The person in the above image is Bertrand Russell. Famous Mathematician and Logician.  An excellent graphic novel, Logicomix, is based around his life. That is a must read.

First thing first. Even if you are not a Math demi-god, it is fine, Digit Math is still for you. In fact my academic track record shows that I barely survived Mathematics! I invented this thing when I was in standard eleventh while trying to create some shortcut formula, because I kept forgetting long formulae and was very slow in arithmetic. Digit Math is a slightly different approach to classical Math. To understand this all you need to know is how to add and multiply, and a little bit of logic. That is it.

What is Digit Math?

Numbers are group of digits. Classical Mathematical operators are not meant to deal with few digits of a number. Digit Math allows that. You can interact with a single digit or a group of digits in a number. You might wonder, what is the use? Well, I would say, for fun. Like any good puzzle, this too forces you to think differently. It is not all pointless fun though.

Can you prove that any number which ends with 5 must be divisible by 5? Well try that using classical Math. I have used Digit Math to prove the same. There some more empirical facts which I have proved using Digit Math. One of that is proving that multiplying any number by \(10^n\) will indeed give us that number but followed by \(n\) zeroes. This might look obvious to you, but think for a moment, why does adding a number like \(6\), \(100\) times is guaranteed to yield the exact number \(600\).

Before we begin…

Before we begin I must list out the notations and symbols I would be using. They were introduced specifically for this, since I don’t know of any existing symbols which convey the same information.

Symbols and Operators

I would be using these symbols throughout the articles on Digit Math, so pay attention to them.

\(\omega\) (With-operator)

This is the main operator, which delimits the digits or groups of digits. I call this the ‘with-operator’. So, 251 can be written as \(2\omega51\) (spoken as two-with-fifty-one) or \(25\omega1\) (spoken as twenty-five-with-one) or \(2\omega5\omega1\) (spoken as two-with-five-with-one) or even \(0\omega251\). They all are the same, only the digit partitioning is different. How you partition depends on the problem at hand.

\(]x[\) (Digit Group operator)

This is used in multiple ways. Its primary use is to group digits into a digits group. While in \(2\omega51\) it is quite clear that 2 and 51 are different digit groups of one and two digits respectively, but sometimes we need to convey more information about a digit group.

• \(]x[\) – This is equal to the number of digits in x, when used on its own. So, \(]25[ = 2\).
• \(]x[^y\) – This means that x should have y digits. Note that I said ‘should’ have. So, it is possible that while solving the problem, at intermediate steps we can have less or more digits. Eventually, the final result must have exactly y digits. This kind of notation is usually used when actual value of x is not known and x is part of a bigger number. So, \(2\omega]x[^2\) means that this is a number which starts with 2 but ends with two digits.
• Underline notation – Take the example of \(2\omega]345[^2\). Here 3 is the extra digit in this digit group which has the maximum capacity of two digits. In this case since all the digits are known, so here we can use a short-hand notation of underlining the excess digit in the group. So, \(2\omega]345[^2 = 2\omega\underline{3}45\). (Note that 3 is underlined.)

\(\omega\sum\) (Digits summation operator)

So,$$\omega\sum_{i=1}^3a_i = a_3{\omega}a_2{\omega}a_1$$

Note that \(a_1\) is the rightmost digit and \(a_3\) the leftmost. Maintaining the sequence in digitized equations are of massive importance.

Precedence of With-operator

\(\omega\) has lower precedence than multiply and division. So the full operator precedence would be:

B – Bracket
O – Orders (powers, roots, etc.)
M – Multiply
D – Divide
W – With (\(\omega\))
A – Add
S – Subtract

Properties of With-operator

• This is not commutative. This is because, \(2\omega5 = 25 \neq 52 = 5\omega2\).
• This is associative, but only when none of the digit groups have excess digits. That is, less number of digits than required is tolerated. So, \(a\omega(b{\omega}c) = (a{\omega}b){\omega}c\).
• This is distributive, but only when all digit groups have the exact number of digits. So, less digits is not tolerated here. So, \(n \times (a{\omega}b) = (na)\omega(nb)\).

