 = 0 → but result is 1 → contradiction.

So maybe we had a borrow from thousands?

Let’s track carefully.

Thousands: 6 - E = 5 → So E = 1? Possibly.

But let’s go step by step.

Let’s assume:

```
6 C 3 D
- E 2 F 7
---------
5 1 8 1
```

We already found D = 8 (from units: D - 7 = 1 → D = 8)

Tens: 3 - F = 8 → impossible without borrow.

So borrow from hundreds → tens becomes 13 - F = 8 → F = 5

So F = 5

Now, hundreds digit: was 3, but we borrowed 1 → now 2

Now: 2 - 2 = 0 → but result is 1 → so we need to borrow from thousands.

So borrow from 6 → thousands becomes 5, hundreds becomes 12

Then: 12 - 2 = 10 → but we need 1 → too big.

Wait, no: after borrow, hundreds is 12, subtract 2 → 10 → but result digit is 1 → not matching.

Wait — maybe I made a mistake.

Let’s reframe: We know:

6C3D - E2F7 = 5181

Try adding 5181 + E2F7 = 6C3D

Let’s suppose E = 1 → then E2F7 = 12F7

So 5181 + 12F7 = 6C3D

Try F = 5 → 1257

5181 + 1257 = 6438

So 6C3D = 6438 → so C = 4, D = 8

Check:
6438 - 1257 = ?

6438 - 1200 = 5238
5238 - 57 = 5181 ✔

Perfect!

So:
- C = 4
- F = 5
- E = 1
- D = 8

So the numbers are:

```
6 4 3 8
- 1 2 5 7
--------
5 1 8 1
```

✔ All digits filled.

---

Problem 3:

```
_ 2 3 _
- 1 _ _ 7
--------
3 7 8 2
```

Let’s denote:

```
A 2 3 B
- 1 C D 7
--------
3 7 8 2
```

So: A23B - 1CD7 = 3782

So A23B = 3782 + 1CD7

Try to solve column by column.

Start from right:

Units: B - 7 = 2 → B = 9? Or B < 7 → borrow.

Case 1: B ≥ 7 → B - 7 = 2 → B = 9

Case 2: B < 7 → B + 10 - 7 = 2 → B = -1 → impossible.

So B = 9

No borrow.

Tens: 3 - D = 8 → impossible → must borrow.

So borrow from hundreds → tens becomes 13 - D = 8 → D = 5

Now, hundreds digit: was 2, but we borrowed 1 → now 1

Now: 1 - C = 7 → impossible unless borrow from thousands.

So borrow from A → thousands becomes A-1, hundreds becomes 11

Then: 11 - C = 7 → C = 4

Now thousands: A - 1 - 1 = 3 → because we subtracted 1 (from 1CD7) and also had borrow

Wait: thousands: (A - 1) - 1 = 3 → A - 2 = 3 → A = 5

So A = 5

Now verify:

Minuend: 5239
Subtrahend: 1457
Difference: 5239 - 1457 = ?

5239 - 1400 = 3839
3839 - 57 = 3782 ✔

Perfect!

So:
- A = 5
- B = 9
- C = 4
- D = 5

Answer:
```
5 2 3 9
- 1 4 5 7
--------
3 7 8 2
```

---

Problem 4:

```
8 _ 1 _
- _ 0 _ 7
--------
4 3 8 9
```

Let’s denote:

```
8 C 1 D
- E 0 F 7
--------
4 3 8 9
```

So: 8C1D - E0F7 = 4389

So 8C1D = 4389 + E0F7

Try solving column by column.

Units: D - 7 = 9 → D = 6? But D < 7 → borrow.

So D + 10 - 7 = 9 → D = 6

Yes → D = 6

Borrow from tens.

Tens: 1 - F = 8 → but we borrowed → so tens digit is 0 (after borrow)

So: 0 - F → impossible → borrow from hundreds.

So: 10 - F = 8 → F = 2

But we borrowed → so hundreds digit reduced by 1.

