Finger Multiplication

Tobias Dantzig makes mention of the lost art of finger multiplication in his book Number: The Language of Science, first published in 1930.  "Finger counting is a lost art among modern civilized people ... only a few hundred years ago finger counting was such a widespread custom in Western Europe that no manual of arithmetic was complete unless it gave full instructions in the method."  Using the method of finger multiplication, you can find the product of the numbers 6, 7, 8, and 9 with just your own ten fingers and a knowledge of the addition and multiplication tables up to five.

To use finger multiplication, hold out your hands palm down, label your thumbs 6, your index fingers 7, your middle fingers 8, your ring fingers 9, and your pinky fingers 10, and touch the tips of the two fingers that correspond with the two numbers you are multiplying together.  Then count all the fingers including and below the two fingers touching together and multiply that number by ten, and add to it the product of the left fingers above the two fingers touching together and the right fingers above the two fingers touching together. 

For example, let’s say you wanted to multiply 8 x 9.  You would hold your hands out palm down, and touch the tip of your left middle finger (which corresponds with 8) with the tip of your right ring finger (which corresponds with 9).  Including and below these two fingers there are a total of 7 fingers, and 7 x 10 = 70.  There are 2 fingers above the two fingers touching together on the left hand and 1 finger above the two fingers touching together on the right hand, and 2 x 1 = 2.  Adding these two numbers together gives 70 + 2 = 72.  So 8 x 9 = 72.

Or let’s say you wanted to multiply 6 x 7.  You would hold your hands out palm down, and touch the tip of your left thumb (which corresponds with 6) with the tip of your right index finger (which corresponds with 7).  Including and below these two fingers there are a total of 3 fingers, and 3 x 10 = 30.  There are 4 fingers above the two fingers touching together on the left hand and 3 fingers above the two fingers touching together on the right hand, and 4 x 3 = 12.  Adding these two numbers together gives 30 + 12 = 42.  So 6 x 7 = 42.

In general, if we want to multiply a x b using finger multiplication, then there are a – 5 fingers and b – 5 fingers including and below the two fingers touching together and 10 – a fingers and 10 – b fingers above the two fingers touching together.  The sum of all the fingers including and below the two fingers touching together and multiplying by ten can be represented as 10(a – 5 + b – 5), and the product of the left fingers above and the right fingers above can be represented as (10 – a)(10 – b).  Adding these together and simplifying would give 10(a – 5 + b – 5) + (10 – a)(10 – b) = (10a – 50 + 10b – 50) + (100 – 10a – 10b + ab) = ab, which is why finger multiplication works.

There is an easier method of finger multiplication, but it only works if one of the factors is 9.  Hold your hands out, count off the other factor you want to multiply by 9, and put that finger down.  Then the product will be the number of fingers to the left times ten plus the number of fingers to the right.

For example, let’s say you wanted to multiply 3 x 9.  You would hold your hands out and put down the third finger from the left (the left hand middle finger).  There are 2 fingers to the left and 7 fingers to the right, and 2 x 10 + 7 = 27.  So 3 x 9 = 27.

Or let’s say you wanted to multiply 8 x 9.  You would hold your hands out and put down the eighth finger from the left (the right hand middle finger).  There are 7 fingers to the left and 2 fingers to the right, and 7 x 10 + 2 = 72.  So 8 x 9 = 72.

In general, if we want to multiply a x 9 using finger multiplication for nines, then there are a – 1 fingers to the left of the finger that is down and 10 – a fingers to the right of the finger that is down.  The number of fingers to the left times ten plus the number of fingers to the right can be represented as 10(a – 1) + (10 – a).  Simplifying would give 10a – 10 + 10 – a = 9a, which is why this method of finger multiplication for nines also works.

As observed by Dantzig, finger multiplication is a dying art.  Few people know how to use finger multiplication for nines, and even fewer know how to use finger multiplication for the numbers six through nine.  It is likely that finger multiplication is no longer taught in schools anymore because for most people it takes just as much effort to memorize the multiplication table up to ten than to remember all the steps involved for finger multiplication.

Calculating a Square Root

In Terry Pratchett’s book Hogfather, a series of hilarious events leads a professor named Ponder Stibbons to wait up for the Disc World’s version of Santa Claus.  “I’m waiting for the Hogfather, thought Ponder Stibbons.  I’m in the dark waiting for the Hogfather.  Me.  A believer in Natural Philosophy.  I can find the square root of 27.4 in my head. (He’d have to admit that the answer would be ‘five and a bit,’ but at least he could come up with it.)  I shouldn’t be doing this.”

