4 Digit Subtraction with regrouping - skoolon.com - Free Printable

Problem: Solve the 4-digit subtraction problems with regrouping.

The worksheet provides several subtraction problems, and we need to solve them step by step. Let's go through each problem systematically.

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#### Problem 1:
\[
\begin{array}{c@{}c@{}c@{}c}
& \text{T} & \text{H} & \text{T} & \text{O} \\
& 9 & 1 & 7 & 3 \\
- & 5 & 2 & 6 & 8 \\
\hline
& 3 & 9 & 0 & 5 \\
\end{array}
\]
This is already solved as an example. The answer is 3905.

---

#### Problem 2:
\[
\begin{array}{c@{}c@{}c@{}c}
& \text{T} & \text{H} & \text{T} & \text{O} \\
& 6 & 5 & 2 & 0 \\
- & 8 & 6 & 8 & \\
\hline
\end{array}
\]

1. Ones place: \(0 - 8\). Since 0 is less than 8, borrow 1 from the tens place (2 becomes 1, and 0 becomes 10).
\[
10 - 8 = 2
\]
2. Tens place: After borrowing, we have \(1 - 6\). Since 1 is less than 6, borrow 1 from the hundreds place (5 becomes 4, and 1 becomes 11).
\[
11 - 6 = 5
\]
3. Hundreds place: After borrowing, we have \(4 - 8\). Since 4 is less than 8, borrow 1 from the thousands place (6 becomes 5, and 4 becomes 14).
\[
14 - 8 = 6
\]
4. Thousands place: After borrowing, we have \(5 - 0 = 5\).

The final answer is:
\[
\boxed{5652}
\]

---

#### Problem 3:
\[
\begin{array}{c@{}c@{}c@{}c}
& \text{T} & \text{H} & \text{T} & \text{O} \\
& 7 & 0 & 2 & 4 \\
- & 3 & 7 & 6 & 8 \\
\hline
\end{array}
\]

1. Ones place: \(4 - 8\). Since 4 is less than 8, borrow 1 from the tens place (2 becomes 1, and 4 becomes 14).
\[
14 - 8 = 6
\]
2. Tens place: After borrowing, we have \(1 - 6\). Since 1 is less than 6, borrow 1 from the hundreds place (0 becomes 9, and 1 becomes 11).
\[
11 - 6 = 5
\]
3. Hundreds place: After borrowing, we have \(9 - 7 = 2\).
4. Thousands place: \(7 - 3 = 4\).

The final answer is:
\[
\boxed{3256}
\]

---

#### Problem 4:
\[
\begin{array}{c@{}c@{}c@{}c}
& \text{T} & \text{H} & \text{T} & \text{O} \\
& 8 & 0 & 0 & 0 \\
- & 2 & 9 & 9 & 1 \\
\hline
\end{array}
\]

1. Ones place: \(0 - 1\). Since 0 is less than 1, borrow 1 from the tens place (0 becomes 9, and 0 becomes 10).
\[
10 - 1 = 9
\]
2. Tens place: After borrowing, we have \(9 - 9 = 0\).
3. Hundreds place: \(0 - 9\). Since 0 is less than 9, borrow 1 from the thousands place (8 becomes 7, and 0 becomes 10).
\[
10 - 9 = 1
\]
4. Thousands place: After borrowing, we have \(7 - 2 = 5\).

The final answer is:
\[
\boxed{5009}
\]

---

#### Problem 5:
\[
\begin{array}{c@{}c@{}c@{}c}
& \text{T} & \text{H} & \text{T} & \text{O} \\
& 5 & 4 & 3 & 2 \\
- & 2 & 3 & 4 & 5 \\
\hline
\end{array}
\]

1. Ones place: \(2 - 5\). Since 2 is less than 5, borrow 1 from the tens place (3 becomes 2, and 2 becomes 12).
\[
12 - 5 = 7
\]
2. Tens place: After borrowing, we have \(2 - 4\). Since 2 is less than 4, borrow 1 from the hundreds place (4 becomes 3, and 2 becomes 12).
\[
12 - 4 = 8
\]
3. Hundreds place: After borrowing, we have \(3 - 3 = 0\).
4. Thousands place: \(5 - 2 = 3\).

The final answer is:
\[
\boxed{3087}
\]

---

#### Problem 6:
\[
\begin{array}{c@{}c@{}c@{}c}
& \text{T} & \text{H} & \text{T} & \text{O} \\
& 9 & 6 & 2 & 8 \\
- & 5 & 7 & 6 & 9 \\
\hline
\end{array}
\]

1. Ones place: \(8 - 9\). Since 8 is less than 9, borrow 1 from the tens place (2 becomes 1, and 8 becomes 18).
\[
18 - 9 = 9
\]
2. Tens place: After borrowing, we have \(1 - 6\). Since 1 is less than 6, borrow 1 from the hundreds place (6 becomes 5, and 1 becomes 11).
\[
11 - 6 = 5
\]
3. Hundreds place: After borrowing, we have \(5 - 7\). Since 5 is less than 7, borrow 1 from the thousands place (9 becomes 8, and 5 becomes 15).
\[
15 - 7 = 8
\]
4. Thousands place: After borrowing, we have \(8 - 5 = 3\).

