Reversal Nines

Relevant wiki: Divisibility Rules (2,3,5,7,11,13,17,19,...)

The digital sum of a number also gives the remainder of that number when it is divided by $9$ (see Casting Out Nines ).

Reversing the order of the digits in a number does not change the digital sum, and therefore does not change the remainder of that number when it is divided by $9$ .

Therefore, when two numbers with the same digits are subtracted, the same remainders are subtracted, which gives a resulting remainder of $0$ , which means the result is always divisible by $9$ .

Relevant wiki: Algebraic Manipulation Problem Solving - Intermediate

Number the digits from right to left as digits 0, 1, 2, etc. Let $m < n$ . The $m$ th digit has value $a_m\cdot 10^m$ and the $n$ th digit has value $a_n\cdot 10^n$ . Swapping the two digits results to a change in the amount $\Delta = (a_n-a_m)\cdot 10^m + (a_m-a_n)\cdot 10^n = (10^n - 10^m)(a_m - a_n) = 10^m(10^{n-m} - 1)(a_m - a_n).$ Now $10^{n-m} - 1 = 9\cdot (10^{n-m-1} + 10^{n-m-2} + \cdots + 10 + 1)$ is a multiple of 9; therefore $\Delta$ is a multiple of a 9, which means that any swapping of digits will result in a change by a multiple of 9. Therefore inverting the order of digits changes the number by a multiple of 9.

Relevant wiki: Divisibility Rules (2,3,5,7,11,13,17,19,...)

If a positive integer $N$ is reversed there we just exchange the place value digits of $N$ in order of $\underbrace{{\color{#3D99F6}10^n} ,{\color{#D61F06}10^{n-1}} ,{\color{#20A900}10^{n-2}}\cdots ,{\color{#E81990}10}, 1}_{\text{original place value in descending order }}\quad \underbrace{{\color{#3D99F6}1}, {\color{#D61F06}10},\cdots , {\color{#20A900}10^{n-2}}, {\color{#E81990}10^{n-1}},{\color{#333333}10^{n}}}_{\text{reversed place value in ascending order }}$ For example $N= 12345 ={\color{#3D99F6} 10^4} , {\color{#D61F06}10^3} , {\color{#20A900}10^2} ,{\color{#E81990}10},1 \\ N_{\text{rv}} = 54321 = 10^4 , {\color{#E81990}10^3} , {\color{#20A900}10^2} , {\color{#D61F06}10} ,{\color{#3D99F6} 1}$ Now performing the subtraction we find it as ${\color{#3D99F6} 10^n-1 }, {\color{#D61F06} 10^{n-1} -10} , \cdots {\color{#20A900}10^{n-2}-10^{n-2}},{\color{#E81990}10-10^{n-1}} , 1 -10^n$ The numbers in above form are always divisible by $9$ even we can show that th numbers are divisible by $9$ by factorization (see) of $a^n-b^n$ .

Proof

If $N$ is a positive integer and $a_0,a_1,a_2\cdots a_{n-1},a_n$ are the distinct or identical digits. $N = 10^na_0 + 10^{n-1}a_1 + \cdots + 10a_{n-1} + a_n$ On reversing its digits $N_{rv}= 10^na_n + 10^{n-1}a_{n-1} +\cdots + 10a_{1} +a_0$ Now performing the subtraction we find $N-N_{rv} = a_0(10^n-1) + a_{1}(10^{n-1} -10) + \cdots + (10-10^{n-1})a_{n-1}+ (1-10^n)a_n \\ N-N_{rv} = 9.a_0.\underbrace{11\cdots 1}_{\text{n 11's}} + 9.10a_1. \underbrace{11\cdots 1}_{\text{n-2 11's}} + \cdots + 9.10a_{n-2}.\underbrace{11\cdots 1}_{\text{n-1 11's}} + 9.a_n\underbrace{11\cdots 1}_{\text{n 11's}} \\ \boxed{N-N_{rv} = {\color{#3D99F6}9}k }$ where $\small{k = a_0.\underbrace{11\cdots 1}_{\text{n 11's}} + 10a_1. \underbrace{11\cdots 1}_{\text{n-2 11's}} + \cdots + 10a_{n-1}.\underbrace{11\cdots 1}_{\text{n-2 11's}} + a_n\underbrace{11\cdots 1}_{\text{n 11's}}}$

Therefore, number created in this way are always divisible by 9 .

