# Multiplication (2 by 2 digits)

The algorithm for multiplying a two-digit number by a one digit number is pretty simple: you multiply the one-digit number by both digits of the second number.  And, if you keep things lined up, your tens digit ends up in the tens column and your ones digit lands in the ones column, and your answer turns out perfect.

Once you start multiplying multi-digit numbers by multi-digit numbers, the process gets a little more complicated.  But, it follows the same logic: each digit of the multiplier must be multiplied by each digit of the multiplicand.  And, each answer value you get must land in the proper place value.  Setting the problem up vertically (lining up place values) helps, but with multi-digit multiplication you will also need some placeholders.  Little kids often call this process "multiplication with stacking."

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

$$\begin{array}{r} &21\\\times\!\!\!\!\!\!&32\\ \hline \end{array}$$

We start by focusing on the ones digit of the multiplier (the number on the bottom).  Some students cover or cross out the tens digit in this phase so that they don't get confused.  If students are struggling to remember which number they are multiplying by, we recommend that they circle the digit that they are working with (rather than crossing anything else out).

Then, we multiply the ones digit of the multiplier (in red below) times each digit in the multiplicand (upper number) and put the answer values in the appropriate columns in the answer line:

$$\begin{array}{r} &21\\\times\!\!\!\!\!\!&3\color{red}{2}\\ \hline &\quad \color{red}{42} \end{array}$$

So far, this is just like multiplying by one digit.   Then comes the trickier "stacking part."

Now, we need to multiply the tens digit of the multiplier (bottom number) times each digit of the multiplicand (top number).  But that digit is in the tens column, so while it looks like a 4, it represents 40.  That means all of our answers are 10 times greater than they look.  So, we slide our answers over by one place value (we'll break this down below).  We slide our answer digits over by putting a place holder in the ones column of the second row of answers (no whole number times 40 will give you an answer less than 10).

$$\begin{array}{r} &21\\\times\!\!\!\!\!\!&32\\ \hline & 42\\&0 \end{array}$$

Then we multiply 4 times each of the numbers on the top row, putting the answers in the first available slots in the second answer row:

$$\begin{array}{r} &21\\\times\!\!\!\!\!\!&\color{purple}{3}\color{black}2\\ \hline & 42\\&\color{purple}{63}0 \end{array}$$

Now we have two rows of answers (one from multiplying the ones digit of the multiplier and one from multipling the tens digit of the multiplier).  We need to add them together... and that's exactly what we'll do.

$$\begin{array}{r} &21\\\times\!\!\!\!\!\!&32\\ \hline & \color{red}{42}\\&+\color{purple}{630} \\\hline & 672\end{array}$$

So, what exactly is going on here?  In short, we multiply each digit in the multiplier by each digit in the multiplicand.  For the first digit of the multiplier, we use the columns to keep our answers in the right place value. Once we start working with digits in the multiplier that represent more than one (so, anything in the tens, hundred, thousands, etc.) columns, we need to account for that.  So, we add place holder zeros into our answer rows.  Note: we have multiple answer rows because each digit of the multiplier represents, essentially, a separate problem.  The algorithm we use above, where we stack the answer rows just keeps everything neat for us.

Think about it the long way:

\eqalign{21 \times\color{purple}{3}\color{red}{2} & = (\color{red}{2}\color{black} \times 20)+ (\color{red}{2}\color{black}\times 1) \mathbf{+} (\color{purple}{30}\color{black} \times 20) + (\color{purple}{30}\color{black}\times 1)\\&=\color{red}{40}\color{black} + \color{red}{2} \color{black}\mathbf{+} \color{purple}{600}\color{black} + \color{purple}{30}\\&=\color{red}{42} \color{black}+ \color{purple}{630}\\&=672}

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!