Vertical Subtraction: Subtraction of 3-digit and 2-digit numbers with regrouping

To solve this problem, we'll follow these steps:

• Step 1: Align the numbers by their place values. The number 132 is written above 35, ensuring the units, tens, and hundreds columns are properly placed.
• Step 2: Start from the rightmost column, which is the units column:
• Subtract the units digit: $2 - 5$ requires borrowing since 2 is smaller than 5. Borrowing from the tens column converts 2 into 12.
• After borrowing, subtract: $12 - 5 = 7$.
• Step 3: Move to the tens column:
• After borrowing one "ten" to the units place, the tens column in the minuend is now $2 - 1 = 1$ (since the tens digit went from 3 to 2 after borrowing).
• Subtract: $2 - 3 = -1$, recognizing the necessity to borrow from the hundreds.
• Borrow 1 hundred, adding it to the tens to make: $12 - 3 = 9$.
• Step 4: Finally, subtract the hundreds column:
• With borrowing adjustment: $0 - 0$ remains unchanged because we lent one hundred to the tens place.
• This happens because we effectively do $1 - 1 = 0$ after resolving borrowing in tens and units.

Therefore, after performing all borrowing and subtractions,

The solution to $132 - 35$ is $97$.

To solve this subtraction problem, we will follow these steps:

• Step 1: Line up the numbers vertically to align the digits by place value.

• Step 2: Start from the units column, note $4 - 6$ requires borrowing from the tens place.

  • Since 4 is smaller than 6, we borrow 1 from the next higher place value. The tens digit (7) becomes 6, and the units digit becomes $14$.

• Step 3: Subtract in the units column: $14 - 6 = 8$.

• Step 4: Move to the tens column, subtract $6 - 8$. Since we borrowed from the hundreds place, it simplifies to $16 - 8 = 8$.

• Step 5: The hundreds digit in the minuend is $2$, reduced by 1 due to previous borrowing, makes the effective hundreds $1$.

• Step 6: Finally, in the hundreds column, subtract $1 - 0 = 1$.

Let's summarize the steps calculations as follows:

$\begin{aligned} 274& \\ - \ 86& \\ \overline{\phantom{2}188} \\ \end{aligned}$

Upon completion, the final result is $188$.

To solve this subtraction problem, we'll perform vertical subtraction with regrouping as necessary:

• Step 1: Align the numbers vertically, ensuring that digits are in the correct place value columns.
• Step 2: Begin with the units column: Subtract 9 from 8. Since 8 is smaller than 9, we need to regroup by borrowing from the tens column.
• We take 1 ten from 5 in the tens column. The tens digit becomes 4, and we add 10 to 8 in the units column, making it 18.
• Now, subtract 9 from 18, which equals 9.
• Step 3: Move to the tens column: Subtract 7 from 4 (after borrowing). Since 4 is smaller than 7, we need to regroup again.
• We borrow 1 hundred from the hundreds column; 3 becomes 2. The tens digit becomes 14.
• Now, subtract 7 from 14, which equals 7.
• Step 4: Finally, handle the hundreds column: Subtract 0 from 2 (after borrowing), resulting in 2.

Therefore, the solution to the subtraction is $279$.

To solve this problem, we'll perform vertical subtraction with regrouping of the numbers 481 and 83.

Follow these steps:

• Step 1: Write the numbers vertically, aligning the digits by place value:

$\begin{array}{c} 481 \\ - ~~83~~~ \\ \underline{\phantom{77633}} \end{array}$

• Step 2: Begin with the units column (rightmost column). The digit in the minuend (1) is less than the digit in the subtrahend (3). Therefore, we need to regroup.

  • Regrouping involves borrowing 1 from the tens column of the minuend, changing 8 in 481 to 7 and increasing the units digit by 10, changing 1 to 11.

• Step 3: Subtract the units column: $11 - 3 = 8$.

