Here I gather together arithmetical results that depend on the base of numeration, in particular results that depend on the base being ten, which is the one normally used. The method of positional numeration and use of different number bases is explained in the Rational Mathematics section under numbers. Personally I feel that far too much is made of incidental effects of the base-ten numeration, simply because it is the system familiar to most people. Of more interest are results that are also possible or have analogues in other bases. The notes here have been in my files for a long time, and need to be more systematised.

On Notation: It should be noticed that multiplication of two digits a and b is shown here by a.b (using a dot on the line) while concatenation of the letters like ab or (a)(b) indicates the number 10.a + b.

In all positional number systems in which the base is an even number, divisibility by 2 is shown by an even final digit. This is a particular case of the general rule that in a system with composite base n = u.v (where u and v are between 1 and n and thus digits) a given number is divisible by u if and only if its final digit is divisible by u. Thus in base ten, 10 = 2.5, numbers ending in 5 are divisible by 5, while in base six, 10 = 2.3, numbers ending in 3 are divisible by 3.

The number produced by adding up the digits of a number, then adding the digits of the sum if necessary,
and so on, until a single digit result is found is sometimes called the *digital root* of the number. It is in
fact the same as the **remainder** when the number is divided by m = n−1 where n is the base, except that where the
remainder is 0 the digital root is taken to be m. If the digital root is m then the number is divisible by m.
This is obvious since for example uvwxy = 10000.u + 1000.v + 100.w + 10.x + y
= u + v + w + x + y + mmmm.u + mmm.v + mm.w + m.x, which is divisible by m if and only if u + v + w + x + y is divisible by m.

Divisibilty tests in Base Ten.

1 is a divisor of any number: n = 1.n = n.1, or in other words n/1 = n.

2 is a divisor of any number whose last digit is 0, 2, 4, 6 or 8.

3 is a divisor if it divides the sum of the digits. This is because 10−1 = 9 = 3.3,
thus abcd = a + b + c + d + 3.(333.a + 33.b + 3.c).

4 is a divisor if the last two digits form a multiple of 4. This is because 100 = 4.25 is divisible by 4.
As a check, the number must be even, and after divison by 2 it is still even.

5 is a divisor provided the number ends in 0 or 5.

6 is a divisor of any number divisible by both 2 and 3.

7 is a divisor of 1001 = 7.11.13. So divisibility by 7 can be tested by splitting the number into groups of three digits
starting from the unit digit. These groups, taken as 3-digit numbers, are added and subtracted alternately until a number of
3 digits or fewer remains. If this final number is divisible by 7 then so is the original number. For example: abcdefghijklmno
= abc – def + ghi –jkl + mno + (10^12 − 1).abc + (10^9 + 1).def + (10^6 − 1).ghi + (10^3 + 1).jkl,
and all the coefficients are divisible by 7.

8 is a divisor if the number formed by the last three digits is divisibile by 8. This can be done in three steps of divison by 2.

9 is a divisor if it divides the sum of the digits.

10 is a divisor if the number ends in 0 (this is a truism in any base).

11 is a divisor if it divides the difference of the sum of digits in the odd and even positions.
This works since abcdef = (b+d+f) &minus (a+c+e) + 100001.a + 9999.b + 1001.c + 99.d + 11.e,
and the coefficients 100001, 9999, etc are divisible by 11, e.g. 10000...001 = 11 + 9999...990,
where there is an even number of 0s and 9s on the left. The test for 7 also works for 11.

12 is a divisor if 3 and 4 are divisors.

13 is tested by the same method as for 7.

**The 1089 trick.** In this well-known trick you take any number of three digits, reverse and subtract;
then reverse the result and add. There must be three digits at each stage, inserting a leading zero if necessary.
In base n with m = n−1 and k = n−2 the result will always be 10km. *Proof:* Stage 1 with
a > c gives: abc − cba = (a−c)0(c−a) = (a−c−1)m(n+c−a). Stage 2 gives
(a−c−1)m(n+c−a) + (n+c−a)m(a−c−1) = m(m+m)m = nkm = 10km. In base ten the
result is always 1089. The same trick can be done in more general systems of units. For example when the trick
is done in the predecimal monetary system of pounds shillings and pence (where 12.d = 1.s, 20.s = 1.£) the result
is always 12.£ + 18.s + 11.d (customarily written £12 18s 11d).

