4-Digit Addition Missing Numbers Challenge 1 worksheet for practicing addition with missing digits.

Let’s solve each problem one by one. We’ll fill in the missing digits so that the addition is correct. Remember: when adding, if a column adds up to 10 or more, we carry over to the next column on the left.

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Problem 1:
```
3 2 _ 5
+ _ 4 8 2
----------
4 _ 9 7
```

Start from rightmost column (units):

- Units: 5 + 2 = 7 → OK, no carry.
- Tens: _ + 8 = 9 → So blank must be 1.
- Hundreds: 2 + 4 = 6 → But answer shows _ → Wait, let’s check again.

Wait — actually, look at hundreds place: top number has “2”, bottom has “4”, sum should be “_” in hundreds place of result. But result says “_97” — so hundreds digit is missing? Actually, looking again:

Actually, the layout is:

Top: 3 2 _ 5
Bottom: _ 4 8 2
Sum: 4 _ 9 7

So aligning properly:

Thousands | Hundreds | Tens | Units
3 | 2 | _ | 5
| _ | 4 | 8 2 ← wait, this doesn’t align! Oh — it's 4-digit numbers, so probably:

Actually, re-aligning with proper columns:

It should be:

Thousands Hundreds Tens Units
3 2 ? 5
? 4 8 2 ← no, that can't be — second number is 3-digit? No, title says 4-digit additions.

Looking back at image description — all are 4-digit additions. So likely:

Problem 1 is:

3 2 _ 5
+ _ 4 8 2 ← but this is only 3 digits? That can’t be.

Wait — perhaps formatting issue. Let me reinterpret based on standard worksheet format.

Actually, looking at problem 1 as written:

"3 2 _ 5" — that’s 4 digits: thousands=3, hundreds=2, tens=?, units=5

Then "+ _ 4 8 2" — that’s also 4 digits? But starts with underscore — so thousands digit missing? Yes!

So:

Number A: 3 2 X 5
Number B: Y 4 8 2
Sum: 4 Z 9 7

Now add column by column from right:

Units: 5 + 2 = 7 → matches → no carry.

Tens: X + 8 = 9 → so X = 1 → no carry.

Hundreds: 2 + 4 = 6 → but sum shows Z → so Z = 6? But then thousands: 3 + Y = 4 → so Y = 1.

Check: 3215 + 1482 = ?

3215 + 1482 = 4697 → yes! Sum is 4 6 9 7 → so Z=6.

✔ Problem 1: Missing digits: tens of first number = 1; thousands of second = 1; hundreds of sum = 6.

But the blanks are:

First number: 3 2 _ 5 → blank is tens → 1
Second number: _ 4 8 2 → blank is thousands → 1
Sum: 4 _ 9 7 → blank is hundreds → 6

So answers for prob 1: 1, 1, 6

But since the question is to "work out the missing numbers", and there are multiple blanks per problem, we need to fill each blank.

We'll go problem by problem and list the missing digits in order they appear.

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Let me make a table for clarity.

I’ll write each problem with blanks labeled A, B, C etc., then solve.

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Problem 1:

A B C D
3 2 A 5
+ B 4 8 2
---------
4 C 9 7

From above: A=1, B=1, C=6

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Problem 2:

5 A 3 1
+ 1 4 B C
---------
D 5 5 8

Add column by column:

Units: 1 + C = 8 → C = 7 → no carry.

Tens: 3 + B = 5 → B = 2 → no carry.

Hundreds: A + 4 = 5 → A = 1 → no carry.

Thousands: 5 + 1 = 6 → so D = 6.

Check: 5131 + 1427 = 6558 → yes.

So A=1, B=2, C=7, D=6

Blanks: first number tens digit = A=1; second number tens and units = B=2, C=7; sum thousands = D=6

---

Problem 3:

6 4 A 2
+ B 2 3 6
---------
8 C 7 D

Units: 2 + 6 = 8 → but sum has D → so D=8? Wait, 2+6=8 → yes, D=8, no carry.