Some Important Formulae

General formula of transformation

This provides the relation between a normal equation and a digitized equation.

So,$$\omega\sum_{i=1}^{n}a_i = \sum_{i=1}^{n}a_i \times {10}^{n-i}$$

Provided \(a_i\) has one digit only (i.e. \(]a_i[ = 1\)). This is not hard to see, how the above is true.

Bi-multiplication formula

This formula is vital. This describes how the groups of digits behave when they are multiplied. This starts with how we usually manually multiply two numbers.

$$
\begin{align}
x\omega y&\\
\underline{\times\hspace1em a\omega b}&\\
bx\omega by&\\
ax \omega ay\hspace1.5em&\\
\overline{ax\omega (ay+bx)\omega by}&
\end{align}
$$

So,

$$\underline{x \omega y \times a \omega b = ax \omega (ay+bx) \omega by}\tag{bi-multiplication formula}$$

The above relation is pretty easy to understand, but it does say how many digits should be in each group (digits separated by \(\omega\)). To understand this, try multiplying 12 by 34, the manual way. This would be like

$$
\begin{align}
1\omega 2&\\
\underline{\times\hspace1em 3\omega 4}&\\
4\omega 8&\\
3 \omega 6\hspace1.5em&\\
\overline{3\omega (6+4)\omega 8}&\\
\Rightarrow 3\omega 10\omega 8&\\
\Rightarrow 4\omega 0\omega 8&\tag{adding carry over 1 from center to left group}\\
\Rightarrow 408&
\end{align}
$$

Now lets try multiplying 111 by 123.

$$
\begin{align}
01\omega 11&\tag{prefixing with 0 to equate digit counts}\\
\underline{\times\hspace1em 01\omega 23}&\\
23\omega 253&\\
1 \omega 11\hspace1.5em&\\
\overline{1\omega (11+23)\omega 253}&\\
\Rightarrow 1\omega (11+23+2)\omega 53&\tag{2 is the carry over from 253}\\
\Rightarrow 1\omega 36\omega 53&\\
\Rightarrow 13653&
\end{align}
$$

So you can observe from the above scenarios that:-

$$
]a\times b[ = n\hspace1em\ldots\text{where ]a[ = ]b[ = n}
$$

Similarly,

$$
]a + b[ = n\hspace1em\ldots\text{where ]a[ = ]b[ = n}
$$

Any excess digits should be carried over to the digit group on the left and added there.

So,

$$
\begin{align}
1&\\
\underline{+\hspace1em9}&\\
0\omega\underline{1}0&\\
\Rightarrow (0+1)\omega0\\
\Rightarrow 1\omega0\\
\Rightarrow 10
\end{align}
$$

If  \(]a[ \neq ]b[\) then I am not sure what should be \(]a \times b[\). So, we need to make sure that all digit groups have equal number of digits. This is restrictive but fortunately this does not cause problem as there is always a way around.

One more example. Let’s multiply 98 by 76.

$$
\begin{align}
9\omega 8&\\
\underline{\times\hspace1em 7\omega 6}&\\
54\omega 48&\\
63 \omega 56\hspace1.5em&\\
\overline{63\omega (56+54)\omega 48}&\\
\Rightarrow 63\omega 110\omega 48&\\
\Rightarrow 63\omega (110+4)\omega 8&\tag{since max size of each group is 1}\\
\Rightarrow 63\omega 114\omega 8&\\
\Rightarrow (63+11)\omega 4\omega 8&\\
\Rightarrow 74\omega4\omega8&\\
\Rightarrow 7448&
\end{align}
$$

Rules for using the bi-multiplication formula

You have already witness many of the rules, but let me summarize them for clearly. For the following assume that we are multiplying \(a\omega b\) by \(x\omega y\).

• Make two fragments of both the numbers, by placing \(\omega\) at the positions suitable for you. Remember, the number of digits in all the four fragments should be equal. That is, \(]a[ = ]b[ = ]x[ = ]y[\).Partitioning \(255\) as \(2\omega55\) is perfectly fine, since \(]02[ = ]55[ = 2\).
• If the number of digits in \(by\) or \(ay+bx\) terms are less than the required number then insert zeroes in front of them to get the exact number of digits.
• If the number of digits in \(by\) or \(ay+bx\) terms are more than the required number then remove the excess digits from the front (left) of them and add these excess digits to the group on left. Note, that shifting of excess digits (carry over) should started from the rightmost group and then progress towards left sequentially.
• When all the above conditions are met then you are good to remove the \(\omega\) (with-operator).