$abc - cba \\ 100a +10b +c - (100c +10b +a) = 99a-99c \\$

Let $N$ be some positive integer and $N_{rev}$ its reverse: $\begin{aligned} N - N_{rev} &=\sum_{k=0}^{n}a_{k}\cdot10^{n-k} - \sum_{k=0}^{n}a_{k}\cdot10^{k} \\ &\equiv \sum_{k=0}^{n}a_{k}\cdot1^{n-k} - \sum_{k=0}^{n}a_{k}\cdot1^{k} \\ &\equiv \sum_{k=0}^{n}a_{k} - \sum_{k=0}^{n}a_{k} = 0~(\text{mod} ~ 9) \end{aligned}$

Remember that in base 10, adding the digits does not change the residue left by 9. Rearranging the digits does not change their addition. When we take the difference, the resulting number becomes divisible by 9.

However, this does not work for numbers with the same digits.

10x + y - 10y - x = 9x - 9y = 9(x-y)

$\color{#D61F06}\text{Case 1 : Two digit positive number .}$

Let us assume that the 2 digit positive number is $10x + y$ . On reversing the number becomes $10y + x$

On subtracting them : $10x + y - 10y - x = 9x - 9y = 9 \times (x - y)$

Therefore , the resultant 2 digit number will always be a multiple of 9 .

$\color{#D61F06}\text{Case 2 : Three digit positive number.}$

Let us assume that the 3 digit positive number is $100x + 10y + z$ . On reversing the digits the number becomes $100z + 10y + x$ .

On subtracting them : $100x + 10y + z - 100z - 10y - x = 99x - 99z = 9 \times (11x - 11z)$

Therefore, the resultant 3 digit number will always be a multiple of 9.

$\color{#D61F06}\text{Case 3 : Four digit positive number.}$

Let us assume that the four digit positive number is $1000a + 100b + 10c + d$ . On reversing the digits the number becomes $1000d + 100c + 10 b + a$ .

On subtracting them : $1000a + 100b + 10c + d - 1000d - 100c - 10b - a = 999a + 90b - 90c - 999c = 9 \times (111a + 10b - 10c - 111c)$

Therefore, The resultant 4 digit number will always be a multiple of 9.

$\color{#D61F06}\text{Final Conclusion : }$

General form of the resultant number :

$\text{Resultant number= }9 \times a$

Where a is the number left when the resultant number is divided by 9 .

$\color{#20A900}\text{Any number which has n numbers when reversed and when both are subtracted the resultant number will always be a multiple of 9 . }$

(10a + b) - (10b+ a) = 9( a - b). When a = b, the answer is 0. Is it valid to consider 0 divisible by 9?

Consider one digit in a 6 digit setting e.g. 050000 - 000050 = $5 \times 10^4 - 5 \times 10^1 = 5 \times 10 \times (10^3 - 1) = 5 \times 10 \times 999$ , clearly divisible by 9. This pattern applies to all the digits of any number as follows.

A number, $N$ , with n + 1 digits denoted $a_i$ , i.e, $a_na_{n-1}...a_0$ can be expressed as the sum $N = a_n10^n + a_110^{n-1} +... + a_010^0$ .

The same number with the digits reversed, $N_r$ , can be expressed as $N_r = a_010^n + a_110^{n-1} +... + a_n10^0$

The difference can be expressed as $N - N_r = a_n (10^n - 10^0) + a_{n-1} (10^{n-1} - 10^2)$ $+ ... a_k(10^{n-k} - 10^k) + ... + a_0( 10^0 - 10^n)$

In the case n-k > k $(10^{n-k} - 10^k) = 10^k(10^{n-2k} - 1)$ , the quantity in parentheses on the right hand side will be all 9's, divisible by 9.

In the case n-k = k $(10^{n-k} - 10^k) = 0$ there is nothing to divide.