Tens: A + 3 = 7 → A=4 → no carry.

Hundreds: 4 + 2 = 6 → but sum has C → so C=6? But wait, thousands: 6 + B = 8 → B=2.

Check: 6442 + 2236 = 8678 → yes.

So A=4, B=2, C=6, D=8

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Problem 4:

4 6 A 2
+ B 1 3 C
---------
5 D 1 9

Units: 2 + C = 9 → C=7 → no carry.

Tens: A + 3 = 1 → impossible without carry. So A + 3 = 11 → A=8, carry 1 to hundreds.

Hundreds: 6 + 1 + carry1 = 8 → but sum has D → wait, sum is 5 D 1 9 → so hundreds digit is D? Let's see:

After tens: we had carry 1 to hundreds.

Hundreds: 6 (from top) + 1 (from bottom) + 1 (carry) = 8 → so D=8? But thousands: 4 + B = 5 → B=1.

Check: 4682 + 1137 = 5819 → yes.

So A=8, B=1, C=7, D=8

---

Problem 5:

6 A 4 0
+ 2 3 7 B
---------
C 6 D 7

Units: 0 + B = 7 → B=7 → no carry.

Tens: 4 + 7 = 11 → so write 1, carry 1 → so D=1? Sum tens digit is D → yes, D=1, carry 1 to hundreds.

Hundreds: A + 3 + carry1 = 6 → A + 4 = 6 → A=2 → no further carry.

Thousands: 6 + 2 = 8 → so C=8.

Check: 6240 + 2377 = 8617 → yes.

So A=2, B=7, C=8, D=1

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Problem 6:

3 7 1 A
+ B 5 C 7
---------
6 D 4 9

Units: A + 7 = 9 → A=2 → no carry.

Tens: 1 + C = 4 → C=3 → no carry.

Hundreds: 7 + 5 = 12 → write 2, carry 1 → but sum has D in hundreds? Sum is 6 D 4 9 → so hundreds digit is D → but 7+5=12, plus any carry? From tens: no carry, so 7+5=12 → so write 2, carry 1 to thousands.

So D=2? But then thousands: 3 + B + carry1 = 6 → 3 + B +1 =6 → B=2.

Check: 3712 + 2537 = 6249 → yes.

So A=2, B=2, C=3, D=2

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Problem 7:

2 0 A 7
+ B 7 3 5
---------
3 C 3 D

Units: 7 + 5 = 12 → write 2, carry 1 → so D=2.

Tens: A + 3 + carry1 = 3 → A + 4 = 3 → impossible → so must be 13 → A + 4 = 13 → A=9, carry 1 to hundreds.

Hundreds: 0 + 7 + carry1 = 8 → but sum has C → so C=8? Thousands: 2 + B = 3 → B=1.

Check: 2097 + 1735 = 3832 → yes.

So A=9, B=1, C=8, D=2

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Problem 8:

7 1 3 5
+ A 7 B 6
---------
1 2 C 5 D

This is 4-digit + 4-digit = 5-digit? Sum is 12C5D — so 5 digits.

Units: 5 + 6 = 11 → write 1, carry 1 → so D=1.

Tens: 3 + B + carry1 = 5 → 3 + B +1 =5 → B=1 → no carry.

Hundreds: 1 + 7 = 8 → but sum has C → so C=8? Thousands: 7 + A = 12? Because sum is 12... so 7 + A must be 12 or more? 7 + A = 12 → A=5, and carry 1 to ten-thousands? But sum starts with 12, so ten-thousands digit is 1, which comes from carry.

Actually:

Thousands column: 7 + A = ? and we have carry from hundreds? Hundreds was 1+7=8, no carry to thousands.

So 7 + A = 12? Then A=5, and write 2, carry 1 to ten-thousands → which becomes the '1' in 12...

Yes.

So A=5, B=1, C=8, D=1

Check: 7135 + 5716 = 12851 → yes.