A related result is: with a > b: ab − ba = (n.a + b) − (n.b + a) = m.a − m.b = m.(a−b). In base ten m = 9 and we have, for example 92 − 29 = 63 = 9.(9−2) = 9.7.

There are many variations on tricks of this type. The effect is that the conjuror quickly finds the hidden numbers from apparently inadequate and garbled information.

**Think of a number**. An audience-member is asked to think of a number, then to perform a series of
specified operations on it before revealing the result. The operations are designed to produce a number of the
form (100.n + abc). The conjurer has then only to subtract abc and ignore the last two zeros to reveal the
original number. For example: multiply by 5, add 6, multiply by 4, add 9, multiply by 5 again giving (100.n + 165).

**Think of three digits**. The customer is asked to select three numbers less than ten, say x, y, z.
Then to multiply the first by 2, add 3, multiply by 5, add 7. Now add the second number, multiply by 2, and add 3,
multiply by 5, add the third number, and disclose the result. This will be 100.x + 10.y + z + 235.
To decode, the conjurer subtracts 235, giving the 3-digit number xyz.

**The deleted digit**. A person chooses a number of five or more digits (say 12345). He adds the digits (= 15)
and deducts the sum from the original number (= 12330). He now deletes one of the digits other than 0 (say 3),
and reveals the remaining digits in any order, deliberately permuting them. The conjuror can now name the deleted digit.
He adds the digits of the result, recursively if necessary, and deducts the resulting digit from 9 (in this case
9−6 = 3). The reason for this is that if we deduct from a number the sum of its digits the result is always
a multiple of 9 (or of n−1 in base n).

**The window reader**. The method of determining a number by the 'window reader' method is well known.
It is based on the binary number system. The conjuror has a set of six cards bearing assorted numbers, up to 63.
The customer indicates on which of the six cards his number appears. This tells the conjuror which powers of 2
appear in the binary numeral, and he adds the appropriate powers of 2. The card bearing 1 has all the odd numbers,
the card bearing 2 has all the pairs of numbers 2, 3, 6, 7, 10, 11, 14, 15 and so on, the card bearing 32 has all
the numbers from 32 to 63. Each card contains 32 numbers. The method can be mechanised so that a card with six holes,
when placed over the chosen set, reveals the powers of two that are to be added.

Repdigit numbers are those consisting of one digit repeated any number of times. The repdigit numbers 55 and 66 are triangular numbers. The only other repdigit triangular number is 666. The number 66 is also hexagonal (in the old sense).

Repunit Numbers are those formed of repeated 1s in any base. In base n the repunit 11 = n+1, and the repunit of length k represents the number 1 + n + n² + n³ + ... + n^(k−1) = (n^k − 1)/(n−1). In bases 2, 4, 6, 10, 12, 16, 18, 22, ... (one less than a prime) the repunit 11 happens to be a prime. Any repunit with an even number of digits > 2 is divisible by 11; in fact 1111 = 11.101, 111111 = 11.10101 and so on in any base. The odd-length repunits may be prime.

In base 2 the repunits form the sequence 1, 3, 7, 15 = 3.5, 31, 63 = 3.21, 127, 255 = 3.85, 511 = 7.73, 1023 = 3.341, 2047 = 23.89, 4095 = 3.1365, 8191, ... with formula 2^k − 1. It can be shown that if 2^k − 1 is prime then k must also be prime. If k = u.v then 2^(u.v) − 1 = (2^u − 1).???? Such numbers are called Mersenne numbers. The known cases (as at 1988) were: n = 2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091.

In base 3 the odd-length repunit numbers are 1, 13, 121 = 11², 1093, 9841 = 13.757, ... In base 4 they are 1, 21 = 3.7, 341 = 11.31, ... and so on In base ten they are: 1, 111=3.37, 11111 = 41.271, 1111111 = 239.4649, ... The next prime occurs when there are 19 units (proved in 1918). The information I have is that primes of this sort occur when n = 2, 19, 23, 317, 1031.

