Kaprekar's Demlo Numbers

ON KAPREKAR’S DEMLO DEMLO NUMBERS Shri D. R. Kaprekar, in the preface to his book “ Demlo Numbers “ says that he coined the word Demlo Number from the name of the town Dombivli, a station 30 miles distant from Bombay on the then G. I. P. Railway. During the year 1923, as he was commuting as a daily passenger in local trains between Dombivli and Bombay, he began to note down numbers like 165, 176, 2553, 1776, 47773, 17776 randomly found on trains, wagons, motors, trams or ticket numbers. These numbers held a certain fascination for him. He discovered many interesting and curious properties of these numbers and unraveled unexpected connections they have with rep-digit numbers, recurring decimals etc. In this article, we will sketch how Demlo numbers arise, deal with only a few of their curious properties, and outline one or two applications. More interested readers can go through the various references of Kaprekar’s work, given elsewhere. What is a Demlo Number?

Look at the number 21555534 rewritten as 21 5555 34 . What strikes us first is the middle portion 5555 consisting of only 5’s. A closer look reveals that the first and last portions, namely, 21 and 34 add up to 55.This has led Kaprekar to define a Demlo number as a number which can be thought of as consisting of three parts, say, M, ( r)n , P where (i) the first part M when added to the last part P we get a number with the same digit r repeated (ii) the middle part ( r)n denotes a number with the same digit r repeated n times.. Usually M and P have equal number of digits. Sometimes M may have one digit less than P, as in 98 7777 679. It is possible that the middle part may altogether missing (n = 0) or that M (and so P) may be absent. To accommodate all such peculiarities Kaprekar has classified Demlo numbers as follows: Linear Demlo numbers: These are of the form (r)n. e. g. 777777 =(7)6 Binary Demlo numbers: These are of the form M(r)nP. e. g. 21(7) 21( 7)4 34. The Binary Demlo numbers are subdivided as Primary Demlo numbers if r = 9 as in 236 99 763 Secondary Demlo numbers if r = 3 or 6 as in 124 3333 209 and 521 66 145 Neutral Demlo numbers in which r is other than 9, 3, and 6.

Complementary Demlo numbers in which n = 0 (the repeated digit is absent as in 24 53).  Just saturated Demlo nubers if n = 1 (only one digit occurs in the middle part, as in 25 7 43). Oversaturated Demlo numbers if n > 1 (r occurs more than once in the middle portion as in 25 777 43). Unsaturated Demlo numbers . An example of  such a number comes from the decimal expansion 1/31 = 03225806456129 Here the blocks 03225, 80645, 16129 add upto 99999 Wonder Demlo numbers are of the type 12…k  (k+1)n k…21 where 1 ≤k ≤8. Here M consists of the first k natural numbers and P consists of the same in the reverse order. order. e. g. 12345654321 Consecutive Demlo numbers are a pair of  Demlo numbers M( r)nP and (M +1) (r)n (P – 1) in which the sum of the first and last parts is unchanged. Now we will indicate some ways of generating new Demlo numbers from given Demlo numbers.

Take any Demlo number numbe r M(r)nP . A simple way of forming a new Demlo number from it is to exchange the parts M and P to get P( r)n M. In doing so care must be taken to first annex a zero in i n case M has one digit less than P. P. Kaprekar calls such pairs of numbers as Exchanged Demlo numbers. Any number when multiplied by a rep-unit (a number consisting of only 1’s) gives a product which whic h is a Demlo number. number. e. g. 274

1111 = 30 44 12.

(3) Start with any number number N. select the block consisting of the first i digits of N and the block containing the last i digits of N, ensuring there is no overlap. Exchange these blocks, retaining the middle portion of N intact. Call this new number as N’. Subtract the smaller of the two numbers N and N’ from the larger of the two. The resulting number is a Demlo number. number. e. g. N = 3472 897 6356. Exchanging the block 3472 of the first four digits of N with the block 6356 of the last four digits of N we get the number N’ = 6356 897 3472. Here N’ > N. So N’ – N = 2883 999 7116, which is a Demlo number. number.

(4) Look at the number 3461.Partition it as 3 461, 34 61, 346 1. Exchange the two blocks in each of these to get 461 3, 61 34, 1 346. thus we have altogether numbers 3461, 4613, 6134, 1346. Adding them we obtain 15554, a Demlo number. More generally genera lly,, given gi ven any n-digited umber, umber, without disturbing the order of the digits, partition it into two blocks in which the number of digits are ordered pairs (1, n – 1), (2, n – 2),…, (n – 2, 2), (n -1, 1) respectively and exchange the two blocks in each case . That way we get n numbers (including the given number) whose sum is always a Demlo number. Kaprekar calls this as cyclic process. (5)Take any number, say, 372. Consider the following sequences of steps: 372 371 3719 and 372 371 628 6280. in the first sequence the number is reduced by 1, then a 9 is annexed. In the second sequence the number is reduced by 1, complement compleme nt w.r.t. w.r.t. 9 (meaning 999 99 9 – 371) is taken, and a 0 is annexed. Now if we add 3719 and 6280 we get the total as 9999, a linear Demlo number num ber.. Kaprekar calls this process as H- theorem stated as follows: “ If N is any number then N + (N – 1) + 8N + 10C =(9)n ” . Here C is the complementary number of N – 1 w.r.t. .r.t. 9 and n is i s the number of digits in 8N.

