Mike Rowe’s most recent podcast (ep. 241 the Maddest March Ever), with Scott Flansburg (aka the Human Calculator), Scott mentioned what he called the “Rule of Nine”.

Take any 2 digit number, for example, \(71\), and:

- Add the digits (in this case that gives: \(8\))
- Subtract that from the original number, giving, in this case: \(71-8 = 63\)
- The result will always be nine times the first digit.

Listening in double speed (and I think Scott may be a fast talker anyways), this sounds impressive, perhaps even mysterious, but it is easy to decode:

Let’s represent that two digit number as ‘ab’. i.e:

$$a b \equiv a * 10 + b.$$

The algorithm gives us:

$$a * 10 + b \,- (a + b) = a * (10 – 1) = 9 * a$$

We see that this “Rule of Nine” algorithm above has a built-in distractor, the second digit. You could express it more simply as:

Take any 2 digit number, for example, \(71\), and:

- Ignore the second digit, giving in this case \(70\)
- Subtract that first digit from the original number, giving, in this case: \( 70-7 = 63\)
- The result will always be nine times that first digit.

But it’s not as cool to point out that \(7*10 \,- 7 = 9 * 7\). It’s kind of cool that adding the digits of the result will again always be nine: \(6 + 3 = 9\), which I am sure was also mentioned in the episode, but we can also decode that secondary rule of nine. That second rule of nine you know from ancient history when you learned your time tables, but algebraically, it is nothing more than:

$$10 * (x \,- 1) + 9 \,- (x-1) = 9 * x – 10 + 10 = 9 * x.$$

For example, for \( x = 7 \)

$$10 * 6 + (9-6) = 63$$

but we can also write that as

$$10 * (7-1) + (9-(7-1)) = 9 * (7-1) + 9 = 9 * 7,$$

which isn’t as cool.

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