The repunit numbers in any base are associated with the cycles of recurring positional fractions. According to Reichmann, in base ten, any number 999...9 with k 9s has a factor q.k + 1 and this will also be a factor of 111...1, e.g. 11111 has 41 = 8.5 + 1 as factor. He also states that a number with k ones is exactly divisible by k+1 if k+1 is a prime except where k+1 equals 2, 3 or 5. e.g. 10101 = 7.1443.

Squares of 99...9: (8+1)² = 81, (98+1)² = 9801, (998+1)² = 998001, (9998+1)² = 99980001. Cubes of 99...9: 9³ = 729, 99³ = 970299, 999³ = 997002999.

In any base, the number of digits required to paginate a book of N pages is k.N − (111...1 − k) where the base n repunit number 111...1 has k digits.

Near-Repunits. Numbers of the form 2^p − 2 are divisible by p if p is a prime. The smallest composite p that divides 2^p − 2 is 341 = 11.31 (found by Sarrus). [Reichmann]. These are numbers of the form 111...1110.

The following are (base ten) prime numbers: 11111117, 11111119, 111111113, 11111111113, and 61111111, 71111111, 31111111111. [Lines]

These are numbers that have many of their digits in numerical order.

The squares of the repunit numbers up to 9 digits are palindromic sequential: 11² = 121, 111² = 12321, 1111² = 1234321, ..., 111111111² = 12345678987654321. (The numbers formed from these palindromes by omitting the initial and final 1s are all products of the form 4.(5 ... 8): thus 232 = 4.58, 23432 = 4.5858, 2345432 = 4.586358, ... 234567898765432 = 4.58641974691358. [Reichmann])

A similar progression is: 1.8 + 1 = 9, 12.8 + 2 = 98, 123.8 + 3 = 987, 1234.8 + 4 = 9876, ..., 123456789.8 + 9 = 987654321.

Primes: 1,234,567,891, 1,234,567,891,234,567,891,234,567,891. The longest prime of this type yet discovered cycles seven times before finishing with 1,234,567. [Lines]

Sequential Divisibility. Is there a longest number in which the first n digits form a number that is exactly divisible by n, for each value of n? There is, a unique solution: 3,608,528,850,368,400,786,036,725, with 25 digits. There are 3 such numbers of length 24, 6 of length 23, 12 of length 22, 18 of length 21, 44 of length 20, ..., 2492 of length 10, ..., 150 with 3 digits, 45 with 2 digits, 10 with 1 digit. The 10-digit solutions include 3,000,060,000. [Lines]

Product of Digits. The *persistence* of a number is the number of steps necessary to reduce it to one digit by repeated
multiplication of digits. The smallest numbers of persistences 1, 2, 3, ... are: 10 (0), 25(10, 0), 39 (27, 14, 4), 77(49, 36, 18, 8),
679 (378, 168, 48, 32, 6), and the sequence continues: 6788; 68,889; 2,677,889; 26,888,999; 3,778,888,999; 277,777,788,888,999.
If there is a number with persistence greater than 11 then it is greater than 10^50. [Lines]

Reversible Primes. A number of type X which remains of type X when its digits are read in the reverse direction is naturally called a
reversible X-number. Reversible primes have been called *emirps*. There are 4 pairs of 2-digit reversible primes in base ten:
13, 31; 17, 71; 37, 73; 79, 97; there are 13 pairs of 3-digits, 102 pairs of 4-digits, 684 pairs of 5-digits. [Lines]

312.221 = 68952, 213.122 = 25986, i.e. (rA).(rB) = r(A.B) where r denotes reversal.

ab² = cde and ba² = edc: 12²=144, 21²=441 and 13²=169, 31²=961

abcd×k = dcba: 2178×4 = 8712, 1089×9 = 9801, 4356×3/2 = 6534, all these are multiples of 1089.

The following result was sent to me by Tom Marlow on a Christmas Card, but it is undated and I cannot recall which year. 641297 and its reversal 792146 both divide exactly by prime 151. It is the only 6-digit non-palindrome to do this.

