The Skiver Method of Division

During a psychological study of his renowned ability to mentally decimalize fractions [1], the mathematician Alexander Craig Aitken was given the problem 1/697. He seems to have found this a challenge, reaching a solution 0.0014 in a time of twelve seconds. His method was traditional long division. A short time later, he recognised 697 as 17 × 41 and attempted to solve the problem again, this time by calculating 1/41 and dividing by 17 as he went. Aitken described this process of parallel division as “severe,” though he did have more success than in his previous attempt, obtaining a solution 0.00143472 in about nine seconds. He remarked, “That’s as far as I can go on that one.”

The proximity of 697 to 700, which was curiously not exploited by Aitken, screams Skiver Method, a handy but little-known technique for simplifying division problems, especially those involving divisors that would not be considered division friendly but which are close to divisors that would be considered division friendly.

The difficulty in dividing A by B is dictated mainly by B. It is logical, therefore, to transform A/B by adjusting B to obtain an agreeable divisor and make the appropriate, compensatory adjustment to A, such that the modified division problem yields the same result as A/B.

We shall consider an additive adjustment to B, to be denoted by b. The following algebra shows that it is always possible to preserve A/B by making an additive adjustment to A.

A/B
= A/B × (B + b)/(B + b)
= A/B × B + A/B × b divided by (B + b)
= (A + A/B × b)/(B + b)

Perhaps not surprisingly, the additive adjustment to A is the adjustment made to B scaled by A/B. Taken at face value, the above equation is quite useless, since A/B appears on both sides. However, the above equation suggests an iterative formula for determining A/B. Denoting A/B by C, the formula is as follows.

Cₙ₊₁ = (A + Cₙ × b)/(B + b) [SK1]

The subscripts n and n+1 are iteration indices. They are used to denote successive approximations C to A/B. We begin with an initial estimate, typically a rough, first impression guess. This is C₀. Then we calculate C₁ by inputting C₀ into the right-hand side of the above formula. Then we calculate C₂ by inputting C₁ into the right-hand side of the above formula. And so on. A little more algebra shows that the estimate improves with each iteration and reveals the degree of that improvement.

Define the error in estimate Cₙ by

eₙ = Cₙ − A/B

Now,

eₙ₊₁ = Cₙ₊₁ − A/B
= (A + Cₙ × b)/(B + b) − A/B
= (A × B + Cₙ × b × B − A × B − A × b)/((B + b) × B)
= b × (Cₙ − A/B)/(B + b)
= (b/(B + b)) × eₙ

It follows that

eₙ = (b/(B + b))ⁿ × e₀

Provided the magnitude of b/(B + b) is smaller than 1, eₙ will converge to zero as n tends to infinity. More importantly for the mental calculator, who is likely interested in only one or two iterations, the error in the estimate is reduced by a factor b/(B + b) with each iteration. So, if the magnitude of b is small compared with the magnitude of B, a significant improvement in the estimate will be seen from one iteration to the next.

By way of illustration, consider 124/489.

Here, A = 124 and B = 489. A natural choice for b is 11, since it transforms 489 into the much more division friendly divisor 500.

We need an initial estimate C₀. Since B is about four times as big as A, we would take C₀ to be 1/4, or 0.25.

Plugging this into formula [SK1] yields

C₁ = (124 + 0.25 × 11)/500
= (124 + 2.75)/500
= 253.5/1000
= 0.2535

Compare this with the actual A/B, which is 0.253578…

Let’s do one more iteration. This time, formula [SK1] yields

C₂ = (124 + 0.2535 × 11)/500
= (124 + 2.7885)/500
= 253.5770/1000
= 0.253577

It is sometimes advantageous to transform a division problem so as to make it more amenable to the Skiver Method. Consider, for instance, 1426/267.

1426/267 × 3/3 = 4278/801

Take C₀ to be 5 and b to be −1. Now,

C₁ = (4278 + 5 × (−1))/800
= 4273/800
= 5.34125

This compares with the actual A/B of 5.34082397…

Proceeding to the next iteration with C₁ rounded to 5.34 yields

C₂ = (4278 + 5.34 × (−1))/800
= 4272.66/800
= 5.340825

Since in this instance b is such a convenient multiplier, we can proceed quite easily to the next iteration. Indeed, we can plug C₂ calculated above into [SK1] without any rounding to obtain

C₃ = (4278 + 5.340825 × (−1))/800
= 4272.659175/800
= 5.340823968 …

Division friendly divisors are not only limited to single-digit numbers with zeroes attached. Consider the conversion of centimetres to inches. This may be achieved by dividing the number of inches by 2.54 (the precise number of centimetres in one inch). It is, however, generally much easier to divide by 2.5, this being one quarter of ten. For example, to convert 130 centimetres to inches, we may apply our formula [SK1] as follows.

A first estimate C₀ of 52 (in inches) is readily obtained by calculating 130 × 0.4 (dividing by 2.5 is the same as multiplying by 0.4). Plugging C₀ = 52 and b = −0.04 into formula [SK1] yields

C₁ = (130 + 52 × (−0.04))/2.5
= (130 − 2.08)/2.5
= 127.92/2.5
= 51.168

The actual result, i.e. the actual number of inches in 130 centimetres, is 51.1811… In practice, the calculation would proceed much more economically, essentially reducing 130 by half the reduction in 254, resulting in 128/250, then doubling 128 to obtain 256 and then doubling 256 to obtain 512. Estimate C₁ is therefore 51.2. The disregard for place value during the calculation is deliberate. The actual result is clearly bigger than 5.12 and smaller than 512. The correct estimate must therefore be 51.2.

Let us now return to Aitken and his headache of a calculation, 1/697. We have A = 1 and B = 697. Naturally we take b to be 3. Since 1000 is about one and half times 697, we’ll take 0.0015 for our initial estimate C₀. Now,

C₁ = (1 + 0.0015 × 3)/700
= 1.0045/700
= 0.001435

This compares favourably with the actual result of 1/697 (= 0.001434720…) and is twice as accurate as Aitken’s first result. A second iteration yields

C₂ = (1 + 0.001435 × 3)/700
= 1.004305/700
= 0.001434721 …

We have matched the accuracy of Aitken’s best result and have done so with relatively little work.

[1] An Exceptional Talent for Calculative Thinking, I. M. L. Hunter, British Journal of Psychology, Volume 53, Issue 3, Pages 243-258, 1962.