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Multi-Digit Addition Without Regrouping

Core Concepts

Practice adding 2-digit and 3-digit numbers where you don't need to carry over any values. This helps master the column addition method before moving on to more complex problems.

CATEGORY
Elementary School (Grades 1-5) / Elementary Mathematics / Basic Arithmetic: Addition and Subtraction / Multi-Digit Addition
Next Steps in Multi-Digit Addition
CONTINUE IN CORE CONCEPTS
Regrouping from Tens to Hundreds
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LEVEL UP: ADVANCED
Adding Three or More Multi-Digit Numbers
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Multi-Digit Addition Without Regrouping

1 of 12
$$550 + 349 = ?$$
$$\begin{array}{rr} & 550 \\ + & 349 \\ \hline & 899 \end{array}$$ 1. Add the ones: \(0 + 9 = 9\). 2. Add the tens: \(5 + 4 = 9\). 3. Add the hundreds: \(5 + 3 = 8\).
$$123 + 245 = ?$$
$$\begin{array}{rr} & 123 \\ + & 245 \\ \hline & 368 \end{array}$$ 1. Add the ones: \(3 + 5 = 8\). 2. Add the tens: \(2 + 4 = 6\). 3. Add the hundreds: \(1 + 2 = 3\).
$$402 + 181 = ?$$
$$\begin{array}{rr} & 402 \\ + & 181 \\ \hline & 583 \end{array}$$ 1. Add the ones: \(2 + 1 = 3\). 2. Add the tens: \(0 + 8 = 8\). 3. Add the hundreds: \(4 + 1 = 5\).
$$621 + 27 = ?$$
$$\begin{array}{rr} & 621 \\ + & 27 \\ \hline & 648 \end{array}$$ Be sure to line up the place values! 1. Add the ones: \(1 + 7 = 8\). 2. Add the tens: \(2 + 2 = 4\). 3. Add the hundreds: \(6 + 0 = 6\).
When adding without regrouping, what is the largest possible sum you can have in any single column?
The largest sum is \(9\). If the sum were \(10\) or more, you would have to regroup.
$$333 + 333 = ?$$
$$\begin{array}{rr} & 333 \\ + & 333 \\ \hline & 666 \end{array}$$ Ones: \(3+3=6\). Tens: \(3+3=6\). Hundreds: \(3+3=6\).
$$714 + 162 = ?$$
$$\begin{array}{rr} & 714 \\ + & 162 \\ \hline & 876 \end{array}$$ Ones: \(4+2=6\). Tens: \(1+6=7\). Hundreds: \(7+1=8\).
$$805 + 190 = ?$$
$$\begin{array}{rr} & 805 \\ + & 190 \\ \hline & 995 \end{array}$$ Ones: \(5+0=5\). Tens: \(0+9=9\). Hundreds: \(8+1=9\).
What is the sum of the digits in the tens place for the problem \(251 + 536\)?
The digits in the tens place are \(5\) and \(3\). Their sum is \(5+3=8\).
$$264 + 315 = ?$$
$$\begin{array}{rr} & 264 \\ + & 315 \\ \hline & 579 \end{array}$$
$$111 + 888 = ?$$
$$\begin{array}{rr} & 111 \\ + & 888 \\ \hline & 999 \end{array}$$
$$42 + 35 = ?$$
$$\begin{array}{rr} & 42 \\ + & 35 \\ \hline & 77 \end{array}$$
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