Palindromic Numbers are those that read the same backwards as forwards. The repdigit numbers are a particularly simple case of palindromic number, and as noted above their squares tend to be palindromic. Here are some palindromic powers of non-palindromic numbers: 26² = 676, 836² = 698896, 2201³ = 10662526601 (2201 is the only non-palindromic number known that has a palindromic cube, checked up to 2.(10^14).

Palindromic powers of palindromes: 111³ = 1367631, 1001^4 = 1004006004001. It has been proposed that there are no palindromes of fifth or higher powers.

A special case: 12345679.99999999 =1234567887654321 (note no 8 in the first number, and there are 8 figures 9 in the repdigit).

Palindromic primes: 151, 727, ...

There is a conjecture that if we take any number, reverse it and add to the original, and repeat the process, we always eventually reach a palindrome. For example 18, 27, 36, 45, 54, 63, 72, 81, 90 all produce 99 in a single step; 68 gives 1111 in 3 steps. The number less than 100 which requires the most steps is 89 which after 24 summations yields 8813200023188. For integers between 100 and 10000 some 249 lead eventually to five separate sequences which have not yet generated a palindrome in the first 10000 addition steps, though none of the other numbers take more than 24 steps. Some cases that yield near palindromes are: 7059 which in six steps gives 4692864; and 879 in 8 steps gives 8884788; 1997 in 15 steps gives 93523232638; 196 in 16 steps gives 897100798; 9999 in 46 steps gives 3863706276576756727063683 (but this is a palindrome in 0 steps). For numbers beyond 10000 some palindromes have been found in more than 24 steps, the largest and most delayed known being: 10911 which in 55 steps gives 4668731596684224866951378664. But 5842 five-digit numbers have no palindrome within 10000 additions, and produce 69 distinct sequences. An example giving near-palindromes is: 80289 which in 22 steps gives 2799886655889972 and in 23 steps 5599772222779944. The palindromic conjecture is known to be false in base 2. [Lines]

Cyclic Numbers. These remain in the same type X when cyclically permuted. Cyclic Reversible Numbers. These combine both the previous properties. The five digit number 11939 is a cyclic reversible prime (cyclic emirp), since 19391, 93911, 39119, 91193 are all emirps (one is even palindromic). The next cyclic emirp is 193939 giving 939391, 393919, 939193, 391939, 919393. No others are known less than 10 million. [Lines]

Cyclic permutations: 142857.2 = 285714, .3 = 428571, .4 = 571428, .5 = 714285, .6 = 857142 and 230769.3 = 692307, 307692.3 = 923076, 153846.3 = 461538, .4 = 615384, These relate to recurring decimals.

Multiplication by moving first digit to last: 142857.5 = 714285, 285714.3 = 857142, (also, trivially 0a...z.10 = a...z0)

Multiplication by moving last digit to first: 102564.4 = 410256, ... 'it is comparatively easy' to find such cases.

Permutable Primes. A permutable prime is one that remains prime if its digits are permuted in any other order. In any number system the digit 0 cannot occur in a permutable prime since it can be permuted to the units column, resulting in divisibility by the base, and the number 10, representing the base itself, when permuted gives 01 = 1 which does not count as a prime. Thus in base 2 permutable primes are 'repunit' numbers, represented by repeated unit digits, i.e. the Mersenne primes (see above).

In base ten the single-digit primes 2, 3, 5, 7 are trivially permutable. The digits 0, 2, 4, 5, 6, 8 cannot occur in any permutable prime of two or more digits since they can be permuted to the units place to produce a number divisible by 2 or 5. All higher permutable primes are therefore formed from the digits 1, 3, 7, 9. The two-digit permutable primes are 11, 13, 17, 31, 37, 71, 73, 79, 97 and the three-digit cases are 113, 131, 199, 311, 337, 373, 733, 919, 991. There are, perhaps surprisingly, no four-digit permutable primes in base ten, since: (a) 1379 = 7.197, eliminating the form abcd, with all four digits different; (b) abab = 101.ab, eliminating the forms aabb and aaaa; (c) 1711 = 29.59, 1333 = 31.43, 3337 = 47.71, 7771 = 19.409, 9919 = 7.1417 and 9997 = 13.769, while aaa3 and aaa9 are divisible by 3, eliminating thr form aaab; (d) 1139 = 17.67, 1339 = 13.103, 1337 = 7.191, 9373 = 7.1339, 7973 = 7.1139, 1939 = 7.277, 1799 = 7.257, 7399 = 7.1057, eliminating the form aabc. The only other permutable primes known are those consisting of repeated ones, known as 'repunit' primes (see above).