The final answer is:
\[
\boxed{3859}
\]

---

#### Problem 7:
\[
\begin{array}{c@{}c@{}c@{}c}
& \text{T} & \text{H} & \text{T} & \text{O} \\
& 4 & 0 & 2 & 0 \\
- & 1 & 0 & 6 & 4 \\
\hline
\end{array}
\]

1. Ones place: \(0 - 4\). Since 0 is less than 4, borrow 1 from the tens place (2 becomes 1, and 0 becomes 10).
\[
10 - 4 = 6
\]
2. Tens place: After borrowing, we have \(1 - 6\). Since 1 is less than 6, borrow 1 from the hundreds place (0 becomes 9, and 1 becomes 11).
\[
11 - 6 = 5
\]
3. Hundreds place: After borrowing, we have \(9 - 0 = 9\).
4. Thousands place: \(4 - 1 = 3\).

The final answer is:
\[
\boxed{2956}
\]

---

#### Problem 8:
\[
\begin{array}{c@{}c@{}c@{}c}
& \text{T} & \text{H} & \text{T} & \text{O} \\
& 6 & 0 & 0 & 0 \\
- & & & & 5 \\
\hline
\end{array}
\]

1. Ones place: \(0 - 5\). Since 0 is less than 5, borrow 1 from the tens place (0 becomes 9, and 0 becomes 10).
\[
10 - 5 = 5
\]
2. Tens place: After borrowing, we have \(9 - 0 = 9\).
3. Hundreds place: \(0 - 0 = 0\).
4. Thousands place: \(6 - 0 = 6\).

The final answer is:
\[
\boxed{5995}
\]

---

#### Problem 9:
\[
\begin{array}{c@{}c@{}c@{}c}
& \text{T} & \text{H} & \text{T} & \text{O} \\
& 3 & 0 & 6 & 0 \\
- & 1 & 5 & 8 & 6 \\
\hline
\end{array}
\]

1. Ones place: \(0 - 6\). Since 0 is less than 6, borrow 1 from the tens place (6 becomes 5, and 0 becomes 10).
\[
10 - 6 = 4
\]
2. Tens place: After borrowing, we have \(5 - 8\). Since 5 is less than 8, borrow 1 from the hundreds place (0 becomes 9, and 5 becomes 15).
\[
15 - 8 = 7
\]
3. Hundreds place: After borrowing, we have \(9 - 5 = 4\).
4. Thousands place: \(3 - 1 = 2\).

The final answer is:
\[
\boxed{1474}
\]

---

#### Problem 10:
\[
\begin{array}{c@{}c@{}c@{}c}
& \text{T} & \text{H} & \text{T} & \text{O} \\
& 6 & 3 & 1 & 2 \\
- & 1 & 2 & 3 & 6 \\
\hline
\end{array}
\]

1. Ones place: \(2 - 6\). Since 2 is less than 6, borrow 1 from the tens place (1 becomes 0, and 2 becomes 12).
\[
12 - 6 = 6
\]
2. Tens place: After borrowing, we have \(0 - 3\). Since 0 is less than 3, borrow 1 from the hundreds place (3 becomes 2, and 0 becomes 10).
\[
10 - 3 = 7
\]
3. Hundreds place: After borrowing, we have \(2 - 2 = 0\).
4. Thousands place: \(6 - 1 = 5\).

The final answer is:
\[
\boxed{5076}
\]

---

#### Problem 11:
\[
\begin{array}{c@{}c@{}c@{}c}
& \text{T} & \text{H} & \text{T} & \text{O} \\
& 4 & 3 & 2 & 1 \\
- & 1 & 2 & 3 & 4 \\
\hline
\end{array}
\]

1. Ones place: \(1 - 4\). Since 1 is less than 4, borrow 1 from the tens place (2 becomes 1, and 1 becomes 11).
\[
11 - 4 = 7
\]
2. Tens place: After borrowing, we have \(1 - 3\). Since 1 is less than 3, borrow 1 from the hundreds place (3 becomes 2, and 1 becomes 11).
\[
11 - 3 = 8
\]
3. Hundreds place: After borrowing, we have \(2 - 2 = 0\).
4. Thousands place: \(4 - 1 = 3\).

The final answer is:
\[
\boxed{3087}
\]

---

Final Answers:

\[
\boxed{5652, 3256, 5009, 3087, 3859, 2956, 5995, 1474, 5076, 3087}
\]