A step on from short multiplication is long multiplication.
The long multiplication approach is like the short multiplication method, but requires a little bit more work.
Long multiplication is the method to use when we encounter multiplication sums involving large numbers that are of size two digits or greater.
It’s quite handy for assisting us in learning how to multiply 3 digits in Math.
Long Multiplication Steps
Prior to observing some examples of how to multiply 3 digits.We can first introduce the method by looking at the sum, 13 × 22.
This multiplication sum can be set up using columns like we do in short multiplication.
Generally the larger number in the sum is placed above.
\begin{array}{r} &2\space2\\ \times &1\space3\\ \hline \end{array}
The first step here is to multiply the whole top number by only the 3 in the lower units column, as if this was a short multiplication sum.
\begin{array}{r} &2\space2\\ \times &\space\space\space3\\ \hline &6\space6 \end{array}
The next step is to multiply the top number by the lower 1 in the column for tens.
Placing the result of this multiplication below the first multiplication result.
However to account for this multiplication being for the tens column, a 0 is placed in the answer section for the units column first.
\begin{array}{r} &2\space2\\ \times &1\space3\\ \hline &6\space6\\ &\space\space\space0 \end{array} => \begin{array}{r} &\space2\space2\\ \times &\space1\space3\\ \hline &\space6\space6\\ &2\space2\space0\space\space \end{array}
To conclude, we add the two separate multiplication results obtained together, which gives the full answer to the entire multiplication sum.
\begin{array}{r} &\space2\space2\\ \times &\space1\space3\\ \hline &\space6\space6\\ &2\space2\space0\space\space\\ \hline &2\space8\space6\space\space \end{array} 13 × 22 = 286
Long Multiplication
and How to Multiply 3 Digits, Examples
(2.1)
12 × 15
Solution
This example isn’t a particularly large multiplication sum, but the approach for long multiplication can be used in order to solve.
The columns are set up as usual.
\begin{array}{r} &1\space5\\ \times &1\space2\\ \hline \end{array} First we multiply the 1 and the 5 on the top row, by the 2 in the lower row.
1)
\begin{array}{r} &{\tiny{1}}\space\space\space\\ &1\space5\\ \times &\space\space\space2\\ \hline &6\space0\\ \\ \\ \end{array} => \begin{array}{r} &\space1\space5\\ \times &\space1\space2\\ \hline &\space6\space0\\ &1\space2\space0\space\space \end{array}
2)
Then for the last step in long multiplication, the two separate multiplication results get added together.
\begin{array}{r} &\space1\space5\\ \times &\space1\space2\\ \hline &\space6\space0\\ &1\space2\space0\space\space\\ \hline &1\space8\space0\space\space\\ \end{array} So, 12 × 15 = 180.
(2.2)
346 × 24
Solution
1)
\begin{array}{r} &{\tiny{1}}\space\space{\tiny{2}}\space\space\\ &\space3\space4\space6\\ \times &\space\space\space\space\space\space\space4\\ \hline &1\space3\space8\space4\space\space \end{array} => \begin{array}{r} &\space\\ &\space\space\space{\tiny{1}}\space\space\\ &\space3\space4\space6\\ \times &\space2\\ \hline &1\space3\space8\space4\space\space\\ &\space6\space9\space2\space0\space\space\space \end{array}
2)
Now adding the two results together:
\begin{array}{r} &\space3\space4\space6\\ \times &\space\space\space\space2\space4\\ \hline &1\space3\space8\space4\space\space\\ &6\space9\space2\space0\space\space\\ \hline &8\space3\space0\space4\space\space\\ &{\tiny{1}}\space\space{\tiny{1}}\space\space\space\space\space\space\space\space\\ \end{array} Thus, 346 × 24 = 8304
(2.3)
435 × 324
Solution
1)
\begin{array}{r} &{\tiny{1}}\space\space{\tiny{2}}\\ &\space\space\space4\space3\space5\\ \times &\space\space\space\space\space\space\space\space\space4\\ \hline &1\space7\space4\space0 \end{array}
2)
\begin{array}{r} &{\tiny{1}}\\ &4\space3\space5\\ \times &2\\ \hline &\space\space\space1\space7\space4\space0\space\space\space\space\space\space\\ &8\space7\space0\space0\space\space\space \end{array}
Now with the 5, we are multiplying by “HUNDREDS”, so two 0‘s are placed in the answer section initially.
\begin{array}{r} &{\tiny{1}}\space\space{\tiny{1}}\space\space\space\\ &4\space3\space5\\ \times &3\\ \hline &\space\space\space1\space7\space4\space0\space\space\space\space\space\space\\ &8\space7\space0\space0\space\space\space\\ &1\space3\space0\space5\space0\space0\space\space\space\space\space\space\space\space \end{array}
4)
Now we add the three results to obtain the answer.
\begin{array}{r} &\space\space\space1\space7\space4\space0\space\space\space\space\space\\ &8\space7\space0\space0\space\space\\ + &1\space3\space0\space5\space0\space0\space\space\space\space\space\space\space\space\\ \hline &1\space4\space0\space9\space4\space0\space\space\space\space\space\space\space\space\\ &{\tiny{1}}\space\space{\tiny{1}}\space\space\space\space\space\space\space\space\space\space\space\space\space\space\\ \end{array}
435 × 324 = 140940