So, finally the bi-multiplication formula can be precisely expressed as,

$$\boxed{x \omega y \times a \omega b = ax\hspace2pt\omega\hspace2pt]ax+by[^n\hspace2pt\omega\hspace2pt]by[^n}\hspace1em\ldots\text{where ]a[ = ]b[ = ]x[ = ]y[ = n}$$

Squaring formula

This is directly derivable from bi-multiplication formula.

$$
\begin{align}
(a \omega b)^2&\\
\Rightarrow& a \omega b \times a \omega b\\
\Rightarrow& \boxed{a^2\omega ]2ab[^n \omega ]b^2[^n}\hspace1em\ldots\text{where ]a[ = ]b[ = n}
\end{align}
$$

General Powering Formula

For now I will provide only the formula. The proof is provided later in this article. The way I derived it was to I manually find the values of \((a\omega b)^2\), \((a\omega b)^3\), \((a\omega b)^4\) and so on. I found a pattern in all these and from there I got this formula. It was later when it struck me as to how to prove it.

$$
\boxed{(a\omega b)^n = \omega\sum_{r=0}^n\hspace1em{]^nC_r \times a^{(n-r)} \times b^r[}^d}\hspace1em\ldots\text{where ]a[ = ]b[ = d}
$$

Notice that the above equation looks very similar to Binomial equation. In fact I found this to be true for many equations. If we replace \(+\) by \(\omega\) then we end up with a ‘digitized’ version of that equation.

Now let me demonstrate how to use this equation. Let’s find cube of 99, i.e. \(99^3\).

$$
\begin{align}
&(9\omega9)^3\\
&= ]^3C_0 \times 9^3 \times 9^0[^1 \omega ]^3C_1 \times 9^2 \times 9^1[^1\\
&= 9^3 \omega 3\times9^3 \omega 3\times9^3 \omega 9^3\tag{for brevity removing ].[ operator}\\
&= 729 \omega \underline{218}7 \omega \underline{218}7 \omega \underline{72}9\\
&= 729 \omega \underline{218}7 \omega \underline{225}9 \omega 9\tag{adding carry overs}\\
&= 729 \omega \underline{241}2 \omega 9 \omega 9\\
&= 970 \omega 2 \omega 9 \omega 9\\
&= 970299
\end{align}
$$

I didn’t underline the excess digits in leftmost group, since that is anyway not going to affect the result. Now let us find \(241^4\).

$$
\begin{align}
&(02\omega41)^4\\
&= \omega\sum_{r=0}^4\hspace1em]^4C_r \times 2^{(4-r)} \times 41^r[^2\\
&= ^4C_0.2^4\hspace0.5em\omega\hspace0.5em^4C_1.2^3.41\hspace0.5em\omega\hspace0.5em^4C_2.2^2.41^2\hspace0.5em\omega\hspace0.5em^4C_3.2.41^3\hspace0.5em\omega\hspace0.5em^4C_4.41^4\\
&= 16\omega\underline{13}12\omega\underline{403}44\omega\underline{5513}68\omega\underline{28257}61\\
&= 16\omega\underline{13}12\omega\underline{403}44\omega\underline{5796}25\omega61\\
&= 16\omega\underline{13}12\omega\underline{461}40\omega25\omega61\\
&= 16\omega\underline{17}73\omega40\omega25\omega61\\
&= 33\omega73\omega40\omega25\omega61\\
&= 3373402561
\end{align}
$$

Note, one important observation. Here you can immediately identify the last couple of digits of the final result by evaluating for the last term. Unlike in Binomial Theorem, where all the terms need to be added. So, there at least last few digits of all the terms need to be added to get the  last few digits of the final result. This is an advantage for General Powering formula, but unfortunately we cannot say the for other digit groups which are at higher place order. The reason is simple, they may have to be added with carry-overs from the groups (terms) on the right.

