# Processes of the skilled arithmetician (or, how to multiply 17 by 23)

By dkl9, written 2023-165, revised 2023-165 (0 revisions)

Here I will show, by detailed examples, those methods I run thru in my mind to produce my weirdly effective mental arithmetic. Hopefully this will help others to learn better mental arithmetic.

## § Multiply 17 by 23

Recall that the product of two numbers a and b is more easily done as m² - r², where m is their mean and r is the distance from the mean to each one. (This is especially helpful when a and b are nearby integers.) Notice that 17 and 23 are each 3 away from 20 — I do this by feeling 7's 3-away-ness from 10. Then m = 20, r = 3. I have 20² memorised — it's 400 — but if not, it could be calculated by splitting 20 into 2·10 and squaring both parts, making 4·100, or 400. One should definitely have 3² (9) memorised. Then 17·23 = 400 - 9. To subtract: notice that 9 is between 0 and 10, so subtract 10 from 400 — giving 390 in what feels like an atomic step — and add back 9's distance from 10, which is 1.

## § Solve for x in x² + x + 1 = 0

Some polynomials, I'd try to factor. That looks impractical here, and it's quadratic, so I apply the formula. Then x = (-1 ± sqrt(1 - 4)) / 2 — that's partially simplified from (-1 ± sqrt(1² - 4·1·1)) / (2·1), as far as I could get by simplifying whilst writing it in a single pass. sqrt(1 - 4) reduces to sqrt(-3). Factor out the -1 and get i·sqrt(3). Recall that sqrt(3) = 1.7320508. If you didn't have that memorised — you probably shouldn't have that memorised, but I do — you could approximate it. One way: notice that 300 is between 17² (289) and 18² (324), so guess 1.75, between 1.7 and 1.8. Another way: 7² (49) is roughly three times 4² (16), so guess 7/4 — also 1.75. Divide each term in the original numerator by the 2 and get -1/2 ± 1.732 / 2·i. To divide 1.732 by 2, operate on the "even" 1.72, splitting it into the "even" 1.6 and 0.12, which halve to 0.8 and 0.06, respectively. Add a 0.006 to correspond to half of the omitted 0.012. -1/2 is -0.5.

Answer: x = -0.5 ± 0.866i

That happens to lie on the unit circle, ±π/3 radians from (-1, 0). There's also a geometric argument for that answer.

## § Evaluate 12 choose 5

Expand the problem, by one definition of binomial coefficients, to 12! / (5!·7!). Cancel factors from 7! in denominator out of 12! in numerator to get (12·11·10·9·8) / 5!. Recall that 5! is 120, so we have (12·11·10·9·8) / 120. Notice that 120 factors into 12 and 10, both of which cancel from the numerator, so we have 11·9·8. I see two ways to go from here. 11·9 is 99 (by digit scaling, or m² - p² as in 17·23). Multiplying by 8 doubles it thrice. Notice 99 is 100 - 1, so double each part thrice: 100 - 1 to 200 - 2 to 400 - 4 to 800 - 8, which is 792. Aliter: 8·9 is 72 (that should be memorised), 11 is 10 + 1. Multiply componentwise to find that 72·11 = 72·10 + 72·1. Each component can be found by digit shifting: 720 + 72. Add each digit in parallel (no carries in this case) to get 792.

## § Largest power-of-3 divisor of the sum of the first 8 factorials

I'd recall all those factorials from memory, but I'll calculate them here to present more methods.

Factorials of 1 to 4 should be trivial (1, 2, 6, and 24, respectively). 5! is 5·4!, which is 5·24. Split the 24 and distribute to 5·20 + 5·4, which converts to 5·2·10 + 5·4, reducing to 10·10 + 20, which is 120. 6! is 6·5!, or 6·120, which is 6·100 + 6·20 = 600 + 5·20 + 20 = 600 + 100 + 20 = 720. 7! factorial proceeds by a similar split: 7·720 = 7·7·100 + 7·20 = 4900 + 140 = 5040. We get 8! by either three doublings (5040 to 10080 to 20160 to 40320) or multiplying components by 8 (8·5000 + 8·40 = 8·5·1000 + 8·4·10 = 40000 + 320 = 40320).

Adding them: the 6 of 3! "fits" into the 24 of 4!, so 3! + 4! = 30, and (adding the first two) 1 + 2 + 6 + 24 = 33. For the greater four factorials, split them by digits: 120 + 720 + 5040 + 40320 = (2 + 2 + 4 + 2)·10 + (1 + 7 + 3)·100 + 5000 + 40000. That reduces to 100 + 1100 + 45000 = 46200. Then the sum of the first 8 factorials is 46233.

To divide it by three, split off a 45000, which divides to 15000 (for 45 / 3 = 15), leaving 1233. Split off a 1200, dividing to 400 (for 12 / 3 = 4), leaving 33. 33 / 3 = 11. The parts add up to 15411 — an integer, so that sum is divisible by at least 3¹. Thirding 15411 goes digit-by-digit: 5000 from 15000, 100 from 300, 30 from 90, 7 from 21, totalling to 5137 — 3² is a divisor, at least. There's a simple test for divisibility-by-three: a number is divisible by three iff the sum of its digits is. 5 + 1 + 3 + 7 = 16, not divisible by three, so 3² is the greatest divisor.

## § Concluding remarks

All that, I think, exposes a large part of my methods. I doubt the main issue at this point is a lack of examples, but improper explanation of examples. If some of my steps don't make sense, ask me to clarify.

As for why I know these methods, rather than how the methods work: I got to this point by (surprise, surprise) vast amounts of practice. I find many opportunities for arithmetic in my life — often making them up for fun — and do it all in my head, accepting approximations when easier. I am easily forced to persist with mental calculation rather than resorting to a calculator. In many of these cases when I do calculations for fun, I can't access a calculator, or it would take nontrivial effort, and my laziness forces me to practise.

Naturally, I don't bring calculators for maths tests. If they're multiple-choice, I can approximate my way to see which answer is closest — often the difference between the answers lies in one of several probable mistakes, which tests understanding and error avoidance, not computation. If they're free-response, I show my methods and hope the teacher accepts a well-justified imperfect answer (usually, they do).