
Multiplication of Numbers made with digit 1
Posted:
Apr 28, 2013 1:40 PM


From: Doctor Nisith Bairagi Subject: Multiplication of Numbers made with Digit 1 Date: April 28, 2013 (alt.math.recreational.independent) >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> It can be shown that, the numbers which are solely written with digit 1 (like, 1, 11, 111, 1111 ?etc.), when multiplied by itself results in wonderful numbers. For example: 1.1 = 1 11.11 = 121 111.111 = 12321 11111111.11111111 = 12345678987654321
Both the individual numbers taking part in multiplication and the resulting product, are all palindromic (i.e., they remain the same number if read from left to right or right to left). Few examples of palindromic numbers are: 4309034, 4224, 222, 43934 etc. However, if the individual numbers on the left side written with digit 1 exceed eight (say, 91?s or 101?s), the symmetry of the resulting numbers are lost totally, i.e, they cease to be palindromic. Let us identify some typical palindromic numbers and represent them symbolically (henceforth to be termed as: pform) as follows.
(A) pform representation of palindromic numbers: Type1 : such as p(4x3), number 3333 which is made of 4 numbers of digit 3.
Type2 : such as p(153), number 153351 which consists of odd number of digits in which the symmetry line passes through the space between 3 and 3, at the middle.
Type3 : such as p(153,7), number 1537351 which consists of odd number of digits in which the symmetry line passes through the central digit 7.
Thus p(1x1) = 1, p(1357) = 1537751), p(1537,7) = 153777351 belong to Type1, Type2 and Type3 category, respectively. A Type3 number say, 937739 written as p(93,77) or alternatively as p(93,2x7) can also be represented by Type2 number as p(937) when the total number of the middle digits written after the comma symbol (viz., ,77 or , 2x7)) are even.
Type4 : Apart from these three, there are numbers which are not truly palindromic, but the first and the last few number of digits read from left to right and right to left respectively, are same. For example, 15378027351 is not symmetrical, and if 802 at the middle is absent, the resulting number would become palindromic. For brevity, this type of number similar to Type3 be called as Type4 and is represented as p(1537,802).
(B) Diamond numbers:
There are various types of pform numbers which can be represented by Dform or Diamond numbers. The beauty of this class of numbers lies in the fact that, starting from a certain digit, the subsequent digits progressively increase in steps of 1 to reach a certain maximum at the middle (which being there, may repeat itself several times), and then progressively decrease in steps of 1 to the same starting one., maintaining symmetry in the pattern of the arrangement of the digits. 1234321 = D(4) = p(123,4) = pure diamond number 3456543 = D(36) = p(345,6) = Vcut diamond number 123444321 = D(3,3x4) = p(123,3x4) = Hcut diamond number 2345666665432 = D(25,5x6) = p(2345,5x6) = VHcut diamond number (V and H stand for vertical, and, horizontal, respectively) Numbers like, 12222 or 12344432 may often be encountered (as will be seen in multiplication of two large numbers each of which is made of digit 1), are not truly symmetrical. They would have been termed as diamond numbers if the last digit in each number, (viz., 1) would be present there. Such types will be termed as Spear number Y, since the arrangement of digits appear as the tip of a spear or arrow, see item (G)].
In the light of the symbolic representation of numbers written with digit 1 and their products (already cited earlier), can be expressed as
p(1x1) . p(1x1) = 1 = D(1) p(2x1) . p(2x1) = p(1,2) = D(2) p(3x1) . p(3x1) = p(12,3) = D(3) p(4x1) . p(4x1) = p(123,4) = D(4) ??. ??? ??.. ???. p(8x1) . p(8x1) = p(12345678,9) = D(9)
(C) Decimal numbers expressed as polynomial:
A number is formed by arranging digits sequentially selected from the set (0, 1, 2,***,9), and the value of the number is understood by summing up the product of the digits and their corresponding place value. In the decimal numbering system the place values are 1 (= 10^0), 10^1, 10^2, etc., when read from right to left. Thus:
4301 = 4x10^3 + 3x10^2+ 0x10^1 + 1x1 12935 = 1x10^4+ 2x10^3 + 9x10^2+ 3x10^1+ 5x1 It is possible to express any number N as polynomial, if the digits forming the number are considered as the coefficients a<m> occupying the position having the corresponding place value as x^m, where x = 10 (the base of the decimal numbering system ), as N = a<m>x^m + a<m1>x^(m1) + a<m2>x^(m2) +,,,,,,,,+ a<2>x^2 + a<1>x + a<0> ,and in view of this , 4301, 12935 and 111111 can be rewritten as polynomials of 3 rd, 4 th and 5 th degree, respectively.
