You may have noticed your child multiplying two- and three-digit numbers in a way that looks foreign to you. It’s called “lattice multiplication” and it’s really just a way of doing it that math teachers have developed for kids with sloppy handwriting.
The old way of multiplying two- and three-digit numbers was to multiply first by the ones digit, then by the tens digit, remembering of course to add the all-important zero in the ones place on the second row. Then, you would have to add two zeros (in the ones and tens place) for the third row. It went something like this:
As you remember, we first multiply each digit of the 28 by the 7 ones in 47, then we add the zero in the ones place on the second line of our work and multiply each digit of the 28 by the 4 tens (i.e., by 40). We can then add the two partial products to get our final answer. We can add them because any number (in this case 28) times 47 = that number times 7 plus that number times 40.
The problem many students had is that they were forgetting to line up the second row, which here would be 112 if the student forgot to write the zero down. This would more often than we care to reminisce about end up with the child adding 196 + 112 and getting the wrong answer of 308.
As a result, teachers developed a way to force students to line everything up properly. It’s called lattice multiplication, and to the uninitiated parent, it looks like a lot more work than it’s worth. However, for kids with less-than-perfect handwriting, it means getting the right answer more frequently.
You start by writing one factor on top of a grid and the other factor along the right-hand side of the grid. If we are to continue with our same example, we will write 28 across the top of the grid and 47 down the right-hand side. Next, from each top right-hand corner in the grid, we draw a diagonal line down and to the left (shown in red):
Now, for each square in the grid, we multiply the single-digit number on the top with the single-digit number at the right and put the product inside the square so that the tens digit goes on top and the ones digit goes on bottom. For example, the product in the upper-right square (8 × 4) is 32. We write a “3” in the top corner of that square and a “2” in the bottom corner, like so:
And likewise with the other squares in the grid, each time putting the two-digit product so that the tens digit goes on top (even if it’s a zero) and the ones digit goes on bottom.
Finally, we do the addition, which was part of our old-fashioned way of doing it as well, but the addition takes on a new flavor here. Since we can ignore the original factors (28 and 47), I’m going to make them lighter in the picture to make it slightly less busy. That way, you can see the numbers you have to add a little more clearly.
Watching only the diagonals, and starting with the lowest diagonal and working upwards, add all the single-digit numbers in each diagonal. For example, there is only one number in the first diagonal, which is 6. So, we write “6” in the space below that diagonal. The second diagonal has three numbers in it: 2, 5, and 4, which we add to get 11. Now, we only write the ones digit down below that diagonal, and we “carry” the “1” to the next diagonal:
This gives us four single-digit numbers in the next diagonal: 1, 8, 3, and the carried 1. The sum of these four numbers is 13, so we write the “3” in the space below the diagonal, which is kind of on the side this time, and we carry the “1” to the next diagonal. That diagonal has only two numbers in it now: the carried “1” and a “0” (if we chose to write the product of 2 and 4 as “08” rather than just “8”. The sum is “1” so we write that in the space below that diagonal:
Finally, reading from top to bottom then left to right, like a capital letter “L”, we have the product 1,316, which is the same answer we obtained by doing it the old-fashioned way.