Antonio said, "To solve 506-288, I can think of 5 hundreds as 50 tens." How might this help him subtract?
He can subtract 28 hundreds from 50 hundreds, giving 22 tens (220), but he will still have to add 6 and subtract 8, giving 218.
Antonio's approach is to decompose the number 506 into its place value parts. By recognizing that 5 hundreds is equal to 50 tens, he can break down the original subtraction problem of 506-288 in a way that makes it easier to subtract.
Here's how Antonio's approach might help him subtract:
1. Start by subtracting the ones place: 6 - 8. Since 6 is less than 8, you need to borrow from the tens place.
2. Convert one of the tens into 10 ones. Antonio's strategy involves recognizing that 5 hundreds (500) can be broken down into 50 tens. So, he can exchange 1 ten for 10 ones.
3. Add the 10 ones to the original 6 ones, giving you a total of 16 ones.
4. Now you have 16 ones and still need to subtract 8 ones. Since 16 is more than 8, you can do this subtraction straightforwardly. 16 - 8 equals 8.
5. Moving on to the tens place: subtract 8 tens (80) from the remaining 50 tens. This gives you 50 - 80, which results in -30.
6. Lastly, subtract 2 hundreds from the original 5 hundreds, which gives you 3 hundreds (300).
Putting it all together, the final result is 318.
So, Antonio's reasoning helps him subtract by breaking down the numbers into their place value parts and using the concept of regrouping to simplify the subtraction process.