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To solve this problem, we'll follow these steps:
Step 1: Align the numbers by their place values, so the digits in the same position are directly on top of each other.
Step 2: Subtract the digits in the ones place and perform borrowing if necessary.
Step 3: Subtract the digits in the tens place after performing any borrowing from the previous step.
Now, let's work through each step:
Step 1: Align the numbers vertically:
Step 2: Starting with the ones place, we see that 3 (in the minuend) is less than 6 (in the subtrahend). So we need to borrow.
Borrow 1 from the tens place of 53, giving the tens place a value of 4. The borrowed number converts the ones place's 3 into 13.
Now subtract: .
The new arrangement looks like this:
Step 3: Move to the tens place. After borrowing, the new value is 4.
Subtract: .
Thus, the final result is:
Therefore, the result of the subtraction is .
17
\( \begin{aligned} &105 \\ -& \\ &~~~~3 \\ &\underline{\phantom{776}} & \\ \end{aligned} \)
Because position matters in subtraction! When you flip numbers, you're changing the problem completely. Always subtract the bottom number from the top number in each column.
Borrowing means trading one ten for ten ones! When you borrow 1 from the tens place (5), it becomes 4 tens, and the ones place (3) becomes 13 ones.
You need to borrow whenever the top digit is smaller than the bottom digit in any column. Look at each column from right to left!
You'll need to borrow from the hundreds place first! Turn 0 into 10, then borrow 1 from there. It's like borrowing through multiple places.
Yes! You can also count up from 36 to 53: 36 + ? = 53. Count: 36 + 4 = 40, then 40 + 13 = 53, so 4 + 13 = 17 ✓
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