Rewriting place value addition is important in subtraction problems that use regrouping (borrowing). By regrouping the tens and ones, we can get a new place value addition for any number.
As shown above, the place value addition for 45 is 40 plus 5 (40 + 5). We can get its new place value addition with regrouping. First, we need to regroup 1 ten from the tens place and add that ten to the digit in the ones place. (You will sometimes see the terms "ones column" and "tens column" used.) In 45, the digit in the tens place is 4. To show that we removed 1 ten from 4, cross out the 4 and replace it with the digit that is 1 less than 4, which is 3.
Next, we add a small 1 in front of the number in the ones place to show we took 10 away from the tens place and added 10 to the ones place: 10 plus 5 equals 15 (10 + 5 = 15).
To get our new place value addition, let's use our new numbers. The new digit in the tens column is 3; 3 tens equal 30. The small 1 in front of the 5 shows us that we have added 1 ten to 5. The sum of 10 plus 5 equals 15 (10 + 5 = 15); 30 plus 15 equals 45 (30 + 15 = 45). Our new place value addition for 45 is 30 plus 15.
Let's try another number. The number 56 has a place value addition of 50 plus 6 (50 + 6). Let's rewrite 56 to get its new place value addition.
Subtract 1 ten from the 5 and cross it out to give us 1 less than 5, which is 4. Then write a small 1 in front of the 6. The small 1 in front of the 6 shows us that we have added 1 ten to 6.
Our new digit in the tens column is 4; 4 tens equal 40. Combining 1 ten with 6 ones, 10 plus 6 equals 16 (10 + 6 = 16). Our new place value addition for 56 is 40 plus 16.