Hundreds digit: was C, now C - 1

Now: (C - 1) - 0 = 3 → so C - 1 = 3 → C = 4

Now thousands: 8 - E = 4 → so E = 4

But wait: did we borrow? From hundreds → yes, we borrowed from C → but C is part of 8C1D, so thousands digit is 8 → we didn’t borrow from thousands, just from hundreds.

So thousands: 8 - E = 4 → E = 4

Now test:

Minuend: 8416
Subtrahend: 4027
Difference: 8416 - 4027 = ?

8416 - 4000 = 4416
4416 - 27 = 4389 ✔

Perfect!

So:
- C = 4
- D = 6
- E = 4
- F = 2

Answer:
```
8 4 1 6
- 4 0 2 7
--------
4 3 8 9
```

---

Problem 5:

```
_ 8 0 _
- 2 _ _ 6
--------
2 2 6 5
```

Denote:

```
A 8 0 B
- 2 C D 6
--------
2 2 6 5
```

So: A80B - 2CD6 = 2265 → A80B = 2265 + 2CD6

Solve column by column.

Units: B - 6 = 5 → B = 11? Impossible → so borrow.

B + 10 - 6 = 5 → B = 1 → B = 1

Borrow from tens.

Tens: 0 - D = 6 → but we borrowed → so tens digit is -1 → need borrow from hundreds.

So: 10 - D = 6 → D = 4

But we borrowed → so hundreds digit reduced.

Hundreds digit: was 8 → now 7

Now: 7 - C = 2 → so C = 5

Now thousands: A - 2 = 2 → A = 4

But check: did we borrow from thousands? Only if hundreds needed borrow.

We borrowed from hundreds → so thousands digit becomes A - 1

So: (A - 1) - 2 = 2 → A - 3 = 2 → A = 5

Wait — conflict.

Wait: thousands digit: A - 2 = 2 → A = 4? But we may have borrowed.

Let’s think:

We had to borrow from hundreds (for tens), so hundreds digit became 7 instead of 8.

But thousands digit is A → no borrow from thousands yet.

So thousands: A - 2 = 2 → A = 4

But then A = 4

Now check total: minuend = 4801
Subtrahend = 2546
Difference: 4801 - 2546 = ?

4801 - 2500 = 2301
2301 - 46 = 2255 ≠ 2265 → too low.

Wait — discrepancy.

Let’s recalculate.

We had:
- B = 1 (from units)
- D = 4 (from tens)
- C = 5 (from hundreds: 7 - 5 = 2)
- A = ? → thousands: A - 2 = 2 → A = 4

But 4801 - 2546 = 2255 → but should be 2265 → off by 10.

Hmm.

Wait — maybe we missed a borrow.

Let’s do full subtraction:

Try A = 5 → minuend = 5801

Subtrahend = 2546

5801 - 2546 = ?

5801 - 2500 = 3301
3301 - 46 = 3255 → too big.

Wait — we want difference 2265

So minuend = 2265 + 2CD6

Try: 2265 + 2546 = 4811

So minuend = 4811

So A80B = 4811 → so A = 4, B = 1

Yes! So A = 4, B = 1

Now subtrahend: 2CD6 = 2546 → C = 5, D = 4

Now check:

4811 - 2546 = ?

4811 - 2500 = 2311
2311 - 46 = 2265 ✔

Perfect.

But earlier we said B = 1, D = 4, C = 5, A = 4

Now verify column-by-column:

```
4 8 0 1
- 2 5 4 6
---------
2 2 5 5?
```

Wait — 4811 - 2546:

Units: 1 - 6 → borrow → 11 - 6 = 5 → but we want 5 → okay

Tens: 0 - 4 → but borrowed → 9 (after borrow?) → wait:

After borrow from units: tens digit is 0 → but we need to borrow from hundreds.

So tens: 10 - 4 = 6 → but result is 6 → good.

Wait — result tens digit is 6 → yes.

But earlier I thought it was 5 — no, result is 2265 → tens digit is 6 → yes.

So:

Units: 1 - 6 → borrow → 11 - 6 = 5 → ok

Tens: 0 - 4 → but borrowed → so 9 (after borrow from hundreds) → 9 - 4 = 5? Wait — no.

