The Great Mental Calculators
By Steven Smith. 1983Chapret Thirteen
Roots Integer Roots
EXTRACTING roots1 of perfect powers (traditionally termed evolution) is one of the most common calculations undertaken by prodigies (Aitken 1954:297):
Or againthis was asked of another calculator"What is the cube root of 327,082,769?" [Answer: 689, since by inspection it must be a little less than 700, for 7 cubed is 343; also it must end in 9, and cannot be 699.) Such cube-rooting (and there is a good deal of it in the records) is almost trivial, since the important digits yield by inspection the first digit, and mere inspection of the last digit of the proposed number gives the last digit of the answer (for example 7 would give 3, while 3 would give 7); the rest is commonsense [sic], having no reference at all to the other digits. Calculators have always preferred extracting cube roots to square roots, the reason being that cube roots are more impressive because a higher power is involved, and cube roots are easier to do, since the last digit of the power unambiguously determines the last digit of the root (not true of squares), according to the following table:
Root
0
1
2
3
4
5
6
7
8
9Cube
0
1
8
7
4
5
6
3
2
9For instance, the cube root of 551,368 must end in 2 (assuming, correctly, that it is a perfect cube). The table is quite easy to rememberfor zero, one, four, five, six, and nine the final digit of the cube and the root are identical; in the other cases the final digit of the root and the final digit of the cube add up to ten. The first digit of the root is almost equally easy to determine. It requires that one memorize the cubes of the integers from one to nine (1, 8, 27, 64, 125, 216, 343, 512, 729, respectively). Compare these to the digits to the left of the leftmost comma in the cube;2 the cube root of the number next below that of the power is the first digit of the root. For example, 512 is the cube next below 551; therefore, 8 (the cube root of 512) is the first digit of the cube root of 551,368. Since the cube roots of numbers less than a million contain at most two digits, the cube root of 551,368 must be 82. As can be seen, no great sophistication or computing skill is required to find cube roots of perfect cubes less than a million. Three-digit roots will cover perfect cubes up to a billion, and the first and last digits of the root can be easily found by the above methods. An extension of the method for finding the units digit can be used to fix the last two digits of roots of perfect powers.3 The method is described by Colburn (1833:180-82), Bidder (1856:266-67), and Mitchell (1907:93-94), all of whom discovered it independently, and it has probably been used by all calculators who were adept at evolution. This method of "two-figure endings" (Mitchell's term) consists in memorizing all the terminating pairs of digits which when cubed (or squared, etc.) will produce a power ending in a particular pair of digits. The last two digits (tens and units) of perfect cubes determine, with some ambiguity, the last two digits of the cube root, in accordance with table 13.1.
TABLE 13.1
Colburn's Table For Cube RootsCube
01
03
04
07
08
09
11
12
13
16
17
19
21
23Root
01
87
34
84
43
02
52
69
71
08
58
17
06
56
73
39
41
47Cube
24
25
27
28
29
31
32
33
36
37
39Root
24
74
05
25
45
65
85
03
12
09
11
18
68
77
46
96
33
79Cube
41
43
44
47
48
49
51
52
53
56
57
59
61
63Root
81
07
14
64
63
22
72
49
51
28
78
37
36
86
93
19
21
67Cube
64
67
68
69
71
72
73
75
76
77Root
04
54
23
32
82
89
91
38
88
97
15
35
55
75
95
26
76
53Cube
79
81
83
84
87
88
89
91
92
93
96
97
99Root
59
61
27
44
94
83
42
92
29
31
48
98
57
16
66
13
99Source: Colburn 1833:182. Colburn gives the following example of the use of his table: Take for instance, 28,094,464. First, seek what number cubed, will end in 64. Ans. 04: then what number cubed, comes nearest under 28. Ans. 3. Combine them, 304, the root required . . . there is not the same latitude for mistaking in the root of a cube number, that there is in extracting the square root. Some difficulty, however, exists; tor instance, it the given cube ends in 08, it will be a question whether 02 or 52 will be correct; and so of all similar cases. As mentioned at the outset of this chapter, square roots have found less favor among calculators, because of the considerably greater uncertainty regarding the correct terminating digits. Nevertheless, calculators of the past, such as Bidder and Colburn, were frequently called upon to say "What number multiplied by itself will give ?" The method used again was that of two-figure endings, in accord with Colburn's table given here as 13.2.