A corollary of the General Formula of Transformation (for two digits)

With to Plus form

$$a\omega b = 10a+b$$

The above equation is true only when \(b\) has one digit. The more general equation is:-

$$\boxed{a\omega b = a.10^d+b}\hspace1em\text{where d}=]b[\hspace1em\ldots\text{Corollary 1}$$

Plus to With form

$$a\omega b = 10a +b \Rightarrow a\omega b = 9a + a + b \Rightarrow \boxed{a+b = a\omega b -9a}\hspace1em\ldots\text{Corollary 2}$$

Negative digits

We all are aware of negative numbers but in digit equations we might end up with negative digits! I see no physical significance of that but let’s try to find out some mathematical meaning of this.

$$
\begin{align}
a\omega(-b) &= 10a + (-b)\\
&= 10a -b\\
&= 10a + b -2b\\
&= a\omega b -2b
\end{align}
$$

Altire,

$$
\begin{align}
a\omega(-b) &= -(-10a + b)\\
&= -(10a+b -20a)\\
&= -(a\omega b – 20a)\\
&= -(a\omega b) +20a
\end{align}
$$

$$
\therefore \boxed{a\omega(-b) = a\omega b-2b = -(a\omega b)+20a}
$$

Remember, \((-a)\omega b \neq -(a\omega b)\). The former means that only the digit \(a\) is negative. The later means that the complete number is negative.

Similarly,

$$
\boxed{(-a)\omega b = a\omega b-20a = -(a\omega b)+2b}
$$

Also similarly evaluating, we get,

$$
\boxed{(-a)\omega(-b) = -(a\omega b)}\hspace1em\ldots\text{Corollary 3}
$$

So, we see that in a negative number all the digits too are negative. Not surprising, since in a positive number all digits are positive.

Proof of General Powering Formula

Since I already know this formula, so I just need to prove that it is correct. For this the best tool is Mathematical Induction. So, the formula to prove is

$$(a\omega b)^n = \omega\sum_{r=0}^n\hspace1em{]^nC_r \times a^{(n-r)} \times b^r[}^d$$

Case 1: n = 0

So,

$$
\begin{align}
L.H.S. &= (a\omega b)^0 = 1\\
\\
\\
R.H.S. &= \omega\sum_{r=0}^0\hspace1em ^0C_r \times a^{(0-r)} \times b^r\\
&= 1\times a^0 \times b^0\\
&= 1\tag{Proved}
\end{align}
$$

Case 2: n = 1

$$
\begin{align}
L.H.S.&= (a\omega b)^1 = a\omega b\\
\\
\\
R.H.S. &= \omega\sum_{r=0}^1\hspace1em ^1C_r \times a^{(1-r)} \times b^r\\
&= ^1C_0. a^1. b^0 \omega ^1C_1.a^0.b^1\\
&= a\omega b\tag{Proved}
\end{align}
$$

Case 3: For n

Suppose that the formula is true for all values of \(n\), then by Mathematical Induction, the formula must be valid for \(n+1\) too.

So,

$$
\begin{align}
&(a\omega b)^{(n+1)}\\
&= (a\omega b)^n \times (a\omega b)\\
&= (a\omega b)^n \times (a\times 10^d+b)\hspace1em\ldots\text{where d}=]b[\tag{By Corollary 1}\\
&= 10^da\times(a\omega b)^n + b\times (a\omega b)^n\\
&= 10^da\times(\omega\sum_{r=0}^n\hspace1em ^nC_r\times a^{(n-r)} \times b^r) +\\
&\hspace2em b\times(\omega\sum_{r’=0}^n\hspace1em ^nC_{r’}\times a^{(n-r’)} \times b^{r’}) \tag{Since formula is correct for n}\\
&= 10^d\times(a^{(n+1)} \hspace2pt\omega\hspace2pt \omega\sum_{r=1}^n\hspace1em ^nC_r\times a^{(n+1-r)} \times b^r) +\\
&\hspace2em(\omega\sum_{r’=0}^{(n-1)}\hspace1em ^nC_{r’}\times a^{(n-r’)} \times b^{(r’+1)} \hspace2pt\omega\hspace2pt b^{(n+1)})\tag{1}
\end{align}
$$

