Subtraction worksheet with missing digits for students to solve.

Problem: Solve the missing digits in the subtraction problems.

We are given six subtraction problems with missing digits. We need to determine the missing digits step by step using logical reasoning and basic arithmetic principles.

---

Problem 1:

\[
\begin{array}{r}
\square 1 \square 6 \\
- 4 \square 1 5 \\
\hline
2 8 1 1 \\
\end{array}
\]

#### Step-by-step solution:
1. Units place:
- The result in the units place is `1`.
- Since `6 - 5 = 1`, the units digit is correct as it is.

2. Tens place:
- The result in the tens place is `1`.
- To get `1` in the tens place, we need to borrow from the hundreds place. This means the original number in the tens place of the top number must be `0` (since `10 - 9 = 1`).
- So, the top number in the tens place is `0`.

3. Hundreds place:
- After borrowing, the hundreds place of the top number becomes `0` (originally `1`).
- The result in the hundreds place is `8`.
- To get `8` in the hundreds place, we need to borrow from the thousands place. This means the original number in the hundreds place of the bottom number must be `2` (since `10 - 2 = 8`).
- So, the bottom number in the hundreds place is `2`.

4. Thousands place:
- After borrowing, the thousands place of the top number becomes `0` (originally `□`).
- The result in the thousands place is `2`.
- To get `2` in the thousands place, the original number in the thousands place of the top number must be `7` (since `7 - 5 = 2`).
- So, the top number in the thousands place is `7`.

#### Final answer for Problem 1:
\[
\begin{array}{r}
7 1 0 6 \\
- 4 5 1 5 \\
\hline
2 8 1 1 \\
\end{array}
\]

---

Problem 2:

\[
\begin{array}{r}
6 \square 3 \square \\
- \square 2 \square 7 \\
\hline
5 1 8 1 \\
\end{array}
\]

#### Step-by-step solution:
1. Units place:
- The result in the units place is `1`.
- Since `□ - 7 = 1`, the units digit of the top number must be `8` (since `8 - 7 = 1`).
- So, the top number in the units place is `8`.

2. Tens place:
- The result in the tens place is `8`.
- To get `8` in the tens place, we need to borrow from the hundreds place. This means the original number in the tens place of the bottom number must be `5` (since `13 - 5 = 8`).
- So, the bottom number in the tens place is `5`.

3. Hundreds place:
- After borrowing, the hundreds place of the top number becomes `2` (originally `3`).
- The result in the hundreds place is `1`.
- To get `1` in the hundreds place, the original number in the hundreds place of the bottom number must be `1` (since `12 - 1 = 11`, and we take `1` after borrowing).
- So, the bottom number in the hundreds place is `1`.

4. Thousands place:
- The result in the thousands place is `5`.
- To get `5` in the thousands place, the original number in the thousands place of the bottom number must be `1` (since `6 - 1 = 5`).
- So, the bottom number in the thousands place is `1`.

#### Final answer for Problem 2:
\[
\begin{array}{r}
6 4 3 8 \\
- 1 2 1 7 \\
\hline
5 1 8 1 \\
\end{array}
\]

---

Problem 3:

\[
\begin{array}{r}
\square 2 3 \square \\
- 1 \square \square 7 \\
\hline
3 7 8 2 \\
\end{array}
\]

#### Step-by-step solution:
1. Units place:
- The result in the units place is `2`.
- Since `□ - 7 = 2`, the units digit of the top number must be `9` (since `9 - 7 = 2`).
- So, the top number in the units place is `9`.

2. Tens place:
- The result in the tens place is `8`.
- To get `8` in the tens place, we need to borrow from the hundreds place. This means the original number in the tens place of the bottom number must be `5` (since `13 - 5 = 8`).
- So, the bottom number in the tens place is `5`.

3. Hundreds place:
- After borrowing, the hundreds place of the top number becomes `1` (originally `2`).
- The result in the hundreds place is `7`.
- To get `7` in the hundreds place, the original number in the hundreds place of the bottom number must be `4` (since `11 - 4 = 7`).
- So, the bottom number in the hundreds place is `4`.

4. Thousands place:
- The result in the thousands place is `3`.
- To get `3` in the thousands place, the original number in the thousands place of the top number must be `5` (since `5 - 1 = 4`, and we take `3` after borrowing).
- So, the top number in the thousands place is `5`.

#### Final answer for Problem 3:
\[
\begin{array}{r}
5 2 3 9 \\
- 1 4 5 7 \\
\hline
3 7 8 2 \\
\end{array}
\]

---

Problem 4:

\[
\begin{array}{r}
8 \square 1 \square \\
- \square 0 \square 7 \\
\hline
4 3 8 9 \\
\end{array}
\]

#### Step-by-step solution:
1. Units place:
- The result in the units place is `9`.
- Since `□ - 7 = 9`, the units digit of the top number must be `6` (since `16 - 7 = 9`).
- So, the top number in the units place is `6`.