TABLE 13.2
Colburn's Table For Square RootsSquare
01
04
09
16
21
Root
01
51
49
99
02
52
48
98
03
53
47
97
04
54
46
96
11
61
39
89Square
24
25
29
36
Root
18
68
32
82
05
15
25
35
45
55
65
75
85
95
27
77
23
73
06
56
44
94Square
41
44
49
56
61
Root
21
71
29
79
12
62
38
88
07
57
43
93
16
66
34
84
19
69
31
81Square
64
69
76
81
84
Root
08
58
42
92
13
63
37
87
24
74
26
76
09
59
41
91
22
72
28
78Square
89
96
Root
17
67
33
83
14
64
36
86Source: Colburn 1833:181.
As can be seen there is much more ambiguity in finding the final two digits of a square root than there is in finding the final two digits of a cube root. Bidder (1856:266-67) describes how he proceeded in difficult cases: In reference to square numbers terminating in 25; although all numbers ending in 5, when squared, give 25 as terminals, I noticed that the squares of numbers ending in 5, 45, 55, and 95, ended in 025, and that those of the numbers terminating in 15, 35, 65, and 85, ended in 225; whereas the squares of those ending in 25 and 75, ended in 625. Hence in extracting the square root of 442,225, I perceived, as before, that 600 must be the first factor, and that the last one must lie between 15, 35, 65, and 85, and judging from the position of 442,225 between 360,000 and 490,000, the squares of 600 and 700 respectively, I saw that 65 was the factor required, and root was therefore 665. Colburn (1833:181) also discusses the difficulties attendant to finding square roots: It is thonght that there is little difficulty, according to the rule presented, in determining the two last figures of the root; it is obvious, however, that it requires a good share of quickness and discernment, in a large sum, to see which of the four roots in ordinary numbers, or which of the ten, where 25 is the termination, is the right one to be employed. Such discernment, however, the writer cannot impart. A traditional method of testing the answers of arithmetic problems, known as the rule of eleven, or the elevens test, can be adapted to find one digit of a root, when the other digits are known.4 Any number can be represented as a multiple of some constant plus a remainder. In this case eleven is taken as the constant; thus a = 11b + R, where R is the remainder upon dividing a by 11 (or, for those familiar with modular arithmetic, a is congruent to R, modulo 11). One can check any arithmetic calculation by carrying out the same calculation on the remainders in place of the numbers in the original calculation. The correct answer to the original problem will have the same remainder as the remainder of the answer obtained in this way; for example, 897 × 764 = 685,308,897 = (81 × 11) + 6; and 764 = (69 × 11) + 5. Multiplying the remainders (6 × 5) gives 30, or (2 × 11) + 8; and 685,308 = (62,300 × 11) + 8. The remainder is eight in both cases. The test checks. This does not mean that the answer is necessarily correct, but if the test does not hold, the answer is certainly wrong. With a little practice, dividing even fairly large numbers by 11 is not so difficult as it might first appear. However, there is another method for finding the elevens remainder recommended by books on practical arithmetic. Subtract the sum of the digits in the even-place positions (counting from the right) from the sum ot those in the odd-place positions. (That is, subtract the sum of the tens, thousands, hundred thousands, etc. digits from the units, hundreds, ten thousands, etc. digits.) If the answer is negative, add 11 to get the remainder; if the result exceeds 11, subtract 11 (one or more times). Take, for example, the cube 640,503,928. Add together the odd place digits: 8 + 9 + 0 + 0 + 6 = 23. Subtract from this sum of the even place digits: 2 + 3 + 5 + 4 = 14 and 23 – 14 = 9. Thus 9 is the remainder upon dividing 640,503,928 by 11. Having established the first and last digits of the cube in the manner described above, the middle digit (or any digit when the others are known) can be deduced by a reversal of the elevens test. The cube remainder unambiguously determines the remainder of the root, according to the table below.