(6) Starting with any any two numbers A and B each containing k digits, letting C and D to be the respective complements of A and B each w.r.t. .r.t. 9 , an interesting relation (named H-Theorem) holds: A B+B + D = (9)2k 

C+C

D+D

A+A+B+C

e. g. Let A = 4537, 4537, B= 7528. So C = 9999 – 4537 = 5462, D = 9999 – 7528 = 2471, 2k = 8 and A B+B C+C D+D A+A+B+C + D = (A + C + 1) (B + D +1) – 1 = (9999 + 1) (9999 + 1) – 1 = 10000 10000 – 1 = 100000000 – 1 = 99999999 (7) Haphazard Process: Consider the sequence 236, 479, 543, 861, 861, 861, 625, 382, 318 where (i) the first four numbers are in increasing order, (ii) the fourth number is repeated three times and (iii) the last three numbers are the differences of each of the first three numbers from 861. We arrange the above numbers in either of the following diagonal pattern and add: 236 479 543 861

236 479 543 861

861

861

861 625 382 318

29028637638 numbers,

861 625 382 318

36340630326

both of which are Demlo

namely, namely, 29028 (6) 37638 and 36340 (6) 30326

More generally, with numbers a < b < c < … e < x if we form any sequence a, b, c, …,e, x, x, …, x, (x – a), (x – b), (x – c), …(x – e) and add them diagonally as illustrated above, the result is a Demlo number. number. As a sample indicator to Kaprekar’s beautiful powers of close observation and insight, we end up this short essay with a detailed account of what he called The Partition and Insertion Process Consider all two-digited numbers of the form ab (a in the tenth place and b in the unit place) subject to the requirement b a+1 and a + b 9. With a little effort, we can see that there are twenty five such numbers, namely,

01, 02, 03, 04, 05, 06, 07, 08, 09; 12, 13, 14, 15, 16, 17, 18; 23, 24, 25, 26, 27; 34, 35, 36; 45 The result of multiplying each of these numbers with nine i.e. numbers P of the form 9

ab is called a partition number. number. We get twenty five such

Partition Numbers P, namely, 9, 18, 27, 36, 45, 54, 63, 72, 81; 108, 117, 126, 135, 144, 153, 162; 207, 216, 225, 234, 243; 306, 315, 324; 405 An easy way of writing the digits A, B, C in the 100th, 10th, and unit places respectively of  each partition number P = ABC is to start with any a and b mentioned above and apply the rule A = a, B = b – a – 1,C = 10 – b . e. g. Consider the partition number P = 9 = 234.

26

Here a = 2, b = 6 . So as verified easily A = 2, B = 6 – 2 1 = 3, C = 10 – 6 = 4. A nice property of every wonder Demlo number W is that it is a square of a repunit. Precisely if W = 12 …(k – 1)k (k – 1)…21 (where k is any digit between 1 and 9), then W = {12 … (k -1)k(k – 1)…21}² = {(1)k }2. e. g. 121 = (11)², 12321 = (111)², …

Now we are ready to state the “Partition and Insertion Process”: If W = {(1) k }2is a wonder Demlo number and P =9 ab = ABC is a partition number Then W P = A (c)k-1B(9-c)k-1C where c = a + b Illustration: Let us find the product 123454321 162 Here k = 5, A = 1, B = 6, C = 2, c = 9 (Note that 162 = 9 18, a = 1, b = 8, c = a + b = 9). So the required product = 1 9999 6 0000 2. Powers of rep-digit numbers (9)n stands for the number in which the digit 9 repeats n times. The notation 9 means the m th power of (9) n. Kaprekar gives a peculiar gap-filling process for finding the m th power of (9)n. We illustrate the method for n < 6. m

n

We need to remember the first five powers of  9. These are 9, 81, 729, 6561, 59049. 3

Suppose it is required to evaluate 9 , the cube of 9999. Remembering 9³ = 729, 4

Here n = 4, m = 3. We introduce m (= 3) gaps thus: …7…2…9

We then start filling these gaps from left to

( )

right with 9

n

−1

( )

and 0

n

−1

alternately, thereby

getting 999 7 000 2 999 9 which is indeed the actual value (9999)³ 5

Another example: To find the value of  9 . Here n = 3, m = 5 = number of gaps. 3

The five gaps in …5…9…0…4…9 are to be filled thus: 99 5 00 9 99 0 00 4 99 9 (note number of digits in each gap = n – 1 = 2). Thus (9999)³ = 995009990004999. Kaprekar has evolved a general theory of  multiplication of Demlo numbers and introduced processes like Demlofication of  Fibonacci numbers, found new methods of  obtaining recurring periods of the reciprocals of certain prime numbers like 7, 13, 37, 73, 97, 109. His discoveries continue to arouse the curiosity of many people, the world over, over, who hold a fascination fascinati on for numbers.