2. Tens place:
- The result in the tens place is `8`.
- To get `8` in the tens place, we need to borrow from the hundreds place. This means the original number in the tens place of the bottom number must be `3` (since `11 - 3 = 8`).
- So, the bottom number in the tens place is `3`.

3. Hundreds place:
- After borrowing, the hundreds place of the top number becomes `0` (originally `1`).
- The result in the hundreds place is `3`.
- To get `3` in the hundreds place, the original number in the hundreds place of the bottom number must be `7` (since `10 - 7 = 3`).
- So, the bottom number in the hundreds place is `7`.

4. Thousands place:
- The result in the thousands place is `4`.
- To get `4` in the thousands place, the original number in the thousands place of the bottom number must be `4` (since `8 - 4 = 4`).
- So, the bottom number in the thousands place is `4`.

#### Final answer for Problem 4:
\[
\begin{array}{r}
8 7 1 6 \\
- 4 0 7 7 \\
\hline
4 3 8 9 \\
\end{array}
\]

---

Problem 5:

\[
\begin{array}{r}
\square 8 0 \square \\
- 2 \square \square 6 \\
\hline
2 2 6 5 \\
\end{array}
\]

#### Step-by-step solution:
1. Units place:
- The result in the units place is `5`.
- Since `□ - 6 = 5`, the units digit of the top number must be `1` (since `11 - 6 = 5`).
- So, the top number in the units place is `1`.

2. Tens place:
- The result in the tens place is `6`.
- To get `6` in the tens place, we need to borrow from the hundreds place. This means the original number in the tens place of the bottom number must be `4` (since `10 - 4 = 6`).
- So, the bottom number in the tens place is `4`.

3. Hundreds place:
- After borrowing, the hundreds place of the top number becomes `7` (originally `8`).
- The result in the hundreds place is `2`.
- To get `2` in the hundreds place, the original number in the hundreds place of the bottom number must be `5` (since `7 - 5 = 2`).
- So, the bottom number in the hundreds place is `5`.

4. Thousands place:
- The result in the thousands place is `2`.
- To get `2` in the thousands place, the original number in the thousands place of the top number must be `5` (since `5 - 2 = 3`, and we take `2` after borrowing).
- So, the top number in the thousands place is `5`.

#### Final answer for Problem 5:
\[
\begin{array}{r}
5 8 0 1 \\
- 2 5 4 6 \\
\hline
2 2 6 5 \\
\end{array}
\]

---

Problem 6:

\[
\begin{array}{r}
9 \square 2 \square \\
- \square 5 \square 8 \\
\hline
7 5 1 8 \\
\end{array}
\]

#### Step-by-step solution:
1. Units place:
- The result in the units place is `8`.
- Since `□ - 8 = 8`, the units digit of the top number must be `6` (since `16 - 8 = 8`).
- So, the top number in the units place is `6`.

2. Tens place:
- The result in the tens place is `1`.
- To get `1` in the tens place, we need to borrow from the hundreds place. This means the original number in the tens place of the bottom number must be `1` (since `11 - 1 = 10`, and we take `1` after borrowing).
- So, the bottom number in the tens place is `1`.

3. Hundreds place:
- After borrowing, the hundreds place of the top number becomes `1` (originally `2`).
- The result in the hundreds place is `5`.
- To get `5` in the hundreds place, the original number in the hundreds place of the bottom number must be `6` (since `11 - 6 = 5`).
- So, the bottom number in the hundreds place is `6`.

4. Thousands place:
- The result in the thousands place is `7`.
- To get `7` in the thousands place, the original number in the thousands place of the bottom number must be `2` (since `9 - 2 = 7`).
- So, the bottom number in the thousands place is `2`.

#### Final answer for Problem 6:
\[
\begin{array}{r}
9 6 2 6 \\
- 2 5 6 8 \\
\hline
7 5 1 8 \\
\end{array}
\]

---

Final Answers:

\[
\boxed{
\begin{array}{r}
7 1 0 6 \\
- 4 5 1 5 \\
\hline
2 8 1 1 \\
\end{array}
\quad
\begin{array}{r}
6 4 3 8 \\
- 1 2 1 7 \\
\hline
5 1 8 1 \\
\end{array}
\quad
\begin{array}{r}
5 2 3 9 \\
- 1 4 5 7 \\
\hline
3 7 8 2 \\
\end{array}
\quad
\begin{array}{r}
8 7 1 6 \\
- 4 0 7 7 \\
\hline
4 3 8 9 \\
\end{array}
\quad
\begin{array}{r}
5 8 0 1 \\
- 2 5 4 6 \\
\hline
2 2 6 5 \\
\end{array}
\quad
\begin{array}{r}
9 6 2 6 \\
- 2 5 6 8 \\
\hline
7 5 1 8 \\
\end{array}
}
\]