## Double Digit Mental Arithmetic

maartens — Sat, 12/04/2010 - 19:02

I have often wondered why mental arithmetic is so hard beyond the simple tables from one to twelve we learned at school. Take for instance something like: 84 x 32. Seems simple enough, but how long will it take you to work it out in your head? Will you even be able to? Can you do it under pressure? Note that I am referring to real world scenarios, for instance doing the calculation while talking to someone, without the aid of any tools such as pen and paper.

The basic stumbling block appears to be the brain's inability to hold on to this information in consciousness, in real-time, for long enough to get going. It is significantly easier even just to write down the initial figures on paper, leaving your brain free to do the calculations. Doing it all in your head, you are soon spending more time trying to remember the original digits than doing the actual calculation!

It follows that any method of complex mental arithmetic should minimise the information your brain needs to keep in consciousness, instead relying on long term memory and existing skills, i.e. your old arithmetic tables.

At school, this is how we learned to do double digit arithmetic:

84

x 32

_____

168 -> starting from the right: 4 x 2 = 8; 8 x 2 = 16; => 168

+2520 -> starting from the right: 0 for x10; 4 x 3 = 12, (write 2 keep 1); 8 x 3 = 24, add 1 left over = 25; => 2520

_____

2688 -> starting from the right: add 8+0 = 8; add 6+2 = 8; add 1+5 = 6; 2 = 2; => 2688

This is straightforward on paper, but as pure mental arithmetic, it requires working with far too many numbers at once: 84, 32, 8, 16, 168 (throw away 8 and 16 at this point), 12, 1, 24, 25 (throw away 12, 1, and 24 at this point), 2520 (throw away everything except 168 and 2520)

By the time you calculate 2520 your consciousness is cluttered with 9 figures, 5 of which you can throw away, but still has the potential to confuse through the brain's policy of indiscriminate retention. 9 is considered just about the maximum number of discrete elements we can keep in consciousness at any one time (bar advanced memorising techniques of course, which is outside scope).

That is why it becomes necessary to *memorise* some digits while in the middle of calculation, and this is a sheer waste of time.

But once you've arrived at both 168 and 2520, it's relatively straightforward to add them and provide the answer: 2688.

In conclusion, given enough time to memorise digits as they are being calculated, all double digit arithmetic is possible without any aids using this method. However, in my experience, it takes far longer than is practical in any real life situation.

This is how I came to see that using a slightly different method, there's far less chance of forgetting things. It takes a bit of practice to get used to, but most double digit arithmetic can then be done in less than 20 seconds.

The trick lies in handling the "middle" digits immediately after the easy right-hand digits, thereby stringing all the results together and making the flow of the calculation more natural.

Let me illustrate with the same example above:

84

x 32

_____

8 -> starting from the right: 4 x 2 = 8

28 -> diagonals: (4 x 3) + (8 x 2) = 28

24 -> now the left: 8 x 3 = 24

_____

2688 -> let the three figures fall into one, adding where more than one digit appears => 2688

The first phase is always more difficult, but this time there are less big figures to remember, making it easier: 84, 32, 8, 12, 16, 28, 24 (throw away 84 and 32).

This time you only needed to keep in mind 7 figures - and what's more, the largest number is only two digits - compared to 168 and 2520 in the standard method. Now keep the three new figures (8, 28, and 24) in their positions, and let them "fall" to the bottom into one figure. In other words, from the right, 8 fall to the bottom, 8 fall to the bottom, 2 collects 4 = 6 fall to the bottom, 2 fall to the bottom => 2688.

Now, as a final enhancement, consider changing the first set of double digits to a single figure of four digits in a predictable order that will always yield the same results. Eg. 84 x 32 becomes 3428, which makes the all-important "middle digits" calculation easier - just work it from left to right: (3 x 4) + (2 x 8) = 28. Then digit 2 and 3, conveniently next to each other: 4 x 2 = 8 => 288; then the outer digits 1 and 4: 3 x 8 = 24 => 2688.

Suddenly it's easy!

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*- Maartens Lourens, 2010*