Cumulative Numbers. These remain of given type when digits are successively removed from one end. The prime 73939133 is a right-cumulative prime (also termed a prime prime), i.e. 7, 73, 739, 7393, ... are all primes. The following is the complete list of (non-extendable) right-cumulative primes: 53, 317, 599, 797, 2393, 3793, 3797, 7331, 23333, 23339, 31193, 31379, 37397, 73331, 373393, 593993, 719333, 739397, 739399, 239333, 7393931, 7393933, 23399339, 29399999, 37337999, 59393339, 73939133. [Lines] The largest left-cumulative prime is 357686312646216567629137 (24 digits), i.e. generating primes ending with ... 9137, 137, 37, 7. [Lines]

Permutation of sequential numbers: (a) 123456789.2 = 246913578, .4 = 493827156, .5 = 617283945, .7 = 864197523, .8 = 987654312, (b) 0987654321.2 = 1975308642, .4 = 3950617284, .5 = 4938271605, .7 = 6913580247, .8 = 7901234568 (c) 987654321 – 123456789 = 864197532 (d) 51249876.3 = 153749628, 16583742.9 = 149253678, 32547891.6 = 195287346, 0675412398.2 = 1350824796, 8745231.96 = 839542176 (e) 8.473 = 3784, 15.93 = 1395, 35.41 = 1435, (f) 12.483 = 5796 = 42.138, 18.297 = 5346 = 27.198

The only integral solution to a^2 = 2.a is a=2. The general solution is: x + x/(x–1) = x . x/(x–1).

Narcissistic Numbers are numbers that can be derived from their own digits in various ways. Examples are Sums of squares: 153 = 1³ + 5³ + 3³, 370 = 3^2 + 7^2 + 0^2, 371 = 3³ + 7³ + 1³, 407 = 4³ + 0³ + 7³, 165033 = 16³ + 50³ + 33³. [Lines] Squares of sum: 81 = (8+1)², 4913 = (4 + 9 + 1 + 3)³. [Lines]

Cases of abcd = (ab + cd)² : 3025 = 55², 9801 = 99², 2025 =45².

Numbers that are repeated at the end of their squares, and therefore higher powers also, are known as **automorphic** numbers.
Examples are 25 and 76. The number 3792 repeats in the middle of its square: 3792^2 = 14379264.

The units digits of squares (in base ten) continually repeat the palindromic series: 0, 1, 4, 9, 6, 5, 6, 9, 4, 1, 0, ... Squares cannot end in 2, 3, 7 or 8. Numbers ending in 0, 1, 4, 5, 6 or 9 in base ten (like the squares) have cubes ending in the same digit. Numbers ending in 2, 3, 7, 8 have cubes ending in the complementary digit 8, 7, 3, 2 respectively. The end digits of fourth powers are 0, 1, 5 or 6. The fifth power of n has the same end digit as n. This is because the difference between any two consecutive fifth powers is a number having 1 as its end digit, i.e. of the form 10.m + 1. Thus sixth powers have the same end digits as squares, seventh powers have the same end digits as cubes, and so on.

In base ten, dividing by 9: The squares have remainders 1, 4, 0, 7, 7, 0, 4, 1, ... repeatedly. The cubes have remainders 1, 8, 0, ... The remainder of any multiple of 3 is 3, 6 or 0. Multiples of 7 have the remainders 7, 5, 3, 1, 8, 6, 4, 2, 0, ... Triangular numbers have the remainders 1, 3, 6, 1, 6, 3, 1, 0, 0, ... Hexagonal numbers have the remainders 1, 7, 1, ...

