3-Digit Addition: Regrouping Hundreds - Second Grade Mathematics

Just as 10 ones bundle together to create 1 ten rod, collecting 10 tens rods creates a brand-new hundred flat. In three-digit addition, when the sum of the tens column reaches 10 or more, you must regroup 10 tens into 1 hundred and send that hundred into the hundreds column attic. In some exciting problems, you will even encounter double regrouping, where you regroup from ones to tens, and then regroup again from tens to hundreds! In this chapter, you will master regrouping tens into hundreds, follow double-regrouping challenges, and understand the place value logic that makes it work.

Regrouping Tens into Hundreds

Consider what happens when the tens column adds up to 10, 12, or 14 tens.

The Trade:
10 Ten Rods (10 x 10)  <====== EQUAL TO ======>  1 Hundred Flat (100)
| | | | | | | | | |                              [   ]

The Step-by-Step Algorithm for 372 + 284

Let us follow the columns for 372 + 284:

          Hundreds   Tens   Ones
             1  <--- Attic Hundred from Tens Column!
             3         7      2
          +  2         8      4
          ---------------------
             6         5      6

Step 1: Add ones: 2 ones + 4 ones = 6 ones. (No regrouping needed in ones, since 6 is less than 10). Write 6 in the ones place. Step 2: Add tens: 7 tens + 8 tens = 15 tens! Ask: Is 15 ten or greater? Yes! 15 tens = 1 hundred and 5 tens (since 15 x 10 = 150). Write 5 in the tens column below the line. Write a small 1 in the attic above the hundreds column! Step 3: Add all hundreds, including the attic hundred: 1 (attic) + 3 hundreds + 2 hundreds = 6 hundreds. Write 6 in the hundreds column below the line. The sum is 656.

Double Regrouping: From Ones to Tens, and Tens to Hundreds

Sometimes an addition problem requires regrouping twice: once in the ones column, and once in the tens column.

Solving 468 + 275 with Double Regrouping:

        1  1  <--- Attic Regroupings in Both Tens and Hundreds!
        4  6  8
     +  2  7  5
     ----------
        7  4  3

Step 1: Add Ones: 8 + 5 = 13 ones.
        13 ones = 1 ten and 3 ones.
        Write 3 in ones place. Carry 1 ten to tens attic.

Step 2: Add Tens: 1 (attic) + 6 + 7 = 14 tens!
        14 tens = 1 hundred and 4 tens.
        Write 4 in tens place. Carry 1 hundred to hundreds attic.

Step 3: Add Hundreds: 1 (attic) + 4 + 2 = 7 hundreds.
        Write 7 in hundreds place.
        Final Sum: 743!

Celebrating the Chain Reaction

Double regrouping is like a mathematical domino chain: the ones column fills up and sends an extra ten to the tens column; that extra ten causes the tens column to fill up and send an extra hundred to the hundreds column! Staying calm and handling one column at a time makes double regrouping completely straightforward.

Base-Ten Model of Double Regrouping

Let us trace the blocks during 468 + 275:

Original Ones: 8 dots + 5 dots = 13 dots.
Trade 1:       Bundle 10 dots into 1 new ten line |.
               3 dots remain in ones.

Original Tens: 6 lines + 7 lines + 1 traded line = 14 lines.
Trade 2:       Bundle 10 lines into 1 new hundred square [ ].
               4 lines remain in tens.

Original Hundreds: 4 squares + 2 squares + 1 traded square = 7 squares.
Total Result:      7 squares (700) + 4 lines (40) + 3 dots (3) = 743!

Practice Exercises

Exercise 1: Solve 382 + 254 vertically. State which column required regrouping and explain how the regrouped amount was recorded.

Exercise 2: Solve 476 + 385 using double regrouping. Show the attic digits above both the tens and hundreds columns.

Exercise 3: Explain why 14 tens is equal to 1 hundred and 4 tens, using base-ten rods and numerical values.

Exercise 4: A sports arena sold 365 tickets on Thursday and 482 tickets on Friday. How many total tickets were sold across the two days?

Solutions and Explanations

Solution to Exercise 1: Align vertically: 382 over 254. Step 1: Add ones: 2 + 4 = 6. Step 2: Add tens: 8 + 5 = 13 tens. Since 13 tens = 1 hundred and 3 tens, write 3 in the tens place and write 1 in the attic above the hundreds place. Step 3: Add hundreds: 1 (attic) + 3 + 2 = 6 hundreds. The final sum is 636. The tens column required regrouping.

Solution to Exercise 2: Align vertically: 476 over 385. Step 1: Add ones: 6 + 5 = 11 (write 1 in ones place, write 1 in tens attic). Step 2: Add tens: 1 (attic) + 7 + 8 = 16 tens (write 6 in tens place, write 1 in hundreds attic). Step 3: Add hundreds: 1 (attic) + 4 + 3 = 8 hundreds. The final sum is 861.

Solution to Exercise 3: A ten rod has a value of 10. Fourteen ten rods have a total value of 14 x 10 = 140. Ten of those ten rods can be laid side by side and fused into 1 hundred flat (value = 100). That leaves 4 ten rods remaining (value = 40). Therefore, 14 tens equals 1 hundred and 4 tens (140).

Solution to Exercise 4: To find the total tickets sold, add 365 + 482. Add ones: 5 + 2 = 7. Add tens: 6 + 8 = 14 tens (write 4 in tens place, regroup 1 hundred to hundreds attic). Add hundreds: 1 (attic) + 3 + 4 = 8 hundreds. In total, 847 tickets were sold across the two days.