Solve Three-Digit Subtraction: 372 minus 185 in Vertical Format

Question

amp;372amp;amp;185amp;776amp; \begin{aligned} &372 \\ -& \\ &185 \\ &\underline{\phantom{776}} & \\ \end{aligned}

Video Solution

Solution Steps

00:00 Solve
00:03 Each time we consider borrowing 2 digits, and then we'll place
00:07 2 is less than 5
00:13 Therefore we'll subtract 1 from the tens and add this amount to the ones
00:17 So now instead of 2 we'll have 12
00:21 We subtract the ones from the ones plus ten
00:24 We'll place in the ones
00:28 6 is less than 8
00:33 Therefore we'll subtract 1 from the hundreds and add this amount to the tens
00:39 So now instead of 6 we'll have 16
00:43 We subtract the tens from the tens plus ten
00:47 We'll place in the tens
00:52 Subtract hundreds from hundreds, and place in hundreds
00:55 And this is the solution to the problem

Step-by-Step Solution

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

  • Step 1: Set up the numbers for vertical subtraction

  • Step 2: Perform subtraction starting from the rightmost column

  • Step 3: Apply regrouping (borrowing) if necessary

Let's work through each step:

Step 1: Set up the numbers as follows:
amp;3amp;7amp;2amp;1amp;8amp;5amp;amp; \begin{array}{c} & 3 & 7 & 2 \\ - & 1 & 8 & 5 \\ \hline & & \\ \end{array}

Step 2: Start with the units column: 252 - 5. Since 2 is less than 5, we need to regroup:

Step 3: Borrow 1 from the tens column, turning 7 into 6 and 2 into 12. Now subtract the units: 125=712 - 5 = 7.

Next, move to the tens column. After borrowing, we have 66 in the tens column, subtract 88 (from the subtrahend):

Since 686 - 8 is not possible without borrowing, we need to regroup.

Borrow 1 from the hundreds column, turning 33 into 22 and 66 into 1616. Now subtract the tens: 168=816 - 8 = 8.

Lastly, move to the hundreds column. After borrowing, it's 21=12 - 1 = 1. Place this value in the hundreds place.

The final subtraction setup and answer is:
amp;2amp;16amp;12amp;1amp;8amp;5amp;1amp;8amp;7 \begin{array}{c} &2 & 16 & 12 \\ - & 1 & 8 & 5 \\ \hline &1 & 8 & 7 \\ \end{array}

Therefore, the solution to the problem is 187187.

Answer

187