Tag: multiplication

An Expanded Multiplication Table

In the Dune stories, mentats are trained to do complex calculations without any mechanical aids. For normal people, instantly knowing the multiplication up to 10 is a useful skill. Extending the multiplication table to twenty seems to require the skills of a mathematical savant.

However, calculating products of up to 20 actually only requires the single digit multiplication table and the ability to add small numbers mentally. My goal is to be able to do these calculations without paper. How?

If one of the numbers is single digit and the other is between 10 and 20, the calculation works like this:

A number between 10 and twenty can be written as \(10 + a\) or \(1a\). For the results you have one single digit multiplication an addition of a number times ten.

$$1a * b = (a*b) + (10 * b)$$

Since \(10 * b\) is just \(b\) shifted left by one, you can get the result by identifying \(a*b\) and then adding \(b\) to the second digit.

$$1\color{blue}{5} * \color{red}{7} = (\color{magenta}{35} + \color{red}{7}0) = 105$$

The mental steps are “calculate \(\color{blue}{5} * \color{red}{7}\)” and then “add \(\color{red}{7}\) to the second digit.”

For the product of a single digit times and a number under twenty, you always use the normal multiplication table and single digit addition.

If both numbers are between 11 and 19, the calculation works out this way.

$$a * 1b = (10 +a) * (10 + b) = a * b + 10 * (a + b) + 100$$

To start, multiply the singles place values. For example \(1\color{blue}{5} * 1\color{red}{7}\) starts as \(\color{magenta}{35}\).

Then add \(\color{blue}{5} +\color{red}{7}\), giving \(\color{orange}{12}\).

Add that in the 10s place \(\color{magenta}{35} + 10 * \color{orange}{12}\) giving \(\color{brown}{155}\) and finally, increment the hundreds digit, giving \(\color{green}{255}\)

This involves multiplying single digit numbers, adding a pair of single digits, adding a digit and a number less than twenty and incrementing a digit. It works out to needing having six or so digits in mind at once which can be less than one’s estimated working memory of 7 digits.

If one of the numbers is 10, add a zero to the right of the other value. For twenty, double the other number and add a zero. Now there’s a simple way to multiply two numbers where both are 20 or less.

With a little practice, I can do this in my head. More importantly, I’m confident of my result so that I’m getting so that I don’t need to double check on a calculator.

Reduced Colors

If you start with the number 6 and organize the integers based on their remainder when divided by six, you have 6 sets. One set contains \(\{0, 6, 12, …, -6, -12, … \}\). Another set contains \(\{1, 7, 13, …, -5, -11, … \}\). A third \(\{2, 8, 14, …, -4, -10, … \}\). The last three are \(\{3, 9, 15, …, -3, -9, …\}\), \(\{4, 10, 16, …, -2, -8, … \}\) and \(\{5, 11, 17, …, -1, -7, -13 \}\).

If I use the lowest non-negative value in each set as its name, I can signify them by placing a bar over the number. The collection of 6 sets can be summarized as \(\{\overline{0}, \overline{1}, \overline{2}, \overline{3}, \overline{4}, \overline{5}\}\).

For the examples here, I’m going to reduce the list by throwing out all of the numbers that are not relatively prime to 6. Relatively prime means that they don’t have a common factor with 6. 2 is not relatively prime to 6 nor is 3 because they divide 6. Also 4 is not relatively prime because 2 divides both. I also discard 0 because all of the numbers in \(\overline{0}\) are have 6 as a common divisor. This leaves a much shorter list: \(\{\overline{1}, \overline{5}\}\)

What’s interesting about this reduced list is that, if you multiply any pair of its numbers, the product is still in the collection. \(1 \times 1 = 1\), \(1 \times 5 = 5\), \(5 \times 5 = 25\). If you simplify 25 modulo 6, you get 1, still within the reduced set.

In general, if you take an integer \(n\) and keep only the numbers that are relatively prime to \(n\), the result of any product is in one of the “residue classes” of the list. This means that a multiplication table of these products will only contain those relatively prime numbers.

For 5, 8, 10 and 12, there are four numbers in each reduced set. 5 has \(\{\overline{1}, \overline{2}, \overline{3}, \overline{4}\}\). 8 has \(\{\overline{1}, \overline{3}, \overline{5}, \overline{7}\}\). 10 has \(\{\overline{1}, \overline{3}, \overline{7}, \overline{9}\}\), and 12 has \(\{\overline{1}, \overline{5}, \overline{7}, \overline{11}\}\).

The product tables that result from these four sets are:

It’s not obvious whether any of these tables are equivalent and follow the same pattern. However, if you rewrite the table by replacing each number by its position in the set. I’m using the natural numeric order of the values. 7 is the third number in 12’s list, so 7 in replaced by 3. A similar substitution for each value, yields a new product table.

From these, it’s a little clearer to see that 5 and 10 are equivalent as are 8 and 12.

However, if you replace the numbers with colors, the information is even easier to see.

The numbers, 7, 9, 15 and 18 all reduce to sets of 6 numbers. When they are transformed into color arrays with similar procedure, you get these diagrams:

This shows that there are 3 different multiplicative groups coming from a reduced set of 6 values. 9 and 18 are equivalent while 7 and 14 are unique.

When you have the reduced sets represented with these color arrays, interesting patterns can become visible that are hidden if you just look at the values numerically.