Question

3 finish the algorithm for 612 - 235

Explanation:

Step1: Subtract units place

We can't subtract 5 from 2, so we borrow 1 ten (which is 10 units) from the tens place. Now the units place becomes \(12 - 5 = 7\).

Step2: Subtract tens place

After borrowing, the tens place was 1, but we gave 1 to the units place, so now it's 0. We can't subtract 3 from 0, so we borrow 1 hundred (which is 10 tens) from the hundreds place. Now the tens place becomes \(10 - 3 = 7\) (wait, no, wait: original tens place was 1, after borrowing 1 from hundreds, it becomes 11 tens? Wait, no, let's correct. The original number is 612. Hundreds: 6, Tens: 1, Units: 2. When subtracting 235, units: 2 - 5, can't do, so borrow 1 from tens: tens becomes 0, units becomes 12. Then tens: 0 - 3, can't do, so borrow 1 from hundreds: hundreds becomes 5, tens becomes 10. Wait, but in the diagram, it's shown as 5 (hundreds), 11 (tens), 12 (units). Oh, maybe the borrowing is from hundreds to tens first? Wait, maybe the algorithm is:

Hundreds: 6, we borrow 1 to make tens 11 (since 1 hundred = 10 tens, so 10 + 1 = 11 tens), then tens borrow 1 to units, making units 12. So:

Units: \(12 - 5 = 7\)

Tens: \(11 - 3 = 8\)? Wait, no, the diagram has 11 and 12. Wait, maybe the calculation is:

612 - 235:

• Units: 2 < 5, so borrow 1 from tens (but tens is 1, so we need to borrow from hundreds. So hundreds: 6 becomes 5, tens: 1 becomes 11 (because 1 hundred = 10 tens, so 10 + 1 = 11 tens), then units: 2 becomes 12 (because 1 ten = 10 units, so 10 + 2 = 12 units).

Now units: \(12 - 5 = 7\)

Tens: \(11 - 3 = 8\)

Hundreds: \(5 - 2 = 3\)

Wait, but the answer in the box is 377, which would be if tens was 7. Wait, maybe I made a mistake. Wait 612 - 235:

Let's do standard subtraction:

6 1 2

• 2 3 5

--------

Units: 2 - 5: can't, borrow from tens: tens 1 becomes 0, units 12. 12 - 5 = 7.

Tens: 0 - 3: can't, borrow from hundreds: hundreds 6 becomes 5, tens 0 becomes 10. 10 - 3 = 7.

Hundreds: 5 - 2 = 3.

Ah! There we go. So:

Units: \(12 - 5 = 7\)

Tens: \(10 - 3 = 7\) (wait, but after borrowing from hundreds, tens is 10, because we borrowed 1 hundred (10 tens) to tens place (which was 0 after borrowing to units). So:

Hundreds: 6 - 1 = 5

Tens: 0 + 10 = 10, then subtract 3: 10 - 3 = 7

Units: 2 + 10 = 12, subtract 5: 12 - 5 = 7

So 5 (hundreds) 100 + 7 (tens) 10 + 7 (units) = 500 + 70 + 7 = 577? Wait, no, 5100=500, 710=70, 7*1=7: 500+70+7=577? But 612 - 235 is 377. Oh! Wait, I messed up the hundreds place. 6 - 2 is 4, but with borrowing:

Wait, correct standard subtraction:

6 1 2

• 2 3 5

--------

Start from units: 2 - 5. Can't do. Borrow 1 from tens: tens becomes 0, units becomes 12. 12 - 5 = 7.

Tens: 0 - 3. Can't do. Borrow 1 from hundreds: hundreds becomes 5, tens becomes 10. 10 - 3 = 7.

Hundreds: 5 - 2 = 3.

So total: 3 (hundreds) 100 + 7 (tens) 10 + 7 (units) = 300 + 70 + 7 = 377.

Ah, there we go. So the steps are:

1. Units: 12 - 5 = 7

1. Tens: 10 - 3 = 7 (after borrowing from hundreds, tens is 10, because we borrowed 1 hundred (10 tens) to the tens place which was 0 after borrowing to units)

1. Hundreds: 5 - 2 = 3

So putting it together: 3 (hundreds) + 7 (tens) + 7 (units) = 377.

Answer:

377