Cube Remainder
0
1
2
3
4
5
6
7
8
9
10Root Remainder
0
1
7
9
5
3
8
6
2
4
10(This table can be easily constructed by cubing the numbers from 0 through 10 and taking the elevens-remainder of the cubes.) After finding the cube remainder, it is only necessary to find the middle digit which will give the appropriate root remainder. In the case of 640,503,928, the first digit of the root must he 8, since 9 cubed is 729 and 8 cubed 512. The last digit will be 2. We now have 8_2. The cube remainder is 9, so the root remainder must be 4. Since 8 + 2 = 10, the middle digit of the root must be 6 (10 – 6 = 4). Therefore, the cube root of 640,503,928 is 862. Fifth roots have also been a traditional favorite among calculators, since the units digit of the power, as in the case of cube roots, unambiguously determines the units digit of the root. Schumacher wrote to Gauss concerning Dase's delight in fifth roots (Gauss and Schumacher 1861 1:382): "He was especially fond of extracting fifth roots in his head, since he had noticed that at the fifth power the units were the same as they were in the root. I saw that in our number system the 4n + 1 power has the same units as the root, a principle of which his result is but a particular case (for n = 1)." This means that the units digit of the root and power are the same for 9th, 13rh, 17th, etc. roots. There are also roots for which the units digit behaves like cube roots: these are any roots of the 4n + 3 power, where n is a natural number; thus 3d, 7th, 11th powers, etc. These two types cover all the odd powers. In recent years calculators have taken to extracting roots of very high powers, and their successes in this area have been reported in the Guinness Book of World Records. The eleventh edition (1972:43) reported that on October 5, 1970, Herbert B. de Grote of Mexico City had mentally extracted the 13th root of a 100-digit number in 23 minutes. Since then the extraction of roots of high powers has become increasingly popular, no doubt owing to the availability of computers, both for providing data from which calculators can create algorithms and for providing problems for performances. By the sixteenth edition (1977), the record was claimed by Wim Klein of the Netherlands, who extracted the 23d root of a 200-digit number in 10.5 minutes. The 1978 edition reported that Klein had extracted the 73d root of a 500-digit number in 2 minutes, 43 seconds. The difficulty of extracting a root of a perfect power has little to do with the size of the power involved. Much more important are the number of digits in the root and the particular power selected (note that in none of the cases above were even powers employed). Mitchel (1907:94) observed: "in general, the higher the root the easier the problem, and square and cube roots are the only ones which often come up." In this regard, times have changed. The Guinness Book of World Records now recognizes the 13th root of a 100-digit number as a standard for testing the ability to extract integer roots. Wim Klein is the current record holder with a time of less than two minutes. Klein comments: "The problem is harder the more digits there are in the result. The 23rd root of 200 digits has nine digits in the result, and the same also for the seventh root of 63 digits. The 13th root of 100 digits has eight digits, but the 19th root of 133 digits and the 73rd root of 500 digits have only seven digits. I don't find seven digits interesting any longer. It is just a game for me. "With eight digits, one digit is not sure, so you have to put in special tests. With nine digits there are two unknown digits in the middle. That makes it much tougher. The cube root of 30 digits gives ten digitsit's a hell of a work." Klein's methods for extracting 13th roots can be illustrated with the following number: 14762420839370760705665953772022217870318956930659
27236796230563061507768203333609354957218480390144
The first five digits of the root are fixed through the use of logarithms, Klein has memorized to five places the logs of the integers up to 150; this, coupled with his ability to factor large numbers, allows him to approximate the log of the first five digits of the power, which is usually sufficient to determine the first five digits of the root, though, as he says "the fifth digit is a bit chancy." Klein began by factoring 1,476 into 36 times 41 and taking the (decimal) log of each: log 36 = 1.55630 and log 41 = 1.61278; adding the mantissas yields 0.16908, but this is, of course, too little. Through various interpolations Klein estimated the mantissa of the log of 147,624 as 0.16925 (it is more nearly 0.16916). Klein now had an approximation of the log of the 100-digit number above99.16925. This must be divided by 13 to obtain the log of the 13th root. Since 99=13 × 7 with a remainder of 8, to obtain the mantissa of the antilog of the 13th root he divided 8.16925 by 13, which is approximately 0.62840. He estimated the antilog to be about halfway between 4.2 and 4.3 and decided to try 4.25. The result was exact, so the first five digits of the root should be 42500, as indeed they are. It is now necessary to determine the last three digits of the root. This he does from an examination of the last three digits of the power. In the case of odd powers, these uniquely determine the last three digits of the root, but in the case of even roots, like this one, this method yields four possibilities; in the case of 144 they are 014, 264, 514, and 764. (The choices always differ by 250.) To select the correct one Klein divides the original number by 13 and retains the remainder. In the case of 13th roots, the root remainder and the power remainder must be the same. The power remainder is 7; only 764 as the final three digits of the root will yield 7 as the remainder. Thus the 13th root is determined to be 42,500,764. As an example of an odd root take: 75185285487713563581947553291145079861723813162341
53935861550997297991815299022662358976308065985831
