[ *September 20, 2004* ]

Numbers and their reversals have

a __difference__ divisible by 9 and a __sum__ divisible by 11

and a palindromic endresult

A demonstration by Andrei Cucuianu (email)

Dear Mr De Geest, I stumbled upon some funny "laws"

while playing with my cell phone calculator.

Abstract. The difference between numbers

and their reversals is always a multiple of 9 (for

example 25 - 52; 346 - 643; 2683 - 3862; etc.). The

number resulting from division by 9 is a

polynomial with **a palindromic series** of coefficients,

often resulting in a genuine palindromic number when

working it out like for example 373, 5885, 47974, etc.

This is not only true for base 10 numbers but

for any other bases. It can be demonstrated by

expressing the base (x) as [(x-1)+1] .

When doing the subtraction (ax^{n} + bx^{n-1} + ... + ix + j)

– (jx^{n} + ix^{n-1} + ... + bx + a), by substituting x

with [(x-1)+1], we always obtain a multiple of (x-1).

The sum of two symmetrical number pairs is

always divisible by 11 if the numbers involved have an

even amount of terms 2, 4, 6, 8, ... (for example 24 + 42;

3567 + 7653; 245798 + 897542; etc.). The result of the

division leaves us a polynomial with **a palindromic series**

of coefficients, often resulting in a genuine

palindromic number when working it out.

This is not only true for base 10 numbers but for any

other bases. It can be demonstrated by expressing

the base (x) as [(x+1)-1] .

When doing the summation (ax^{n} + bx^{n-1} + ... + ix + j)

+ (jx^{n} + ix^{n-1} + ... + bx + a), if the number of

terms is even, by substituting x with [(x+1)-1], we

always obtain a multiple of (x+1).

Expressing the number base (x) as [(x+1)-1], also

explains a well-known fact, namely, that even-numbered

palindromes are always divisible by 11.

ALGEBRAIC DEMONSTRATION

The number 9 law

We denominate the numerical base as 'x'.

The numbers we use currently are in base 10.

Any number in any base can be defined as (ax^{n} + bx^{n-1} + ... + ix + j).

We express artificially 'x' as [(x-1)+1]

Then we perform the binomial multiplications

according to Pascal's triangle:

(a+b)^{2} = a^{2} + 2ab + b^{2};

(a+b)^{3} = a^{3} + 3a^{2}b + 3b^{2}a + b^{3};

(a+b)^{4} = a^{4} + 4a^{3}b + 6a^{2}b^{2} + 4ab^{3} + b^{4};

etc.

where (x-1) = a and 1 = b

Therefore, numbers, and the difference obtained by

subtracting their reversals, can be written like this:

1. Second order numbers (ax+b, respectively bx+a).

If we express in the right hand part

of the equation 'x' as [(x-1)+1] , we have:

(ax + b) – (bx + a) =

{a[(x-1)+1] + b} – {b[(x-1)+1] + a} =

a(x-1) + a + b – b(x-1) – b – a =

a(x-1) – b(x-1) =

(a-b)(x-1);

if we divide by (x-1) we get a-b

**a singular palindromic term**

2. Third order numbers (ax^{2} + bx + c):

(ax^{2} + bx + c) – (cx^{2} + bx + a) =

(a-c)x^{2} + c - a =

(a-c)[(x-1)+1]^{2} + c - a =

(a-c)[(x-1)^{2} + 2(x-1) + 1] + c - a =

(a-c)(x-1)^{2} + 2(a-c)(x-1) + a - c + c - a =

(a-c)(x-1)^{2} + 2(a-c)(x-1),

obviously divisible by (x-1).

When divided by (x-1) it becomes (a-c)(x+1) + 2(a-c),

which becomes (a-c)x - a + c + 2a - 2c =

(a-c)x + (a-c), **which is a palindromic polynomial**

3. Fourth order numbers (ax^{3} + bx^{2} + cx + d):

(ax^{3} + bx^{2} + cx + d) – (dx^{3} + cx^{2} + bx + a) =

(a-d)x^{3} + (b-c)x^{2} + (c-b)x + (d-a).