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Problem 9:

A 4 B 5
+ 3 0 1 8
---------
1 1 C 8 D

Units: 5 + 8 = 13 → write 3, carry 1 → D=3.

Tens: B + 1 + carry1 = 8 → B + 2 = 8 → B=6 → no carry.

Hundreds: 4 + 0 = 4 → but sum has C → so C=4? Thousands: A + 3 = 11? Because sum is 11... so A + 3 = 11 → A=8, and carry 1 to ten-thousands → which is the first '1'.

Check: 8465 + 3018 = 11483 → yes.

So A=8, B=6, C=4, D=3

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Problem 10:

A 1 3 B
+ 7 5 C 3
---------
1 2 D 2 5

Sum is 5 digits: 12D25

Units: B + 3 = 5 → B=2 → no carry.

Tens: 3 + C = 2 → impossible → so 3 + C = 12 → C=9, carry 1.

Hundreds: 1 + 5 + carry1 = 7 → but sum has D → so D=7? Thousands: A + 7 = 12? Because sum is 12... so A + 7 = 12 → A=5, carry 1 to ten-thousands → which is the '1'.

Check: 5132 + 7593 = 12725 → yes.

So A=5, B=2, C=9, D=7

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Problem 11:

7 A 4 B
+ C 8 3 6
---------
1 1 5 D 8

Sum is 5 digits: 115D8

Units: B + 6 = 8 → B=2 → no carry.

Tens: 4 + 3 = 7 → but sum has D → so D=7? No carry.

Hundreds: A + 8 = 5 → impossible → so A + 8 = 15 → A=7, carry 1.

Thousands: 7 + C + carry1 = 11? Because sum is 11... so 7 + C +1 = 11 → C=3, and carry 1 to ten-thousands → which is the first '1'.

Check: 7742 + 3836 = 11578 → yes.

So A=7, B=2, C=3, D=7

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Problem 12:

9 A 7 1
+ 6 5 B 5
---------
1 C 8 1 D

Sum is 5 digits: 1C81D

Units: 1 + 5 = 6 → but sum has D → so D=6? No carry.

Tens: 7 + B = 1 → impossible → so 7 + B = 11 → B=4, carry 1.

Hundreds: A + 5 + carry1 = 8 → A + 6 = 8 → A=2 → no carry.

Thousands: 9 + 6 = 15 → write 5, carry 1 → but sum has C in thousands? Sum is 1 C 8 1 D → so thousands digit is C → 9+6=15, so write 5, carry 1 to ten-thousands → which is the '1'. So C=5.

Check: 9271 + 6545 = 15816 → yes.

So A=2, B=4, C=5, D=6

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Problem 13: This one has three numbers!

4 1 4 6
+ 2 A B 3
+ 5 2 7 C
---------
D E 9 2 3

Sum is 5 digits: DE923

Add step by step.

First, units: 6 + 3 + C = 3 or 13 or 23? Must end with 3.

6+3=9, so 9 + C ends with 3 → C=4 (since 9+4=13), carry 1.

Tens: 4 + B + 7 + carry1 = 2 or 12 or 22? Sum tens digit is 2.

4 + B + 7 +1 = 12 + B → must end with 2 → so 12 + B = 12 or 22? If 12, B=0; if 22, B=10 invalid. So B=0, and carry 1 to hundreds? 12 +0 =12 → write 2, carry 1.

Hundreds: 1 + A + 2 + carry1 = 9 or 19? Sum hundreds digit is 9.

1 + A + 2 +1 = 4 + A → set equal to 9 → A=5 → no carry? 4+5=9, yes.

Thousands: 4 + 2 + 5 = 11 → but sum has E in thousands? Sum is D E 9 2 3 → so thousands digit is E, ten-thousands is D.

4+2+5=11 → write 1, carry 1 to ten-thousands → so E=1, D=1 (from carry).

Check: 4146 + 2503 + 5274 = ?

4146 + 2503 = 6649; 6649 + 5274 = 11923 → yes, matches DE923 with D=1,E=1.