The first five digits of the power are 75185, which is nearly 7519, and 7519 is 73 times 103. The mantissa of the log of 73 is 0.86332 and that of 103 is 0.01284. Their sum is 0.87616. Dividing 8.87616 by 13 yields 0.68278. This falls between the mantissas of the logs of 48 and 49, but is much closer to 48. Since 481 is 13 (mantissa 0.11394) times 37 (mantissa 0.56820), the mantissa of its log will be 0.68214; dose, but still a bit low; 4,816 can be factored into 16 (mantissa 0.20412) times 7 (mantissa 0.84510) times 43 (mantissa 0.63347). This gives a mantissa of 0.68269. Then 4,818 factors into 66 (mantissa 0.81954) times 73 (mantissa 0.86332), which yields a mantissa of 0.68286. Thus, in the interpolation we want 9/17 of 20 which is about 10 1/2. The first five digits of the root should be 48170 (48160 + 10). This, in fact, is correct. When Klein actually did the calculation he made a minor error (he was looking for the antilog of 0.68277 instead of 0.68278) and first took 48169 for the first five digits of the root. In this case, however, since the root is odd, the last three digits are uniquely fixed-since the power ends in 831, the root must end in 311. Upon dividing the power by 13 Klein got a remainder of 7. But dividing 48,169,311 by 13 gives a remainder of 8. To make these two remainders come into line he changed his solution to 48,170,311, which is correct. Noninteger Roots
If a root of an integer is not itself an integer, it will be irrational, so that only an approximation to some number of decimal places is possible. For example, the cube root of 9 is not an integer; it is approximately 2.0800838, but the calculation can be carried out forever without reaching a point at which the digits repeat. Calculators, in the past, were rarely called upon to extract roots which were not integers. One reason is that most people do not know how to extract roots. (Today pocket calculators have made such questions more popular.) Bidder remarked (1856:266): "the numbers submitted to me were almost invariably perfect squares, or cubes, arising from the circumstance that, in order to save themselves trouble, those who questioned me, squared, or cubed a number, as the readiest mode of testing the accuracy of my reply, which being found correct, they were satisfied and so was I." Extracting noninteger roots is naturally far more difficult than extracting roots of perfect powders. Aitken said (1954:297): "The real test of ability to do square, cube or any other root is, in my view, to have a number proposed that is not an exact power, and to be asked to give the answer to several decimals; but this type of question you will hardly find in the published records." Bidder (1856:267) indicates that he was occasionally confronted with such problems, but gives no specifics about how he went about solving them: "if I suspected that the number was not a perfect square, or cube, I tested it by 'casting out the nines' a process familiar to arithmeticians; and in such cases, the results were approximated to by a tentative process. . . ." Aitken begins his discussion of extracting noninteger square roots with the remark (1954:298): "The central idea here is Newton's; but, remarkable to state, the Babylonians have it in their cuneiform inscriptions, and so did the wonderful Archimedes, the Syracusan Greek." Aitken took as an example the square root of 51. The answer is, of course, somewhat more than 7, so 7 can be taken as a first approximation. On the other hand, it is somewhat less than 51/7, so that taking the mean of the quotients (49 and 51) yields a much better approximation: 50/7=5.1428571. ... To get even closer divide 51 by 50/7, which gives 7.14, so that the mean of the divisor and the quotient is 7.1414285 . . . and the correct answer to eight decimal places is 7.1414284. Algebraically the method is as follows: let x be the number whose square root is sought. Choose for the first approximation the integer (call this y) nearest the square root of x. Then x/y is another approximation. Take the mean of the two:
x + y²
2y= z. Now z becomes the new approximation and is substituted for y in the formula to get a still better approximation. The better the value initially selected for y, the better the value for z will be, thus reducing the calculations required to obtain a satisfactory approximation. Aitken (1954:298-99) continues: But the resources of mental division are not exhausted even here. An expert would know very well that 7.14141414... is 707/99, and dividing 51 by this we have 5049/707, easily accomplished by dividing by 101 first, yielding 49.990099009900... and then by 7, so that we have 7.141442715700... , and the mean of this and 7.14141414141414... is 7.14142842857 to 12 digits, whereas the true value of the square root of 51 is 7.14142842854 to that degree of accuracy. In other words, in this case the initial value for y was taken as 707/99, and since this is a better approximation of the square root of 51, we get a very good approximation in z. Aitken pointed to other methods which might be employed by a skilled mental calculator (1954:299): To revert to 51. How near is 50/7 to the square root? It was got from a divisor 7 and quotient 51/7. These are in ratio 49 : 51. The halfway mark between them is 50; I will say therefore that both 7 and 51/7 "deviate" by 1 in 50. The square of this is 1 in 2500. I double and say, 1 in 5000. Now 50/7 reduced by 1 in 5000 is remarkably near the true value 7.1414284..., and in fact identical with the second approximation given above. It is clear by this time that we have several methods to choose from. There is yet another. Looking again at the first pair, 7 and 51/7, we note once more their ratio 49 : 51. Quarter the distance between 49 and 51, and take the first and third quarter, namely 49½, 50½, their ratio being 99 : 101. I say then that
4999/700 = 7.1414285.... is a good approximation to the square root of 51. Alternatively that
7 by 102/99 = 7.14141414... is equally good; but we have met both of these before, and have seen that their average is spectacularly good. There are even subtler and more powerful approximations still. There is one known to me, quite simple, which I may illustrate by saying that in our example here (rather hard worn by now) we could correct 50/7 by reducing it not by 1 in 5000, but by 1 in 4999½. The result is
51/7 by 99/101 = 7.141442715700... and so committing an error of 1 in 500 000 000 000. This is an extreme approximation for square root; and I have never gone beyond it in mental calculation.