• Step 4: Move to the tens column. Since we borrowed 1, the tens digit in 481 is now 7. Subtract 8 from 7, borrowing again from the hundreds digit:

  • Regroup by borrowing 1 from the hundreds column, changing 4 to 3 and increasing the tens digit by 10, changing 7 to 17.

• Step 5: Subtract the tens column: $17 - 8 = 9$.

• Step 6: The hundreds column remains to be calculated: $3 - 0 = 3$ (since there’s nothing to subtract except the result of our previous borrow).

This process results in:

$\begin{array}{c} 481 \\ - ~~83~~~ \\ \underline{\phantom{776444}} \\ 398 \end{array}$

Therefore, the solution to this problem is $398$, which corresponds to choice four.

Let's solve the problem using vertical subtraction. We have:

$\begin{array}{c} & 5 & 7 & 2 \\ - & & 9 & 4 \\ \hline \end{array}$

Step 1: Subtract the rightmost digits (units):

We have $2 - 4$, which is not possible without borrowing. So, we need to regroup:

• Borrow 1 from the tens place (7 becomes 6), making the 2 a 12.

Now calculate:

$12 - 4 = 8$

Step 2: Subtract the tens digits:

We are left with $6 - 9$, which again requires us to regroup:

• Borrow 1 from the hundreds place (5 becomes 4), making the 6 a 16.

Now calculate:

$16 - 9 = 7$

Step 3: Finally, subtract the hundreds digits:

We now have $4 - 0$, since there is no digit above in the subtrahend:

$4 - 0 = 4$

Putting it all together, the subtraction yields:

$\begin{array}{c} & \phantom{1} 5 & 7 & 2 \\ - & \phantom{1} & 9 & 4 \\ \hline & \phantom{1} 4 & 7 & 8 \\ \end{array}$

Therefore, the answer to the problem is $\textbf{478}$.

To solve $667 - 89$, we perform vertical subtraction by aligning the numbers by place value:

• Step 1: Start with the units column. Subtract $9$ from $7$. Since $7 < 9$, we need to borrow from the tens column. Change $6$ in the tens column to $5$ and add $10$ to the units column, making it $17$. Now, subtract: $17 - 9 = 8$.
• Step 2: Move to the tens column, where we borrowed from $6$, changing it to $5$. Now subtract $8$ from $5$. Since $5 < 8$, we need to borrow from the hundreds column. Change $6$ in the hundreds column to $5$ and add $10$ to the tens column, making it $15$. Now, subtract: $15 - 8 = 7$.
• Step 3: Finally, in the hundreds column: Subtract $0$ from $5$, leaving $5$.

Therefore, the solution to $667 - 89$ is $\mathbf{578}$.

Given the multiple-choice options, the correct choice is option 2: $578$.

To solve this problem, we'll follow these steps:

• Step 1: Write the numbers in column form.

• Step 2: Subtract starting from the rightmost (units) column.

• Step 3: Apply borrowing/regrouping if necessary.

Now, let's work through each step:

Step 1: Write the numbers vertically.
$\begin{array}{c} 713\\ -\phantom{0}58~~~ \\ \underline{\phantom{776}} \\ \end{array}$

Step 2: Starting with the units column:
Since 3 < 8 , we need to borrow 1 from the tens column (1 becomes 0, and 3 becomes 13).

Step 3: Perform the subtraction with borrowing:
- Units: $13 - 8 = 5$
- Tens: Re-evaluated 0 (after borrowing) minus 5, need another borrow so 0 turns to 10. - After borrowing from hundreds column, $10 - 5 = 5$.
- Hundreds: After borrowing, left with 6. Thus, $6 - 0 = 6$

Therefore, put all together, the difference is:
$\begin{array}{c} 713\\ -\phantom{0}58~~~ \\ \underline{655} \\ \end{array}$

Therefore, the solution to the problem is $655$.