When substituting x with [(x-1)+1] we get

(a-d)[(x-1)+1]^{3} + (b-c)[(x-1)+1]^{2} + (c-b)[(x-1)+1] + (d-a),

which, after doing the binomial multiplications according

to Pascal's triangle becomes:

(a-d)(x-1)^{3} + (3a-3d+b-c)(x-1)^{2} + (3a-3d+b-c)(x-1);

when divided by (x-1) we get:

(a-d)(x-1)^{2} + (3a-3d+b-c)(x-1) + (3a-3d+b-c) which becomes:

(a-d)x^{2} + (a+b-c-d)x + (a-d), **which is a palindromic polynomial**

4. Fifth order numbers (ax^{4} + bx^{3} + cx^{2} + dx + e):

(ax^{4} + bx^{3} + cx^{2} + dx + e) – (ex^{4} + dx^{3} + cx^{2} + bx + a) =

(a-e)x^{4} + (b-d)x^{3} + (c-c)x^{2} + (d-b)x + (e-a).

When substituting x with [(x-1)+1] we get

(a-e)[(x-1)+1]^{4} + (b-d)[(x-1)+1]^{3} + (d-b)[(x-1)+1] + (e-a),

which, after doing the binomial multiplications according

to Pascal's triangle becomes:

(a-e)(x-1)^{4} + [4(a-e) + (b-d)](x-1)^{3} + [6(a-e) + 3(b-d)](x-1)^{2} +

[4(a-e) + 2(b-d)](x-1) =

which is obviously divisible by (x-1), resulting:

(a-e)(x-1)^{3} + (4a – 4e + b – d)(x-1)^{2} + (6a – 6e + 3b – 3d)(x-1) +

(4a – 4e + 2b – 2d), that is:

(a-e)x^{3} + (a-e+b-d)x^{2} + (a-e+b-d)x + (a-e)

**which is a palindromic polynomial**

Etc.

Predictions for the palindromes obtained by

dividing the difference of symmetrical number pairs

(ab – ba, abc - cba, abcd - dcba, etc.) with

'x-1' (meaning numerical base -1):

a-b

a-c a-c

a-d a+b-c-d a-d

a-e a+b-d-e a+b-d-e a-e

a-f a+b-e-f a+b+c-d-e-f a+b-e-f a-f

a-g a+b-f-g a+b+c-e-f-g a+b+c-e-f-g a+b-f-g a-g

a-h a+b-g-h a+b+c-f-g-h a+b+c+d-e-f-g-h a+b+c-f-g-h a+b-g-h a-h

a-i a+b-h-i a+b+c-g-h-i a+b+c+d-f-g-h-i a+b+c+d-f-g-h-i a+b+c-g-h-i a+b-h-i a-i

a-j a+b-i-j a+b+c-h-i-j a+b+c+d-g-h-i-j a+b+c+d+e-f-g-h-i-j a+b+c+d-g-h-i-j a+b+c-h-i-j a+b-i-j a-j

or, better,

a-b

(a-c)x +(a-c)

(a-d)x^{2} + (a+b-c-d)x + (a-d)

(a-e)x^{3} + (a+b-d-e)x^{2} + (a+b-d-e)x + (a-e)

Etc.

In case of 'base 10' numbers:

a-b

(a-c) x 10 +(a-c)

(a-d) x 100 + (a+b-c-d) x 10 + (a-d)

(a-e) x 1000 + (a+b-d-e) x 100 + (a+b-d-e) x 10 + (a-e)

Etc.

The number 11 law

The sum of two symmetrical number pairs is always divisible by 11

if the numbers involved have an even amount of terms 2, 4, 6, 8, ...

(for example 24 + 42, 3567 + 7653, 245798 + 897542).

The division leaves **a palindromic polynomial** of a lesser degree.

