# Multiplication (2 by 1 digit)

Times tables are great for multiplying one-digit (and select two-digit) numbers. But, once we get into larger numbers, it's important to understand how multiplication works when numbers have multiple digits.  We typically use a number of algorithms for doing multiplication -- namely we multiply all digits of multipliers by all digits of multiplicands.  The processes become automatic, but it's very helpful for students to understand why we have to multiply each digit by each each digit.

First, think about a typical multiplication problem in which we multiply a one-digit multiplier times a two digit multiplicand:

$$\begin{array}{r} &21\\\times\!\!\!\!\!\!&3\\ \hline \end{array}$$

We put the two digit number on the top (remember, in multiplication, it doesn't matter what order you multiply in, so it's usually more efficient to put the number with more digits on top) and the one-digit number on the bottom.

Then, we multiply the 3 times 1:

$$\begin{array}{r} &21\\\times\!\!\!\!\!\!&3\\ \hline &\quad 3 \end{array}$$

Then we multiply 3 times 2:

$$\begin{array}{r} &21\\\times\!\!\!\!\!\!&3\\ \hline & 63 \end{array}$$

We get a final answer of 63.

We go through this process because each digit in a two-digit number represents a separate value.  In the case of 21, the two represents 2 tens (or 20) and the 1 represents 1 one (or 1).  We multiply the 3 times both 20 and 1, in a vertical format as a short cut.

Think about it the long way:

\eqalign{21 \times 3 & = (20 \times 3)+ (1 \times 3)\\&=60 + 3\\&=63}

When we multiply in a vertical format, we bring the 6 (of the 60) down in the tens column, eliminating the need to write out 60.  We know that the 6 represents 60 because it is in the tens column.

Essentially, we use vertical formats in multiplication to keep our place values in order.  These processes become even more important (and a little more complicated) as we move into multiplying two multi-digit numbers!