With, x = 10, d = total number of digits, n = degree of polynomial = d 1): 4301 = 4.x^3 + 3.x^2 + 0.x + 1, (d = 4, n = 3) 12935 = 1.x^4 + 2.x^3 + 9.x^2 + 3.x + 5 ; (d = 5, n= 4) 111111 = 1.x^5 + 1.x^4 +1.x^3+1.x^2 + 1.x +1 ; (d = 6, n= 5)
If the coefficients in the polynomial are all equal to 1, the numbers formed only by digit 1 can be very easily expressed in polynomial form, which can be identified to be in G.P series (in ascending powers of x), the sum up to nth term of which is given by ( x^d ? 1) / ( x ? 1), which for x = 10 is: ( 10^d ? 1) / 9. (D) Numbers with digit 1 as X function:
Thus considering the place value x = 10, it is possible to represent any palindromic number written with digit 1 only, as p(d x 1) by a function of x , henceforth to be termed as X function, as: X<n> = p(dx1) = Sum [x^k, (k = 0 to n)] = (x^d1) /(x1) = (10^d1) /9, d = n + 1)
Thus the various numbers with only digit 1, when presented as X function form, appear as (Table 1):
[columns: Number = N, Total no of digit = d, pform = p(dx1), Degree of polynomial = n, X<n> = (x^d1)/(x1)] >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Number Total no pform Degree of Xfunction of digit polynomial (N) (d) p(dx1) (n) X<n> = (x^d1) /(x1) >>>>>>>>>+>>>>>+>>>>>>>>+>>>>>>>>>>+>>>>>>>>>>>>>>>>>>>>>>>>> 1, 1, p(1x1), 0, X<0> = 1. 11, 2, p(2x1), 1, X<1> = x +1. 111, 3, p(3x1), 2, X<2> = x^2 +x +1.
1111, 4, p(4x1), 3, X<3> = x^3 + x^2+x +1.
1111111111, 10, p(10x1), 9, X<9> = x^9 + x^8 + x^7 +*** + x^2 +x +1.
111***m times, m, p(mx1), (m1), X<m1> = x^(m1) + x^(m2) +***+ x^2 + x + 1. >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Table 1: X function representation of numbers made of digit 1 >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> (E) Operations with X<m> and X<n> : With two numbers X<m> = p(m + 1 x 1); m > = n, multiplication process has been framed and is presented with numerical examples.
Multiplication rule
Two cases are considered viz., (1) the smaller number having total digits up to 9 (i e., n < = 8), and (2) equal to or more than 10 (i e., n > = 9), respectively. The general multiplication formula remains same for both the cases as:
X<m>.X<n> = p(123?n , (mn+1).(n + 1)) = D(n, (mn+1).(n + 1))
The product number is symmetrical, if both m and n are either even or odd, in which case mn+1 is always odd, resulting in a number of Type 3 category ; otherwise, if any one of m or n is even and the other is odd, the resulting number conforms to Type ?2 category, since, mn+1 is always even.
Example 1. Multiply 1111111 and 11111 1111111 = p(7 x 1) = X<6> , and 11111 = p(5 x 1) = x<4> Here m = 6, n= 4, n +1 = 5, mn+1 = 3 (odd) and, d of the smaller number < = 9. Using multiplication formula : X<6>.X<4> = p(12?4 , 3 x 5) = p(1234,555) = D(4,3 x 5) = 12345554321 (Type3)
Example 2. Multiply 111111111111 and 111111111 111111111111 = p(12 x 1) = X<11> , and 111111111 = p(9 x 1) =X<8> Here m = 11, n= 8, n +1 = 9, mn+1 = 4 (even) and, d of the smaller number > = 9 Using multiplication formula : X<1>.X<8> = p(12?8 , 4x 9) = p(12345678,9999) = D(8, 4 x 9) = 12345678999987654321 (Type3)
Examples 1 and 2 clearly show that, so long as one of the numbers in multiplication possesses total number of digits d less than or equal to 9, there arises no problem in representing the final product through the proposed multiplication formula, which turns out to be truly palindromic. Depending on whether, mn+1 is even or odd, the number conforms to Type2 or Type3, respectively. Moreover if, mn+1 = 1, which occurs for m = n only, the resulting numbers thus obtained as the product have already been displayed as wonderful numbers, such as: 111 . 111 = 1234321 = p(123,4) = D(4), 111111 . 111111 = 1234567654321 = p(12***6,7) = D(7), etc., and belong to Type3 category. This class of numbers has been identified earlier as Diamond numbers of various types.