In the case n-k < k $(10^{n-k} - 10^k) = 10^{n-k}(1 -10^{2k-n})$ the quantity in parentheses on the right hand side will be all 9's, divisible by 9.

Thus every product in the expression is divisible by 9 therefore $N - N_r$ is divisible by 9.

The difference between any two-digit numbers whose numbers are reversed can be represented by the following expression: (10x + y) - (10y + x) If we simplify this expression, we obtain: 9x-9y, or 9(x-y), meaning that any number obtained is a multiple of 9. The same goes for three, four, five digit numbers, etc, considering that the simplification always involves subtracting two numbers that are powers of ten, which will always yield a multiple of nine.

My solution is just for some kids.... Lets start from the lowest pairs of preferably 3 :

10 - 1= 9 =1•9 , 11-11=0 =0•9 , 12-21=-9=-9•9 ,

Then, to a bit higher pairs:

42-24=18=2•9 , 43-34=18=2•9 , 44-44=0=0•9 ,

It can write as $10^{n} \times k+10^{n-1} \times q+...+10^{1} \times j+10^{0} \times r-10^{0} \times k-10^{1} \times q-10^{n-1} \times j-...-10^{n} \times r$ . Infact, every $10^{\alpha} \times z-10^{\beta} \times z$ is multiples of 9. And we can write $10^{n} \times k+10^{n-1} \times q+...+10^{1} \times j+10^{0} \times r-10^{0} \times k-10^{1} \times q-10^{n-1} \times j-...-10^{n} \times r$ as sum of many $10^{\alpha} \times z-10^{\beta} \times z$ . So it is multipies of 9.

Casting out nines

We can generalize and rewrite any of the numbers in the form 10A + B. We can then express the Frankenine process as (10A + b) - (10B + A), which simplifies to: 9A -9B, dividing by 9 we get A - B. From this we can confirm that with this process the numbers will always be divisible by 9. We also get the interesting result that for two digit numbers after the entire process the final number is the tens digit minus the ones digit (Although this doesn't work for numbers with more digits)

For any number，we could treat them as many pairs (a, b). a and b will swap position. So their subtraction result is equal to $10^n(a-b)+(b-a) = (10^n-1)(a-b)$ . Obviously $10^n-1$ could be divided by 9. So the statement is finally true.

Let the number be 10(x)+y. So if it's digits are reversed, the number becomes 10(y)+x. If we subtact the two, we get:- 10x+y -(10y+x) So this becomes -9y+9x Taking 9 common, we get 9(x-y). Therefore the number will be always divisible by 9

• lets say the number is abc , Their place values are = 100a + 10b + 1c ----------(1)
• on reversing the number as cba , The resulting place values are = 100c + 10b +1a ----------(2)
• On subtracting (1) -(2) = 99a - 99c , Which is divisible by 9
• This holds for all the numbers

$x = d_m d_{m-1} \cdots d_2 d_1 = d_m \times 10^{m-1} + d_{m-1} \times 10^{m-2} + \cdots + d_2 \times 10^1 + d_1 \times 10^0$

$y = d_1 d_2 \cdots d_{m-1} d_m = d_1 \times 10^{m-1} + d_2 \times 10^{m-2} + \cdots + d_{m-1} \times 10^1 + d_m \times 10^0$

$x - y = \displaystyle \sum_{i = 1}^{\lfloor \frac{m+1}{2} \rfloor} (d_{(m+1)-i} - d_i )(10^{m-i} - 10^{i-1}) = \displaystyle \sum_{i = 1}^{\lfloor \frac{m+1}{2} \rfloor} (d_{(m+1)-i} - d_i )10^{i-1}(10^{(m+1)-2i} - 1)$

Since $10^{(m+1)-2i} - 1$ is one less than a power of 10, it must be divisible by 9 $\forall i \in \{1, \cdots, m\}$ . Therefore, $9 | x-y$

let integer a be the sum of a i (i=0 to n) times 10 to the power of i, and then integer b (digits reversed) be the sum of a (n-i) (i=0 to n) times 10 to the power of i.

Then a-b = the sum of (a i - a (n-i)) times 10 to the power of i.