This begs the question: how do you find the square root of any number in your head?  Like Ponder Stibbons, we can easily find the closest square number and get a good estimate: 27.4 is just over 25, which is 5², so the square root of 27.4 is “5 and a bit.”  But is there an algorithm for finding the square root of any number mentally, or with paper and a pencil?  These days it’s so easy to mindlessly plug in the problem into a nearby calculator (on a cell phone, tablet, or computer) that we don’t stop to think about how it can be done.

There are actually several different ways of calculating a square root of a number, but we will focus on one particular method.  First, start by guessing an answer.  (The closer the guess is to the correct answer, the less steps we will need to take in the long run.)  Let’s guess that the square root of 27.4 is 5.  Second, we divide our guess into the number being square-rooted.  In our example, 27.4 divided by 5 is 5.48.  Therefore, we know that the square root of 27.4 is somewhere between 5 and 5.48.  Third, we will take the average of those two numbers (5.24) and repeat the process with 5.24 as our new guess.  Here is the result of a few rounds of calculations:

Guess	Divide	Average
5	27.4 / 5 = 5.48	5.24
5.24	27.4 / 5.24 = 5.229007634	5.234503817
5.234503817	27.4 / 5.234503817 = 5.234498046	5.234500931
5.234500931	27.4 / 5.234500931 = 5.234500931	5.234500931

In this case, after the second round of calculations we have an answer that is accurate to five decimal places, and after the third round of calculations we have an answer that is accurate to at least nine decimal places.

Some of us may be able to do the first round of calculations for finding a square root in our heads, but even Professor Ponder Stibbons would have a difficult time with subsequent rounds.  In our example above, dividing 27.4 by 5.24 is no easy task (although it can be done).  Now the question becomes: how can we program a computer to find a square root using only the basic mathematical functions?

Algebraically, we can express this process as an iterated function.  Let p equal the number to be square rooted, x equal the guess, and y equal the square root.  Then y is the average of ^p/_x and x, or y = (^p/_x + x) / 2, or y = ^{p + x^2}/_2x.  In our case above, p = 27.4 and x₀ = 5, which means x₁ = ^{27.4 + 5^2}/_2∙5 = 5.24, x₂ = ^{27.4 + 5.24^2}/_2∙5.24 = 5.234503817, and so on.  The actual square root would be the fixed point of this iterated function, which can be found by letting y = x.  Therefore, y = ^{p + y^2}/_2y, and multiplying both sides by 2y, 2y² = p + y², and subtracting y² from both sides, y² = p or y = √p.  Since we have expressed this process as an iterated function, the first round’s solution can be represented as x₁ = (p + x₀²)/(2x₀), the second round’s solution as x₂ = (p + x₁²)/(2x₁), the third round’s solution as x₃ = (p + x₂²)/(2x₂), and so on.  Using substitution and a little bit of algebra, we can deduce that x₂ = (p² + 6px₀² + x₀⁴)/(4x₀(p + x₀^2)) and x₃ = (p⁴ + 28p³x₀² + 70p²x₀⁴ + 28px₀⁶ + x₀⁸)/(4x₀(p + x₀²)(p² + 6px₀² + x₀⁴)), where each equation is a closer and closer approximation to y = √p.  In fact, using x₀ = 5, the graph of x₃ and √p are nearly indistinguishable from about 0 < p < 250, and x₃ > √p when p > 250.

Graph of x₃ when x₀ = 5 and y = √p

(0 < p < 900, 0 < y < 30)

The equation for x₂ or x₃ may give a good approximation of a square root for numbers less than 250, but we ought to have something that works well for all numbers.  Fortunately, iterated functions are loops, and pretty much all computer programs can loop the same set of commands until a certain criteria is met.  In this case, we would want the computer to loop through the formula ^{p + x^2}/_2x for subsequent terms of x until the same decimal approximation is reached for two subsequent terms.  The following is a program that does exactly that in the computer language Python:

