## Finger Multiplication

A digital computer is one who adds on his fingers, and this definition reminds me of a quote by Tom Lehrer.

Base eight is just like base ten, really… if you’re missing two fingers!

Many students know the trick for multiplying by 9 on their fingers, which I mentioned in yesterday’s post. However, most folks don’t know the following finger trick for multiplying two numbers that are greater than 5 but less than or equal to 10.

1. On each hand, raise a number of fingers to indicate the difference between each factor and 5. For instance, 8 – 5 = 3 and 7 – 5 = 2, so to multiply 8 × 7, raise three fingers on the left hand and two fingers on the right hand, as shown below.

1. Count the total number of fingers that are raised on both hands, and append a 0. As shown above, there are 3 + 2 = 5 fingers raised, so this gives an intermediate result of 50.
2. Count the number of fingers that are folded on each hand, and find their product. Again using the example above, there are two fingers folded on the left and three fingers folded on the right, and 3 × 2 = 6.
3. Add the results of Steps 2 and 3. For our example, 50 + 6 = 56, so 8 × 7 = 56.

We can use algebra to understand why this trick works.

Let L = the number of fingers raised on the left hand, and let R = the number of fingers raised on the right hand. Then the numbers we are multiplying are (5 + L) and (5 + R).

The total number of fingers raised is L + R. Consequently, the result from Step 2 is equal to 10(L + R).

The number of fingers not raised on the left hand is 5 – L, and the number of fingers not raised on the left hand is 5 – R. Their product, which is required in Step 3, is (5 – L)(5 – R) = 25 – 5(L + R) + LR.

Finally, combining them in Step 4 gives

10(L + R) + 25 – 5(L + R) + LR = 25 + 5(L + R) +LR = (5 + L)(5 + R),

which is the product we initially wanted to compute.

I recently learned one other method of multiplication. It does not require fingers, and I’m not even sure it’s useful, but it is interesting.

To find the product of two numbers, a and b, do the following:

Plot the points A(0, a), B(b, 0), and C(0, 1).

Draw segment CB, and then construct segment AD parallel to CB.

As a result, point D(d, 0) is a point on the horizontal axis such that a × b = d.

For instance, to compute 6 × 3, plot A(0, 6), B(3, 0), and C(0, 1). Then draw a line through A that is parallel to CB. This line will intersect the horizontal axis at the point D(18, 0), so 6 × 3 = 18.

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