Multiplication by reduction. To multiply numbers a.b larger than 5, say 8.9. (i) deduct 5 and add (e.g. 8–5 = 3, 9–5 = 4, 3+4 = 7). (ii) deduct the results of (i) from 5 and multiply (5–3 = 2, 5–4 = 1, 2.1 = 2). (iii) Product is first digit (i) second digit (ii), i.e. 72. [Reichmann]

Multiplication by diagonal addition: To multiply abc by def (say 652.318). Enter the numbers along the top and right of a 3 by 3 array. Cut the squares in two by diagonals from top right. Enter the two-digit products a.d, a.e, a.f in the top row, and so on. Now add the diagonals, starting from the lower right, carrying if necessary, writing the totals at the bottom and left. This gives the product (207336). [Reichmann]

6 5 2 2 1/8 1/5 0/6 3 0 0/6 0/5 0/2 1 7 4/8 4/0 1/6 8 3 3 6

Multiplication by doubling and halving (also called Russian Multiplication). Put the numbers to be multiplied at the head of two columns. Repeatedly halve the number in the first column, ignoring remainders, and repeatedly double the number in the second column. Stop when 1 is reached in the first column. Cross off all rows containing an even number in the first column (including the first if that is even). Add the remaining numbers in the second column (including the first). This is the required product. [Reichmann]

Example 397 00196 1 198 00392 0 099 00784 1 049 01568 1 024 03136 0 012 06272 0 006 12544 0 003 25088 1 001 50176 1 total 77812

This is explained by expressing the number in the first column in binary notation. The binary digits correspond to the lines included (1) or omitted (0), read from the bottom upwards. Each 1 represents a power of two multiplying the second number. Each 0 has no multiplicative effect. The method can also be applied to division. For example 99/7 = 99.(1/7), so we put doublings of 1/7 in the right column.

Reduction or augmentation. If a number is near to another number that is simpler to multiply with, such as a power of ten, it is often easier to operate with this value and make adjustments after. For example if X or Y are easily multiplied numbers, like multiples of ten, we can make use of identities like the following. (X–a).Y = X.Y – a.Y, thus to multiply by 9, multiply by 10 and subtract the original number: 9.N = (10 – 1).n = 10.n – n = n0 – n. Similarly to multiply by 99 etc. (X+a)² = X² + 2.a.X + a² = X.(X + 2.a) + a², say 35² = 30.40 + 25 = 1225. (X–a)² = X² – 2.a.X + a² = X.(X – 2.a) + a². (X+a).(X–a) = X² – a² (X+a).(X+b) = X² + (a+b).X + a.b, (X+a).(Y–b) = X.Y + a.Y – b.X – a.b, say 31.39 = 30.40 + 10 – 1 = 1209 (X+a).(Y+b) = X.Y + a.Y + b.X + a.b

Calendar calculations. To find on what day of the week a date falls in the twentieth century 1900 - 1999. (i) Take the last two digits of the year, but if in the first two months of a leap year, reduce by 1. (ii) Add a quarter of this same figure, ignoring fractions. (iii) Add the code number for the month: January - December: 0,3,3,6,1,4,6,2,5,0,3,5. (iv) Add the number of the day in the month. (v) Divide by 7 and note the remainder. his gives the day: Sunday = 0, ..., Saturday = 6. For 1800-1899 add 2 to the final total. In any year other than a leap year the first and last days of the year fall on the same weekday, since 364 = 7.52.

Weights problem. What are the least number of different weights to enable us to weigh any integral number of pounds from 1 to 40. If only one pan may be used then the weights are 1, 2, 4, 8, 16, 32 [Tartaglia 1556]. If both pans may be used simultaneously 1, 3, 9, 27 [Bachet 1624].