7.141428428557..., as against 7.141428428543 Klein (Aarts 1974:10) gives an example of essentially the same method as the first one discussed above for extracting a noninteger square root, but applied in a somewhat different manner. He considers the square root of 38, which is somewhat more than 6. He first divides 38 by 6 and obtains 6.3333. . . . Midway between 6.3333 and 6 is 6.1666. This number squared is 38.02695556. He then takes the excess and divides it by twice 6.1666; that is, 0.02695556/(2 × 6.1666), and takes the answer as 0.0022 (it is 0.0021856. . .)[I suspect Klein does some rounding off before making the calculation. For example, 0.027/(2 × 6.17) is 0.002188 . . . , which is sufficiently accurate for his purposes.] He then subtracts 0.0022 from 6.1666 to obtain 6.1644. This squared is 37.99982736, or 0.00017264 too little. Dividing by 2 × 6.1644 yields 0.000014, which added to 6.1644 gives 6.164414. This last squared is 37.99999996386, a very close approximation. Klein's iterative formula is where, as before, x is the number whose square root is sought, y is initially the nearest integer to the square root (or some better approximation), and z is the approximation to the square root. In subsequent iterations z is substituted for y.
x + y²
2y+ y = z. Noninteger cube roots can also be approximated. Klein makes use of his knowledge of logarithms in extracting noninteger roots. The following example is also from Teachware (Aarts 1974). What is the cube root of 3,721? Klein knows the logs to five places of the integers up to 150, but since 3,721 considerably exceeds that, he must do some calculations to obtain a good value for the log. In this case it is quite simple, as 3,721 is 61²; the log of 61 is 1.78533, and doubling this gives the log required 3.57066. The log of the cube root will be one-third of 3.57066, or 1.19022. The cube root lies between 15 and 16, so Klein tries 15.5 (since 15.5 = 5 × 3.1, and he knows the logs of 5 and 31, it is not difficult for him to calculate). This gives 1.19033, which is too much. He tries 15.48 (4.3 × 3.6), obtaining 1.18977; then 15.488, the log of which is 1.18999. Now 1.19022 is about two-thirds of the way between 1.18999 and 1.19033. The interval between 15.488 and 15.500 is 0.012, two-thirds of which is 0.008; by interpolation he gets 15.496 as the cube root of 3,721. Klein then checks by multiplication: 15.496² is a bit more than 240.126, and 240.126 × 15.496 = 3,720.992496. Thus 15.496 is a good approximation to the cube root of 3,721. Aitken (1954:299) described his methods for extracting cube roots: I will take for illustration the cube root of 128. You can see that it must be near 5, since the cube of 5 is 125. Trisect the interval from 125 to 128. The "middle third" yields the ratio 126:127. I assert that is very close to the required cube root, which is in fact 5.0396842 eight significant digits. This method of "thirding" is here ever so slightly in defect; it is the business of the algebraist to ascertain the formula for the small error committed. I will not go into these delicate refinements. Here of course a proposed number may prove rather intractable, being remote from any suitable cube of an integer, or of a fraction with small denominator. I may mention also a similar method which I call "sixthing." Here for example we have that 128 is 5 by 5 by 5.12. I divide that former interval, 125 to 128, into six parts, and taking the first and the fifth of these I form the ratio 251:255. We have then
5 by 127/126 = 5.0396825... an approximation almost as good as the other, this time slightly in excess. I am not aware that this approximation has ever received notice.
5.12 by 251/255 = 5.0396863,
1 Throughout this discussion, unless otherwise indicated, roots will be assumed to be integers. 2 That is, group the digits in threes starting from the right and consider the leftmost such group (which may contain less than three digits). 3 It is also used to find factors (see chapter 15). 4 A better-known method, "casting out nines," is of limited usefulness in fixing a missing digit of a cube root, since it gives three possible choices.