Example 3. Multiply X<m>.X<9> = p((m + 1) x 1).p(10 x 1) First consider the case : X<9>.X<9> = p(10 x 1).p(10 x 1) ; here m = 9, n = 9, mn+1 = 1. Using multiplication formula, the result can still be expressed as : p(12?9, 1 x 10). This number cannot be directly written since overlapping of one10, is to be incorporated with the last digit on the left. Such overlapping does not occur when the smaller number viz., X<n> = p((n + 1) x 1), is restricted to 9 digits(i.e., n < = 8). Thus, p(12?9,1 x 10) = 123456789 ********* 10******** 987654321 , , , , , , , , , , , , , , , , , , , , 1234567900987654321 = p(12***7, 90098) = D(7, 90098) This number is not truly palindromic and belongs to Type4 category.
Proceeding this way, the multiplication of two numbers, one number being 9 digit one, and the other of 9 or more digits using the same multiplication rule and keeping in view the overlapping of 10?s yields: X<9>.X<9> = p(10 x 1) . p(10 x 1) = p(12?9, 1 x 10) = p(12***7, 90098) X<10>.X<9> = p(11 x 1) . p(10 x 1) = p(12?9, 2 x 10) = p(12***7, 901098) X<11>.X<9> = p(12 x 1) . p(10 x 1) = p(12?9, 3x 10) = p(12***7, 9011098) X<12>.X<9> = p(13 x 1) . p(10 x 1) = p(12?9, 4 x 10) = p(12***7, 90111098) X<13>.X<9> = p(14 x 1) . p(10 x 1) = p(12?9, 5x 10) = p(12***7, 901111098) and so on.
In all these products, it can be noticed that, when read separately, the first and the last seven digits are palindromic conforming to Type2 category, and rare Diamond numbers as D(7) = p(1234567). The number formation is : (123***7, 98***098 ,7***21). The central number is (shown by ***) viz., /90098,901098,*** 901111098/ ctc., can be quickly written down as follows: Calculate values of m and n each in excess of 9, and add them, which is same as (mn). The resulting number decides the number of times the digit 1 be inserted in between 98 and 098 in the central number (shown by ***).Thus the pform result for X<m>.X<9>, is given by : X<m>.X<9> = p(12***7,90 p( [(m9) + (99) ] x 1) 098)
It can be checked that for m = 9, insertion of 1 in between 90 and 098 is not required. The total number of insertion of the digit 1, is 1, 2, 3 and 4 (i.e., 1, 11, 111 and 1111) for m = 10, 11, 12, and 13, respectively.
(F) Multiplication of X<m>.X<n>, (where m > = n, and n > 10) : They follow beautiful pattern. Since essentially they are large numbers, they can be conveniently described by the pform, Diamond numbers (Dform), and Spear numbers (Yform), either individually or in combination. The general pattern of the resulting number after multiplication can be described as: p(12***7,90Y098) = D(7,Y098) = 123456790Y0987654321, which is a Type4 number.
(G) In these, Y in the middle represents a cluster of digits as Spear number Y: [They are called so since the arrangement of digits appear as the tip of a spear]
(1) m= 10, n = 10, mn+1 = 1, n8 = 2, Y = 12 [symbolically, (mn+1)x(n8) next to 1, is (1,1x2)]
(2) m= 20, n = 10, mn+1 = 11, n8 = 2, Y = 122222222222 [symbolically, (mn+1)x(n8) next to 1, is (1,1x2)
(3) m= 14, n = 11, mn+1 = 4, n8 = 3, Y = 12333332 [symbolically, (mn+1)x(n8) next to 1, is (1,D(23,10x3))]
(4) m= 19, n = 12, mn+1 = 8, n8 = 4, Y = 1234444444432 [symbolically, (mn+1)x(n8) next to 1, is (1,D(24,18x4))]
(5) m= 22, n = 15, mn+1 = 8, n8 = 7, Y = 1234567777777765432 [symbolically, (mn+1)x(n8) next to 1, is (1,D(27,8x7))] >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Dear Mathematicians/readers, Please comment on this topic, and suggest modification, if any, for betterment of the proposed topic. . Thanks to All.
Doctor Nisith Bairagi Uttarpara, West Bengal, India >>>>>>>>>>>>>>>>>>>>>>