Now, 10 to the power of i can be written as 1 + 9 times the sum of 10 to the power of j (j=0 to i-1). (For example, 1000 = 1 + 9 times (100+10+1))

So, a-b = 9 times the sum of (a i - a (n-i)) times the sum of 10 to the power of j (j=0 to i-1) + the sum of a i - a (n-i) (i=0 to n) where the latter term is equal to zero.

The first term is obviously divisible by 9.

Sorry, I have to learn LaTeX again to make these more readable.

Assuming that the ten's place digit to be 'y' and unit's place digit to be 'x',then Any two digit number can be written as {10y+x} Reversing the numbers. {10x+y} Subtracting them (10y+x)- (10x+y)=9y-9x =9(y-x). It is divisible by 9. Same like this for other many digits numbers.

Remainder of a number when divided by 9 is simply the summation of digits. Reversing the order doesn't change the sum of digits. Let the number be r(mod 9) . Remainder of the New number(reverse digits) is also r. Difference of numbers leaves remainder of r-r=0 Hence always divisible by nine

Sample 69 and 96 let x=6 ; y=9 ; Z=difference 69=60+9
or 10x+y

96=90+6
or 10y+x

Then subtract Z= 96-69 Z=10x+y - (10y+x) Z= 9x-9y Z=9(x-y)

Therefore Z is always divisible by 9

General case is (10x + y) - (10y+x) = (10x-x)+ (y-10y) = 9x-9y = 9*(x-y) Therefore the equation is always divisible by 9.

There's some serious proofs here. I'm just going to put my thinking too.

(10X)+(y) is the first double digit number. Therefore (10y)+(X) is the reverse So the whole sum is (10x+y)-(10y+x) = z And the question asks whether z is a multiple of 9, well simplifying I got 9x-9y = z Or 9(x-y) = z and as X and y are integers so will x-y.

I did the same with 3 digits (100w+10x+y)-(100y+10x+w)= 99w-99y= z And seeing as the coefficient is 99 is a multiple of 9 I figured it was right again and it would continue like that. (1000p+100w+10x+y)-(1000y+100x+10w+p)= 999p+90w-90x-999y=z 9(111p+10w-10x-111y) = z etc etc

I know it's not proof but I think it should get partial marks.

Lets think about a 3 digit number ABC We know that ABC is:100A+10B+C We can write ABC-CBA as this: 100A+10B+C-100C-10B-A=99A-99C And we can get the common factor of 9 from there This also works for numbers with more/less digits than 3

For 2 digit no. ab (10a+b)-(10b+a)=9(a-b)

For three digit no. abc (100a+10b+c)-(100c+10b+a)=99(a-c)

And so on so it will be divisible by 9 always

Specificando che i numeri sono in base 10 quindi a base pari (2n) invertendo le cifre si avrà un numero divisibile per la base pari -1 Se sottraggo 203 a 302->302-203 però in base 4 otterrò un numero divisibile per 3 in base 4 ovvero 33

We know that:

$\overline{d_1d_2\cdots d_{n-1}d_n}-\overline{d_nd_{n-1}\cdots d_2d_1}=\sum_{k=1}^n 10^{n-k}d_k-\sum_{k=1}^n 10^{k-1}d_k$

A more "explicit" form would be:

$\overline{d_1d_2\cdots d_{n-1}d_n}-\overline{d_nd_{n-1}\cdots d_2d_1}=10^{n-1}d_1+10^{n-2}d_2+\cdots+ 10d_{n-1}+d_n-10^{n-1}d_n-10^{n-2}d_{n-1}-\cdots-10d_2-d_1$

This can be further simplified to:

$\overline{d_1d_2\cdots d_{n-1}d_n}-\overline{d_nd_{n-1}\cdots d_2d_1}=\sum_{k=1}^n d_k(10^{n-k}-10^{k-1})$

And since the difference between two powers of $10$ is always a multiple of $9$ , are are summing $n$ multiples of n, hence obtaining a multiple of $9$ as well. Conclusion: Yes, the numbers Dr. Frankenine creates in this way are always multiples of $9$ .