This is not only true for base 10 numbers

but for any other bases. It can be demonstrated by expressing

the base (x) as [(x+1)-1]. We perform the binomial multiplications

according to Pascal's triangle:

(a+b)^{2} = a^{2} + 2ab + b^{2};

(a+b)^{3} = a^{3} + 3a^{2}b + 3b^{2}a + b^{3};

(a+b)^{4} = a^{4} + 4a^{3}b + 6a^{2}b^{2} + 4ab^{3} + b^{4};

etc.

where (x+1) = a and (-1) = b

When doing the summation

(ax^{n} + bx^{n-1} + ... + ix + j) + (jx^{n} + ix^{n-1} + ... + bx + a),

if the number of terms is even, by substituting x with [(x+1)-1],

we always obtain a multiple of (x+1). The result of the division

by (x+1) is a polynomial with a palindromic series of coefficients,

often resulting in a genuine palindromic number when working it out.

For example:

1. Second order numbers:

(ax + b) + (bx + a) =

(a+b)[(x+1) – 1] + a + b =

(a+b)(x+1) – a – b + a + b =

(a+b)(x+1)

when divided by (x+1) leaves a+b

**a singular palindromic term**

2. Fourth order numbers:

(ax^{3} + bx^{2} + cx + d) + (dx^{3} + cx^{2} + bx + a) =

(a+d)x^{3} + (b+c)x^{2} + (c+b)x + (d+a);

by substituting x with [(x+1)-1] we get:

(a+d)[(x+1)-1]^{3} + (b+c)[(x+1)-1]^{2} + (c+b)[(x+1)-1] + (d+a).

After doing the binomial multiplications

according to Pascal's triangle we get:

(a+d)(x+1)^{3} - (3a+3d-b-c)(x+1)^{2} + (3a+3d)(x+1);

which is obviously divisible by (x+1);

when divided by (x+1) we get:

(a+d)(x+1)^{2} - (3a+3d-b-c)(x+1)+ (3a+3d) =

(a+d)x^{2} + (b+c-a-d)x + (a+d)

**note the palindromicy of the coefficients**

3. Sixth order numbers:

(ax^{5} + bx^{4} + cx^{3} + dx^{2} + ex + f) + (fx^{5} + ex^{4} + dx^{3} + cx^{2} + bx + a) =

... =

following the same procedures as above,

when divided by (x+1) we get

(a+f)x^{4} + (b+e-a-f)x^{3} + (a+f+c+d-b-e)x^{2} + (b+e-a-f)x + (a+f)

**note the palindromicy of the coefficients**

Etc.

Predictions for the palindromes obtained by dividing

the sum of even-termed symmetrical number pairs

(ab + ba, abcd + dcba, abcdef + fedcba, etc.)

with 'x+1' (meaning numerical base +1):

a+b

a+d b+c-a-d a+d

a+f b+e-a-f a+f+c+d-b-e b+e-a-f a+f

Even-numbered palindromes

A well known fact is that even-numbered palindromes

(abba, abcddcba, etc.) are always divisible by 11 (base 10 + 1).

This can also be shown by substituting the numerical

base (x) by [(x+1)-1], demonstrated as follows:

The palindrome "abba" (1221, 2332, 4554, etc..) can be expressed as

(ax^{3} + bx^{2} + bx + a). If we substitute x with [(x+1)-1], we get,

after a series of binomial multiplying :

a(x+1)^{3} - (3a-b)(x+1)^{2} + (3a-b)(x+1)

which is obviously divisible by (x+1), 11 for base 10 numbers.

The division yields: a(x+1)^{2} - (3a-b)(x+1) + (3a-b),

which after binomial multiplying results in ax^{2} + (b-a)x + a,

another palindromic polynomial.

It seems that 11 and 9, meaning x+1 and x-1, if we consider x as the

numerical base, act as key connection numbers between palindromes.

I can't help chasing the thought that the 9/11 plotters had a penchant

for mathematics, the Twin Towers being a palindrome themselves.

Some links

All palindromes with an even number of digits are divisible by 11

proof by induction from Shareef Bacchus