In the problem \(402 - 157\), how do you start regrouping?
Go to the hundreds place. Change the \(4\) to a \(3\). The \(0\) in the tens place becomes a \(10\).
In \(402 - 157\), after changing the tens place to \(10\), what is the next step?
Now you can regroup for the ones place. Change the \(10\) in the tens place to a \(9\). The \(2\) in the ones place becomes a \(12\).
What does the top number \(402\) look like after all the regrouping is done?
The \(4\) is crossed out and becomes a \(3\). The \(0\) is crossed out and becomes a \(9\). The \(2\) is crossed out and becomes a \(12\).
$$ 402 - 157 = ? $$
After regrouping: Ones: \(12 - 7 = 5\). Tens: \(9 - 5 = 4\). Hundreds: \(3 - 1 = 2\). The answer is \(245\).
Describe the pattern for regrouping across multiple zeros, like in \(1000 - 567\).
You borrow from the first non-zero digit (the \(1\)). All the zeros you cross over become \(9\)s, and the last digit you need to add to becomes \(10\).
In \(1000 - 567\), what do the top digits become after regrouping?
The \(1\) becomes a \(0\). The first two \(0\)s become \(9\)s. The last \(0\) in the ones place becomes a \(10\).
What makes subtracting across zeros different, like in \(402 - 157\)?
You cannot borrow from the tens place because its value is zero. You must go to the next place value over (the hundreds place) to start regrouping.
Regroup from the \(6\). The hundreds place becomes \(5\), the tens place becomes \(9\), and the ones place becomes \(10\). The answer is \(366\).
$$ 503 - 129 = ? $$
Regroup from the \(5\). Hundreds place becomes \(4\), tens place becomes \(9\), ones place becomes \(13\). The answer is \(374\).
$$ 705 - 88 = ? $$
Line up as \(705 - 088\). Regroup from the \(7\). Hundreds becomes \(6\), tens becomes \(9\), ones becomes \(15\). The answer is \(617\).
True or False: A shortcut for subtracting from a number like \(500\) is to subtract \(1\) from both numbers first.
True. \(500 - 234\) is the same as \(499 - 233\), which gives \(266\). This is a great mental math trick that avoids regrouping, but it's important to understand the standard way first.