So A=5, B=0, C=4, D=1, E=1

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Problem 14:

4 3 7 A
+ B 7 0 6
+ 6 2 C 0
---------
D 4 E 0 8

Sum is 5 digits: D4E08

Units: A + 6 + 0 = 8 → A=2 → no carry.

Tens: 7 + 0 + C = 0 or 10 or 20? Sum tens digit is 0.

7 + 0 + C = 10 → C=3, carry 1.

Hundreds: 3 + 7 + 2 + carry1 = 13 → write 3, carry 1 → but sum has E in hundreds? Sum is D 4 E 0 8 → so hundreds digit is E → 3+7+2+1=13 → E=3, carry 1.

Thousands: 4 + B + 6 + carry1 = 14? Because sum has 4 in thousands? Sum is D 4 E 0 8 → thousands digit is 4.

So 4 + B + 6 +1 = 11 + B → must equal 14? Because we need to write 4 and carry 1 to ten-thousands? 11 + B = 14 → B=3, and carry 1 to ten-thousands → D=1.

Check: 4372 + 3706 + 6230 = ?

4372 + 3706 = 8078; 8078 + 6230 = 14308 → yes, matches D4E08 with D=1,E=3.

So A=2, B=3, C=3, D=1, E=3

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Problem 15:

2 0 5 7
+ 5 A 7 2
+ 7 1 B 8
---------
C D 6 5 E

Sum is 5 digits: CD65E

Units: 7 + 2 + 8 = 17 → write 7, carry 1 → E=7.

Tens: 5 + 7 + B + carry1 = 5 or 15 or 25? Sum tens digit is 5.

5+7+B+1=13+B → must end with 5 → so 13+B=15 → B=2, carry 1.

Hundreds: 0 + A + 1 + carry1 = 6 or 16? Sum hundreds digit is 6.

0 + A + 1 +1 = A + 2 → set to 6 → A=4 → no carry.

Thousands: 2 + 5 + 7 = 14 → write 4, carry 1 → but sum has D in thousands? Sum is C D 6 5 E → thousands digit is D → 2+5+7=14 → write 4, carry 1 to ten-thousands → D=4, C=1.

Check: 2057 + 5472 + 7128 = ?

2057 + 5472 = 7529; 7529 + 7128 = 14657 → yes, matches CD65E with C=1,D=4,E=7.

So A=4, B=2, C=1, D=4, E=7

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Now, compiling all answers. Since the problem asks to "work out the missing numbers", and each problem has blanks, I'll list the missing digits for each problem in the order they appear in the problem.

For example, in problem 1: blanks are in first number (tens), second number (thousands), sum (hundreds) → so 1, 1, 6

Similarly for others.

Let me list them:

Problem 1: 1, 1, 6
Problem 2: 1, 2, 7, 6
Problem 3: 4, 2, 6, 8
Problem 4: 8, 1, 7, 8
Problem 5: 2, 7, 8, 1
Problem 6: 2, 2, 3, 2
Problem 7: 9, 1, 8, 2
Problem 8: 5, 1, 8, 1
Problem 9: 8, 6, 4, 3
Problem 10: 5, 2, 9, 7
Problem 11: 7, 2, 3, 7
Problem 12: 2, 4, 5, 6
Problem 13: 5, 0, 4, 1, 1
Problem 14: 2, 3, 3, 1, 3
Problem 15: 4, 2, 1, 4, 7

To present clearly, I'll write each problem's missing digits in sequence.

Final Answer:

1) 1, 1, 6
2) 1, 2, 7, 6
3) 4, 2, 6, 8
4) 8, 1, 7, 8
5) 2, 7, 8, 1
6) 2, 2, 3, 2
7) 9, 1, 8, 2
8) 5, 1, 8, 1
9) 8, 6, 4, 3
10) 5, 2, 9, 7
11) 7, 2, 3, 7
12) 2, 4, 5, 6
13) 5, 0, 4, 1, 1
14) 2, 3, 3, 1, 3
15) 4, 2, 1, 4, 7