Wait: tens digit is 0 → we need to borrow from hundreds → so tens becomes 10 → but we already borrowed for units? No — borrow happens from higher digits.

Standard borrowing:

Start from units:

- Units: 1 < 6 → borrow from tens → tens is 0 → can't → borrow from hundreds → hundreds is 8 → so:

- Hundreds: 8 → becomes 7
- Tens: 0 → becomes 10
- But we need to borrow for units → so tens gives 1 to units → tens becomes 9, units becomes 11

Now:
- Units: 11 - 6 = 5
- Tens: 9 - 4 = 5 → but result is 6 → conflict!

Wait — result tens digit is 6 → but we get 5 → contradiction.

But earlier calculation: 4811 - 2546 = 2265 → let’s compute:

4811
-2546
------
Start:
1 - 6 → borrow → 11 - 6 = 5
Tens: 0 - 4 → but borrowed → so 9 (since 10 - 1 = 9) → 9 - 4 = 5
But result tens digit is 6 → not matching.

Wait — 4811 - 2546:

Do it properly:

```
4811
- 2546
------
```

Start from right:

1 - 6 → borrow → 11 - 6 = 5
Tens: 0 → but borrowed → so 9 → 9 - 4 = 5
Hundreds: 8 → but we borrowed → 7 → 7 - 5 = 2
Thousands: 4 - 2 = 2

So result: 2255 → but we wanted 2265 → off by 10.

But earlier we said 2265 + 2546 = 4811 → 2265 + 2546 = ?

2265 + 2500 = 4765
4765 + 46 = 4811 → yes

But 4811 - 2546 = 2265? Let’s compute:

4811 - 2546:

4811 - 2500 = 2311
2311 - 46 = 2265 → yes

But manual subtraction shows 2255? No — let’s do it:

```
4811
- 2546
------
```

Align:

```
4811
-2546
-----
```

Step by step:

- Units: 1 - 6 → can't → borrow from tens → tens is 0 → can't → borrow from hundreds → hundreds is 8 → so:

- Hundreds: 8 → 7
- Tens: 0 → 10
- Now borrow for units: tens gives 1 → tens becomes 9, units becomes 11

Units: 11 - 6 = 5

Tens: 9 - 4 = 5

Hundreds: 7 - 5 = 2

Thousands: 4 - 2 = 2

Result: 2255 → but should be 2265 → contradiction.

Wait — what's 4811 - 2546?

Let’s calculate directly:

4811 - 2546 = (4811 - 2500) = 2311
2311 - 46 = 2265 → yes

But why does manual subtraction give 2255?

Wait — I see the error: when we borrow from hundreds, hundreds becomes 7, tens becomes 10, then we borrow from tens for units → tens becomes 9, units becomes 11 → correct.

Tens: 9 - 4 = 5 → but in the result, tens digit is 6 → but we have 5 → inconsistency.

But 4811 - 2546 = 2265 → so tens digit should be 6 → not 5.

So where is the mistake?

Wait — 4811 - 2546:

Let me do it on paper:

```
4811
-2546
------
```

Start:

- Units: 1 - 6 → borrow → 11 - 6 = 5 → carry over → tens digit reduced by 1
- Tens: 0 - 4 → but we borrowed → so 0 - 1 = -1 → can't → borrow from hundreds
- Hundreds: 8 → becomes 7, tens becomes 10 → but we already borrowed → so tens becomes 9
- Tens: 9 - 4 = 5
- Hundreds: 7 - 5 = 2
- Thousands: 4 - 2 = 2

Result: 2255

But 4811 - 2546 = 2265? Let's add 2255 + 2546 = ?

2255 + 2546 = 4801 → not 4811

Oh! So 4811 - 2546 = 2265 is wrong.

2265 + 2546 = ?

2265 + 2500 = 4765
4765 + 46 = 4811 → yes

So 2265 + 2546 = 4811 → so 4811 - 2546 = 2265 → correct

But manual subtraction gives 2255 → so I must have made a mistake.

