# Multiplication (by 3+ digits)

We discuss the algorithm for multiplying multi-digit numbers in detail in the lesson Multiplication (2x2 digits). Once you start working with longer numbers, the process continues:

**Every time you move to a new digit in the multiplier, you are multiplying by a digit that is larger than the previous one by a factor of 10. So, you add a row in your answer section and add one additional place holder zero** (so, the ones digit of the multiplier gets no place holder; the tens digit in the multiplier gets one placeholder zero; the hundreds digit in the multiplier gets three placeholder zeros; etc.) As your multiplier gets larger, your answer section gets taller, so make sure to keep your digits lined up so it's not too confusing to multiply and then add your answer rows!

*Example*:

$1734 \times 231 = $

First, set the problem up vertically. It's most efficient to put the number with the most digits on top:

$$\begin{array}{r} &1734\\\times\!\!\!\!\!\!&231\\ \hline \end{array}$$

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 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 (don't forget to carry as necessary -- we're going to leave those marks out of the example for clarity):

$$\begin{array}{r} &1734\\\times\!\!\!\!\!\!&23\color{red}{1}\\ \hline &\quad \color{red}{1734} \end{array}$$

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 although it is 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 placeholder zero** in the ones column of the second row of answers (no whole number times 40 will give you an answer less than 10).

$$\begin{array}{r} &1734\\\times\!\!\!\!\!\!&231\\ \hline & 1734\\&0 \end{array}$$

Then we multiply 3 times each of the numbers on the top row, putting the answers in the first available slots in the second answer row:

$$\begin{array}{r} &1734\\\times\!\!\!\!\!\!&3\color{purple}{3}\color{black}1\\ \hline & 1734\\&\color{purple}{5202}0 \end{array}$$

Now we have two rows of answers (one from multiplying the ones digit of the multiplier and one from multiplying the tens digit of the multiplier). Next we multiply the hundreds place of the multiplier by each number in the multiplicand. First, we have to start a new answer row, and because this is the hundreds column, we start the row with **two placeholder zeros**.

$$\begin{array}{r} &1734\\\times\!\!\!\!\!\!&\color{green}{2}\color{black}31\\ \hline & 1734\\&52020\\&\color{green}{3468}00 \end{array}$$

Finally, we need to add all of the answer rows together!

$$\begin{array}{r} &1734\\\times\!\!\!\!\!\!&\color{green}{2}\color{black}31\\ \hline & \color{red}{1734}\\&+\color{purple}{52020}\\&+\color{green}{3468}00\\\hline &\mathbf{400554} \end{array}$$

You can see the pattern. Every additional digit in the multiplier yields another row in the answer section and an additional place holder zero in that row. Know this rule, you can multiple numbers of any size, just make sure to line up the digits in the answer rows that addition doesn't get to complicated!

**Hint**: Students often find it helpful to put a an X or a line through their placeholder zeros so that it's easy to see how many they've written down. If students struggle to keep track of their placeholders, try having them put lines or Xs through those placeholder zeros to make them distinct from regular zeros that result multiplication that equals zero.

$$\begin{array}{r} &1734\\\times\!\!\!\!\!\!&\color{green}{2}\color{black}31\\ \hline & \color{red}{1734}\\&+\color{purple}{5202}\!\bcancel{0}\!\!\\&+\color{green}{3468}\!\bcancel{0}\!\!\!\bcancel{0}\!\!\!\\\hline &\mathbf{400554} \end{array}$$