Printable math worksheet for students to practice solving missing number problems with basic operations and exponents.

Let's solve each of these missing number problems step by step. We'll isolate the missing number (represented by a box) and solve for it using order of operations (PEMDAS: Parentheses, Exponents, Multiplication/Division, Addition/Subtraction).

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1)

$$
\boxed{} + 2^2 + (4 \times 4) = 7
$$

Step 1: Evaluate exponents and parentheses:
- $2^2 = 4$
- $4 \times 4 = 16$

Now:
$$
\boxed{} + 4 + 16 = 7
\Rightarrow \boxed{} + 20 = 7
\Rightarrow \boxed{} = 7 - 20 = -13
$$

✔ Answer: -13

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2)

$$
2^2 + (\boxed{} - 6) \times 4 = 44
$$

Step 1: $2^2 = 4$

So:
$$
4 + (\boxed{} - 6) \times 4 = 44
$$

Subtract 4 from both sides:
$$
(\boxed{} - 6) \times 4 = 40
$$

Divide by 4:
$$
\boxed{} - 6 = 10
\Rightarrow \boxed{} = 16
$$

✔ Answer: 16

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3)

$$
8 + 21 + \boxed{} \times (2 - 3^3) = -13
$$

Step 1: Evaluate exponent and parentheses:
- $3^3 = 27$
- $2 - 27 = -25$

So:
$$
8 + 21 + \boxed{} \times (-25) = -13
\Rightarrow 29 + (-25)\boxed{} = -13
$$

Subtract 29:
$$
-25\boxed{} = -13 - 29 = -42
$$

Divide:
$$
\boxed{} = \frac{-42}{-25} = \frac{42}{25} = 1.68
$$

But let’s keep it as a fraction: $ \frac{42}{25} $

Wait — check if this is correct.

Wait: Is the expression:
$$
8 + 21 + \boxed{} \times (2 - 3^3)
$$

Yes. So:
- $3^3 = 27$, so $2 - 27 = -25$
- $8 + 21 = 29$
- So: $29 + \boxed{} \times (-25) = -13$

Then:
$$
-25\boxed{} = -13 - 29 = -42
\Rightarrow \boxed{} = \frac{-42}{-25} = \frac{42}{25}
$$

✔ Answer: $ \frac{42}{25} $ or $ 1.68 $

But maybe we made an error? Let’s double-check:

$ \frac{42}{25} \times (-25) = -42 $, then $29 - 42 = -13$. ✔ Correct.

So answer is $ \frac{42}{25} $

But perhaps the problem expects an integer? Let’s see.

Wait — could it be that the expression is:
$$
8 + 21 + \boxed{} \times (2 - 3)^3 ?
$$

No, original says: $ (2 - 3^3) $. That means $3^3 = 27$, so $2 - 27 = -25$. So correct.

So answer is $ \boxed{\frac{42}{25}} $

But let’s write it as a mixed number: $ 1 \frac{17}{25} $

We’ll keep it as $ \frac{42}{25} $

✔ Answer: $ \frac{42}{25} $

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4)

$$
\boxed{} + (4^2 + 3) = 3
$$

$4^2 = 16$, so:
$$
\boxed{} + (16 + 3) = 3
\Rightarrow \boxed{} + 19 = 3
\Rightarrow \boxed{} = 3 - 19 = -16
$$

✔ Answer: -16

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5)

$$
49 + (3 + 4) - 9^2 + \boxed{} = -72
$$

Evaluate:
- $3 + 4 = 7$
- $9^2 = 81$

So:
$$
49 + 7 - 81 + \boxed{} = -72
\Rightarrow (56 - 81) + \boxed{} = -72
\Rightarrow -25 + \boxed{} = -72
\Rightarrow \boxed{} = -72 + 25 = -47
$$

✔ Answer: -47

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6)

$$
(11 + 6^2) \times \boxed{} - 7 = 40
$$

First: $6^2 = 36$, so $11 + 36 = 47$

So:
$$
47 \times \boxed{} - 7 = 40
\Rightarrow 47 \times \boxed{} = 47
\Rightarrow \boxed{} = 1
$$

✔ Answer: 1

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7)

$$
\boxed{} + (4 - 5) \times 5^3 = -115
$$

Evaluate:
- $4 - 5 = -1$
- $5^3 = 125$

So:
$$
\boxed{} + (-1) \times 125 = -115
\Rightarrow \boxed{} - 125 = -115
\Rightarrow \boxed{} = -115 + 125 = 10
$$

✔ Answer: 10

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8)

$$
36 + \boxed{} + 3^4 - (5 - 12) = 97
$$

Evaluate:
- $3^4 = 81$
- $5 - 12 = -7$

So:
$$
36 + \boxed{} + 81 - (-7) = 97
\Rightarrow 36 + \boxed{} + 81 + 7 = 97
\Rightarrow \boxed{} + 124 = 97
\Rightarrow \boxed{} = 97 - 124 = -27
$$

✔ Answer: -27

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9)

$$
7^2 + (\boxed{} + 5) = 49
$$

$7^2 = 49$, so:
$$
49 + (\boxed{} + 5) = 49
\Rightarrow \boxed{} + 5 = 0
\Rightarrow \boxed{} = -5
$$

✔ Answer: -5

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10)

$$
\boxed{} - 7 + (2^3 \times 3) = 52
$$

Evaluate:
- $2^3 = 8$
- $8 \times 3 = 24$

So:
$$
\boxed{} - 7 + 24 = 52
\Rightarrow \boxed{} + 17 = 52
\Rightarrow \boxed{} = 52 - 17 = 35
$$

✔ Answer: 35

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✔ Final Answers:

| Problem | Answer |
|--------|--------|
| 1) | -13 |
| 2) | 16 |
| 3) | $ \frac{42}{25} $ (or 1.68) |
| 4) | -16 |
| 5) | -47 |
| 6) | 1 |
| 7) | 10 |
| 8) | -27 |
| 9) | -5 |
| 10) | 35 |

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Let me know if you'd like the answers filled in on the worksheet format!