01	#Square Root Function
02	#Python 2.7.3
03
04	#function to determine if a string is a number
05	def is_number (n):
06	#if the number can successfully change into a float then return true
07	try:
08	float(n)
09	return True
10	#otherwise return false.
11	except:
12	return False
13
14	#function to determine if a number is positive
15	def is_positive (n):
16	#first check to make sure n is really a number
17	if (is_number(n)):
18	#if the number is less than zero then return false
19	if (float(n) < 0):
20	return False
21	#otherwise return true
22	else:
23	return True
24	#if n is not a number return false
25	else:
26	return False
27
28	#function to find the square root without using a built-in square root function
29	def square_root (p):
30	#first check to make sure p is a positive number
31	if is_positive(p):
32	#assign variables
33	p = float(p)
34	r = 9 #number of decimal places to round to
35	x_old = 5 #starting guess for x0
36	#loop values in y = (p + x^2)/(2x) until x_old = x_new (rounded)
37	while True:
38	#find x_new using y = (p + x^2)/(2x)
39	x_new = (p + x_old * x_old) / (2 * x_old)
40	#round x_old and x_new
41	x_old_round = "{:.{prec}f}".format(x_old, prec=r)
42	x_new_round = "{:.{prec}f}".format(x_new, prec=r)
43	#if x_old and x_new are equal, then stop and return value
44	if (x_old_round == x_new_round):
45	return float(x_new_round)
46	break
47	#make x_old the x_new value and start the loop over again
48	x_old = x_new
49	#if p is not a positive number then return an error message
50	else:
51	return "ERROR"

Running the above program and typing in the command to use the square_root function for 27.4 results in the following answer:

>>	square_root(27.4)
5.234500931

This accurate and quick result only needed about 50 lines of code (and only 30 lines if you take out all the comments).

Therefore, there is an easy algorithm for finding the square root of a number.  If you are good at mental math and can remember the formula ^{p + x^2}/_2x, you could even impress your friends by finding the square root to the nearest hundredth place (well, at least Ponder Stibbons would be impressed).  But the best approximation for a square root requires several iterations of the formula ^{p + x^2}/_2x which can be done quickly with a short computer program.

Repeating Decimals

In our monetary system, different coins represent different fractions of a dollar, and most of us have memorized a few basic fraction to decimal conversions.  For example, one quarter is worth ¼ of a dollar or $0.25, so we know that ¼ = 0.25.  Similarly, one dime is worth ¹/₁₀ of a dollar or $0.10, one nickel is worth ¹/₂₀ of a dollar or $0.05, and one penny is worth ¹/₁₀₀ of a dollar or $0.01.  Therefore, we know that ¼ = 0.25, ¹/₁₀ = 0.1, ¹/₂₀ = 0.05, and ¹/₁₀₀ = 0.01.

One of the reasons these particular values were picked for coins was because the decimals for these fractions terminate, despite the fact that most fractions do not.  For example, a quarter can be represented with long division as the following:

	0.	2	5
4	1.	0	0
	–	8	↓
		2	0
	–	2	0
			0

The remainder reaches zero and so the decimal terminates, which makes for easier calculations for adding and subtracting amounts in cents.  But imagine a coin that was worth ¹/₃ of a dollar.  Using long division, we would get:

	0.	3	3	3	…
3	1.	0	0	0	…
	–	9	↓
		1	0
		–	9	↓
			1	0
			–	9
				1

The remainder never reaches zero, and so the threes carry on infinitely, which is why we don’t have a coin worth ¹/₃ of a dollar.  In mathematical terms, the three is repeated and the decimal can be expressed as 1/3 = 0.333…

The following table shows the decimal notation of the first twenty integers as denominators with a numerator of one:

Fraction	Decimal	Repeated Part	Period
1/1	1	(terminates)	0
1/2	0.5	(terminates)	0
1/3	0.333…	3	1
1/4	0.25	(terminates)	0
1/5	0.2	(terminates)	0
1/6	0.1666…	6	1
1/7	0.142857…	142857	6
1/8	0.125	(terminates)	0
1/9	0.111…	1	1
1/10	0.1	(terminates)	0
1/11	0.090909…	09	2
1/12	0.08333…	3	1
1/13	0.076923…	076923	6
1/14	0.0714285…	714285	6
1/15	0.0666…	6	1
1/16	0.0625	(terminates)	0
1/17	0.0588235294117647…	0588235294117647	16
1/18	0.0555…	5	1
1/19	0.052631578947368421…	052631578947368421	18
1/20	0.05	(terminates)	0

There does not seem to be any pattern to the length of the repeated part of the decimal, and as we will see later that is because there is a relationship between it and prime numbers, which has no known pattern.  However, there are certain rules that can be applied.