To solve the problem of subtracting 68 from 827, we will perform vertical subtraction with regrouping (borrowing) where necessary.

First, let's write the numbers vertically aligned by place value:

$\begin{aligned} &amp;827 \\ -&amp;\phantom{0}68 \\ &amp;\underline{\phantom{776}} \\ \end{aligned}$

Next, we'll perform subtraction starting from the rightmost digit (units place):

• The units column: 7 (from 827) minus 8. Since 7 is less than 8, we need to borrow 1 from the tens column. After borrowing, the 7 becomes 17.

• 17 minus 8 equals 9. Write 9 directly under this column.

Now, move to the tens column:

• Since we borrowed 1, the 2 (from 827) is now 1. Subtract 6 (from 68) from 11 (the borrowed number).
  11 minus 6 equals 5. Write 5 directly under this column.

Finally, the hundreds column:

• Since no numbers exist below the hundreds place in 68 and we only borrowed from the tens place, we subtract nothing from 8. So, 8 stays the same.

• Write 8 directly under this column.

Putting it all together, the complete difference is:

$\begin{aligned} &amp;827 \\ -&amp;\phantom{0}68 \\ &amp;\overline{759} \\ \end{aligned}$

Therefore, the solution to the problem is $759$.

To solve this problem using vertical subtraction, follow these steps:

• Step 1: Align the Numbers
  Write the numbers with their digits aligned by place value:
  $\begin{array}{c} 945 \\ - 67 \\ \underline{\phantom{776}} \end{array}$
• Step 2: Subtract the Units Place
  In the units place, we have $5 - 7$. Since 5 is smaller than 7, we need to borrow from the tens place.
  Borrowing from the tens, the 4 becomes 3, and the 5 becomes 15.
  So, $15 - 7 = 8$.
• Step 3: Subtract the Tens Place
  Now, subtract the tens place: $3 - 6$. Since 3 is smaller than 6, borrow from the hundreds place.
  The 9 becomes 8, and the 3 becomes 13 after borrowing.
  So, $13 - 6 = 7$.
• Step 4: Subtract the Hundreds Place
  Finally, subtract the hundreds place: $8 - 0 = 8$, as there is no number in the hundreds place of 67.

Therefore, after performing the subtraction, we find that the result is $878$.

To solve the problem $968 - 89$ using vertical subtraction, follow these steps:

• Align the numbers vertically by place value:

$\begin{array}{c} ~ &amp; 9 &amp; 6 &amp; 8 \\ - &amp; ~ &amp; 8 &amp; 9 \\ \hline \end{array}$

• Subtract the rightmost column (units):
  Since $8 - 9$ is not possible directly, we borrow 1 from the tens column.

• After borrowing, the units column becomes $18 - 9 = 9$.

$\begin{array}{c} ~ &amp; 9 &amp; \cancel{\textcolor{red}{6}}~{5} \to &amp; 18 \\ - &amp; ~ &amp; 8~~ &amp; 9 \\ \hline ~ &amp; ~ &amp; ~ &amp; 9 \\ \end{array}$

• Subtract the tens column:
  After borrowing, the tens column becomes $5 - 8$ not possible, so borrow 1 from the hundreds column.

• After borrowing, the tens column becomes $15 - 8 = 7$.

$\begin{array}{c} ~ &amp; \cancel{\textcolor{red}{9}}~ 8\to &amp;{15} &amp; 8 \\ - &amp; ~ &amp; 8 &amp; 9 \\ \hline ~ &amp; ~ &amp; 7 \\ \end{array}$

• Subtract the hundreds column:
  Now simply $8 - 0 = 8$ since there is nothing to subtract.

$\begin{array}{c} ~ &amp; 8 &amp; ~ &amp; ~ \\ - &amp; ~ &amp; ~ &amp; ~ \\ \hline &amp;8 &amp; 7 &amp; 9 \\ \end{array}$

Therefore, the solution to the problem is $879$.