Benford's Law: Statistical Properties. An unusual phenomenon that is said to go back to an observation that logarithm tables in public libraries appeared to be dirtier at the beginning than at the end. This seems to imply thta the numbers which science and engineering students had to deal with tended to start with the lower numerals. Why should this be? Surely any of the nine numerals 1 to 9 should be equally likely to occur? The physicist Frank Benford (1938) investigated twenty statistical tables, ranging from the surface areas of lakes and rivers to the molecular weights of thousands of chemical compounds. His conclusion was that the frequency of the larger numerals fell off in a regular fashion. The same aplies to house numbers in addresses listed in directories. Benford proposed the 'law of anomalous numbers': P(n) = log10[(n+1)/n] where P(n) is the probability that the first digit will be n. The exact numerical values, to three places of decimals, are: P(1) = 0.301, P(2) = 0.176, P(3) = 0.125, P(4) = 0.097, P(5) = 0.079, P(6) = 0.067, P(7) = 0.058, P(8) = 0.051, P(9) = 0.046. It has been shown that a change in units does not afect the result, even though all the numbers change as a result. Prof. Pinkham of Rutgers University has shown that any set of numbers which is 'scale invariant' (that is it reproduces the same frequencies of the nine first digit numerals when they are all multiplied by the same arbitrary number) must conform to Benford's distribution. Thus nature somehow provides us with scale-invariant lists. Benford's distribution can be reproduced by forming a list in which each term is obtained by multiplying the previous one by the same number, i.e. geometrical sequences. [Lines]

Kaprekar Numbers. These were first studied by the Indian mathematician D. R. Kaprekar (1940s). Think of a number of more than one digit, not all the same. Arrange the digits in decreasing order so as to make the largest possible number. Similarly arrange the digits to form the smallest possible number. Now find the difference of these two numbers. Now repeat the process with the number thus determined. Starting with 90 the differences generated are: 81, 63, 27, 45, 9, and if we interpret the reverse of this as 90 we are back at the start. Starting with 900 the differences are: 891, 792, 693, 594, 495, 495, ... repeating for ever, 495 is self-reproducing. This is in fact the self-reproducing end-point for all 3-digit numbers. Starting with 2359 the differences are: 7173, 6345, 3087, 8352, 6174, 6174, ... Again this is the self-reproducing end-point for all 4-digit numbers. It is now known that any 5-digit series ends in one of three possible loops, namely (a) 71973, 83952, 74943, 62964 (b) 75933, 63954, 61974, 82962, (c) 59994, 53955. Loops have also been found for 6, 7, 8 and 9 digit numbers. [Lines]

Self-Born Numbers. This is another idea of Kaprekar (1949). The process to consider is: take any number, add to it the sum of its digits. Thus 58 generates 71, 79, 95, 109, 119, 130, 134, 142, 149, 163, 173, ... Numbers that are not generated by this process are called self-born numbers. They are: 1, 3, 5, 7, 9; 20, 31, 42, 53, 64, 75, 86, 97; 108, 110, 121, 132, 143, 154, 165, 176, 187, 198, 209, 211, 222, 233, 244, 255, 266, 277, 288, 299, ... (mostly arithmetic sequences with difference 11). The smallest powers of 10 that are self-numbers are 10^6 and 10^16. Kaprekar's method of testing: add up the digits = D, add up the digits again = d. If odd, add 9 to give even number. In either case divide by 2 = k. Now subtract k and check whether this is a generator. Now proceed by subtracting 9s, for as many steps as there are digits in the original number. If no generator is found it is self-born. The smallest number with two generators is 101 = 91 + 10 = 100 + 1. The smallest number with three geenrators is 10 000000 000001 = 9 999999 999892 + 109 = 9 999999 999901 + 100 = 10 000 000 + 1. The smallest number with four generators is 1 000000 000000 000000 000102 = 1000000 000000 000000 000100 + 2 = 1 000000 000000 000000 000091 + 11 = 999999 999999 999999 999902 + 200 = 999999 999999 999999 999893 + 209.

*The Fascination of Numbers*by W. J. Reichmann (Methuen 1957, reprint 1963)*Numbers*by L. F. Taylor (Faber and Faber 1970)*Think of a Number*Malcolm E. Lines (Institute of Physics, Taylor & Francis 1990)- The Number of the Beast by Mike Keith
- On Repdigit Polygonal Numbers by Mike Keith
- Wild Narcissistic Numbers by Mike Keith
- How many zeros in base ten representation of 2^n? by G. E. Davis