First of all, all denominators comprised of only factors of 2 or 5 terminate.  For example, 2, 4 (2²), 5, 8 (2³), 10 (2∙5), 16 (2⁴), 20 (2²5), and so on all terminate.  This is due to our numbering system being in base 10, which is 2 times 5.

Secondly, if the denominator is a prime number (other than 2 or 5), then the period is a factor of one less than the denominator.  For example, 7 is a prime number, and the period for ¹/₇ is 6, which is a factor of 7 – 1 = 6.  Also, 11 is a prime number, and the period for ¹/₁₁ is 2, which is a factor of 11 – 1 = 10.  The reasoning behind this is somewhat involved.  You may recall from your high school math class that the sum of an infinite geometric series is ^t1/_{1 – r} (when |r| < 1), where t₁ is the first term and r is the ratio.  This means that any repeating decimal (such as 0.090909…) can be expressed as a fraction by first rewriting it as an infinite geometric series (0.090909… = 0.09 + 0.0009 + 0.000009 + …) and then using the infinite geometric sum formula (^t1/_{1 – r} = ^0.09/_{1 – 0.01} = ^0.09/_0.99 = ⁹/₉₉ = ¹/₁₁).  In general, the numerator is the repeated part of the decimal and the denominator must contain a string of nines as long as the period of the decimal (before simplifying).  Since 7 has a period of 6, 7 must divide evenly into a string of 6 nines (999999 / 7 = 142857).  Since 11 has a period of 2, 11 must divide evenly into a string of 2 nines (99 / 11 = 9).  Mathematically, the string of nines n digits long can be represented as 10ⁿ – 1.  Now, Fermat’s Little Theorem states that “if p is prime and a is coprime to p, then p divides a^p−1 – 1.”  So if a = 10 and p is a prime number (other than 2 or 5 so that a and p are coprime) and n = p – 1, then p must divide into 10ⁿ – 1, which matches the string of nines needed in the denominator for a repeating decimal, meaning that all prime numbers must have a period that repeats itself every p – 1 digits.  That is why the prime number 7 has a period of 6.  So why doesn’t the prime number 11 have a period of 10 instead of 2?  Technically, it does have a period of 10, but only if you include five sets of two (¹/₁₁ = 0.0909090909…).  Because of the repetition of numbers in the period of 10 itself, the final period is short-circuited to a factor of 10, in this case 2.

Third, if the denominator d has a period of one less than itself, then d is prime.  Such prime numbers are called long primes.  For example, 7 must be a long prime because it has a period of 7 – 1 = 6, 17 must be a long prime because it has a period of 17 – 1 = 16, and 19 must be a long prime because it has a period of 19 – 1 = 18.  The reason behind this is due to Lehmer’s Theorem, which states that “if there exists an a where p divides into a^{p - 1} – 1, but p does not divide into a^{p – 1/q} – 1 for all primes q dividing p – 1, then p is prime.”  Because of our argument in the second point dealing with infinite geometric sums, since 7 has a period of 6, 7 cannot evenly divide into 9, 99, 999, 9999, or 99999, but finally must evenly divide into 999999.  In other words, if p = 7 and a = 10, p divides into a^{p - 1} – 1, but p does not divide into a^{p – 1/q} – 1 for all primes q dividing p – 1, so by Lehmer’s Theorem p = 7 must be prime.  With the help of a computer, other long primes include 7, 17, 19, 23, 29, 47, 59, 61, 97, 109, 113, 131, 149, 167, 179, 181, 193, 223, 229, 233, 257, 263, 269, 313, 337, 367, 379, 383, 389, 419, 433, 461, 487, 491, 499, 503, 509, 541, 571, 577, 593, 619, 647, 659, 701, 709, 727, 743, 811, 821, 823, 857, 863, 887, 937, 941, 953, 971, 977, 983, and so on. 