Wait — 2265 + 2546:

2265 + 2546:

5+6=11 → write 1, carry 1
6+4+1=11 → write 1, carry 1
2+5+1=8
2+2=4 → 4811 → yes

So 4811 - 2546 = 2265 → correct

But my subtraction gave 2255 → so where is the error?

Ah! In the tens place:

When we borrow:

- Units: 1 - 6 → borrow from tens → tens is 0 → can't → borrow from hundreds → hundreds is 8 → so hundreds becomes 7, tens becomes 10
- Now, borrow from tens for units: tens gives 1 to units → tens becomes 9, units becomes 11 → 11 - 6 = 5
- Tens: 9 - 4 = 5
- Hundreds: 7 - 5 = 2
- Thousands: 4 - 2 = 2

Result: 2255

But it should be 2265 → so something is wrong.

Wait — perhaps the subtrahend is not 2546.

Let’s go back.

We have:

A80B - 2CD6 = 2265

We tried A=4, B=1 → minuend = 4801

Then 4801 - 2265 = 2536 → so subtrahend = 2536

So 2CD6 = 2536 → C=5, D=3

Try that:

```
4801
- 2536
------
```

Units: 1 - 6 → borrow → 11 - 6 = 5
Tens: 0 - 3 → but borrowed → so 9 - 3 = 6
Hundreds: 8 → borrowed → 7 - 5 = 2
Thousands: 4 - 2 = 2

Result: 2265 ✔

Yes!

So:
- Minuend: 4801 → A=4, B=1
- Subtrahend: 2536 → C=5, D=3

So answer:

```
4 8 0 1
- 2 5 3 6
--------
2 2 6 5
```

Now verify:

4801 - 2536 = ?

4801 - 2500 = 2301
2301 - 36 = 2265 ✔

Perfect.

So missing digits:
- A = 4
- B = 1
- C = 5
- D = 3

---

Problem 6:

```
9 _ 2 _
- _ 5 _ 8
--------
7 5 1 8
```

Denote:

```
9 C 2 D
- E 5 F 8
--------
7 5 1 8
```

So: 9C2D - E5F8 = 7518

So 9C2D = 7518 + E5F8

Try column by column.

Units: D - 8 = 8 → D = 16? Impossible → so borrow.

D + 10 - 8 = 8 → D = 6

So D = 6

Tens: 2 - F = 1 → but we borrowed → so tens digit is 1 (after borrow)

So: 1 - F = 1 → F = 0

But could be borrow.

If 1 - F = 1 → F = 0

Or if F > 1 → borrow from hundreds.

Try F = 0 → 1 - 0 = 1 → good

No borrow.

Hundreds: C - 5 = 5 → C = 10? Impossible → so borrow.

So: (C + 10) - 5 = 5 → C = 0

But we borrowed → so C was 0, now becomes 0 after borrow? Wait.

After borrow: hundreds digit becomes C - 1 (because we borrowed for tens? No — we didn't borrow for tens, since F=0.

But we are at hundreds: C - 5 = 5 → impossible → so must borrow from thousands.

So borrow from 9 → thousands becomes 8, hundreds becomes 10 + C

Then: (10 + C) - 5 = 5 → 10 + C = 10 → C = 0

So C = 0

Now thousands: 8 - E = 7 → E = 1

Now check:

Minuend: 9026
Subtrahend: 1508
Difference: 9026 - 1508 = ?

9026 - 1500 = 7526
7526 - 8 = 7518 ✔

Perfect!

So:
- C = 0
- D = 6
- E = 1
- F = 0

Answer:
```
9 0 2 6
- 1 5 0 8
--------
7 5 1 8
```

---

✔ Final Answers:

1.
```
3 1 2 6
- 4 1 5
-------
2 8 1 1
```

2.
```
6 4 3 8
- 1 2 5 7
-------
5 1 8 1
```

3.
```
5 2 3 9
- 1 4 5 7
-------
3 7 8 2
```

4.
```
8 4 1 6
- 4 0 2 7
-------
4 3 8 9
```

5.
```
4 8 0 1
- 2 5 3 6
-------
2 2 6 5
```

6.
```
9 0 2 6
- 1 5 0 8
-------
7 5 1 8
```

All problems solved!