It’s pretty amazing that long primes even exist, especially with larger numbers.  If you think about it, using long division on a long prime must result in every single remainder between one and the period before being repeated (otherwise the period would be too short to be a long prime).  For example, using long division on the long prime 7 into 1 looks like the following:

	0.	1	4	2	8	5	7	…
7	1.	0	0	0	0	0	0	…
	–	7	↓
		3	0
	–	2	8	↓
			2	0
		–	1	4	↓
				6	0
			–	5	6	↓
					4	0
				–	3	5	↓
						5	0
					–	4	9
							1

The remainders are 3, 2, 6, 4, 5, and 1, which are all the numbers between 1 and 6 exactly once, before cycling through again.  That means that dividing a larger long prime number such as 983 would result in every single number between 1 and 983 appearing as a remainder exactly once before the cycle is repeated again!

The repeated part of the quotient of long primes, called cyclic numbers, also have fascinating properties.  First of all, multiplying a cyclic number by any integer less than its long prime number results in a number with the same digits in the same order.  For example, the cyclic number of the long prime 7 is 142857.  Multiplying 142857 by any integer less than 7 results in a number with the same digits in the same order:

142857 x 3 = 428571

142857 x 2 = 285714

142857 x 6 = 857142

142857 x 4 = 571428

142857 x 5 = 714285

142857 x 1 = 142857

Note that to make the diagonal number pattern, the multipliers are in the same order as the remainders of dividing ¹/₇ by long division.  Changing the numerator shifts the numbers of the remainders and the digits in the quotient.  So if you divide ³/₇ by long division, your first remainder is 2, then 6, then 4, then 5, then 1, then 3, which is the same order of the remainders in ¹/₇, therefore resulting in the same order of digits in the quotient in ¹/₇.

	0.	4	2	8	5	7	1	…
7	3.	0	0	0	0	0	0	…
–	2.	8	↓
		2	0
	–	1	4	↓
			6	0
		–	5	6	↓
				4	0
			–	3	5	↓
					5	0
				–	4	9	↓
						1	0
						–	7
							3

Since ¹/₇ x 3 = ³/₇, 0.142857… x 3 = 0.428571… which means 142857 x 3 = 428571. 

A second fascinating property is that the first half of the digits of a cyclic number added to the second half of the digits results in a string of nines.  For example, for the cyclic number 142857, the first half of this number is 142 and the second half is 857.  Adding these two numbers results in:

142

+ 857

999

For another example, the cyclic number of the long prime 17 is 0588235294117647 and

05882352

+ 94117647

99999999

The reason for this also quite involved.  From our rules of repeating decimals, ¹/₇ = ¹⁴²⁸⁵⁷/₉₉₉₉₉₉, and since 999999 = 10⁶ – 1 = (10³ – 1)(10³ + 1) = (999)(1001), ¹/₇ = ¹⁴²⁸⁵⁷/₉₉₉₉₉₉ can be rearranged to ¹⁰⁰¹/₇ = ¹⁴²⁸⁵⁷/₉₉₉.  As mentioned above, since 7 has a period of 6, 7 cannot evenly divide into 9, 99, 999, 9999, or 99999, but finally must evenly divide into 999999.  Applying the difference of squares, 999999 = 10⁶ – 1 = (10³ – 1)(10³ + 1), so 7 must evenly divide into (10³ – 1)(10³ + 1).  However, we already said that 7 cannot evenly divide into 999 or 10³ – 1, so 7 must divide into 10³ + 1, which means that 1001 divided 7 must be an integer.  Therefore, since ¹⁰⁰¹/₇ = ¹⁴²⁸⁵⁷/₉₉₉, ¹⁴²⁸⁵⁷/₉₉₉ must be an integer.  Now, ¹⁴²⁸⁵⁷/₉₉₉ = ¹⁴²⁰⁰⁰/₉₉₉ + ⁸⁵⁷/₉₉₉ or 142.142142142… + 0.857857857…  Since this sum must be an integer, each digit after the decimal must be 9, because 0.999… = 1, meaning the first half of the digits of this cyclic number added to the second half of the digits of this cyclic number must result in a string of nines.  (A quick proof that 0.999… = 1 is that 0.999… = (0.333…)3 = (¹/₃)3 = 1.)   A similar argument can be made for any cyclic number, since the period will always be even, which means the difference of squares can always be applied.

Therefore, using long division to turn fractions into decimals leads to finding fascinating properties of terminating and repeating decimals.  The period of repeating decimals do not seem to follow any sort of pattern, but certain rules can be applied, especially concerning prime numbers.  Investigating a little deeper leads to the discovery of long primes and cyclic numbers, which also have some amazing properties.