Since \(]a[\hspace2pt=\hspace2pt]b[\hspace2pt=\hspace2pt d\), so every term of \(\omega\sum\) too will have \(d\) digits, as per General Powering Formula. This implies that term \(b^{(n+1)}\) in equation (1) too has \(d\) digits. So, if we can rewrite (1) as below:-

$$
\begin{align}
a^{(n+1)} \hspace2pt\omega\hspace2pt&\omega\sum_{r=1}^n\hspace1em ^nC_r\times a^{(n+1-r)} \times b^r\tag{2}\\
+\hspace2em\hspace2em &\omega\sum_{r’=0}^{(n-1)}\hspace1em ^nC_{r’}\times a^{(n-r’)} \times b^{(r’+1)} \hspace2pt\omega\hspace2pt b^{(n+1)}\tag{3}\\
a^{n+1} \hspace2pt\omega\hspace2pt & \overline{\omega\sum_{r=1}^n\hspace1em {^nC_r\times a^{n+1-r} \times b^r \brace + ^nC_{(r-1)}\times a^{n-(r-1)} \times b^{(r-1)+1}} \hspace2pt\omega\hspace2pt b^{n+1}}\\
\Rightarrow a^{n+1} \hspace2pt\omega\hspace2pt &\bigg( \omega\sum_{r=1}^n\hspace1em ^nC_r\times a^{n+1-r} \times b^r + ^nC_{r-1}\times a^{n+1-r} \times b^{r}\bigg) \hspace2pt\omega\hspace2pt b^{n+1}\tag{4}
\end{align}
$$

Notice that in the above, I have removed the \(10^d\) factor since all it did is to shift its term by \(d\) digits to left, that is, by exact number of digits in the \(b^{(n+1)}\) group. After removing the first digit group from equation (2) and last digit group from equation (3), \(\omega\sum\) in both the equations have \(n-1\) digit groups left. Each of these digit groups will get added term by term, i.e. group \(1\) of (2) will get added to group \(0\) of (3) and so on. So this implies \(r’ = r – 1\).

Further simplifying,

$$
\begin{align}
(4)&= a^{n+1} \hspace2pt\omega\hspace2pt \bigg(\omega\sum_{r=1}^n\hspace1em a^{n+1-r}.b^r(\frac{n!}{r!(n-r)!}+\frac{n!}{(r-1)!(n-r+1)!})\bigg) \hspace2pt\omega\hspace2pt b^{n+1}\\
&= a^{n+1} \hspace2pt\omega\hspace2pt \bigg(\omega\sum_{r=1}^n\hspace1em a^{n+1-r}.b^r.\frac{(n+1)!}{r!(n+1-r)!}.(\frac{n+r-1}{n+1}+\frac{r}{n+1})\bigg) \hspace2pt\omega\hspace2pt b^{n+1}\\
&= a^{n+1} \hspace2pt\omega\hspace2pt \bigg(\omega\sum_{r=1}^n\hspace1em a^{n+1-r}.b^r.^{n+1}C_r.(1)\bigg) \hspace2pt\omega\hspace2pt b^{n+1}\\
&= \bigg(C_0^{n+1}.a^{n+1-0}.b^0\bigg) \hspace2pt\omega\hspace2pt \bigg(\omega\sum_{r=1}^n\hspace1em C_r^{n+1}.a^{n+1-r}.b^r\bigg) \hspace2pt\omega\hspace2pt \bigg(C_{n+1}^{n+1}.a{(n+1-(n+1))}.b{n+1}\bigg)\\
&= \omega\sum_{r=0}^{(n+1)}\hspace1em ^{(n+1)}C_r.a^{(n+1)-r}.b^r\\
\end{align}
$$

So, proved that \((a\omega b)^{(n+1)} = \omega\sum_{r=0}^{(n+1)}\hspace1em ^{(n+1)}C_r.a^{(n+1)-r}.b^r\) when \((a\omega b)^n = \omega\sum_{r=0}^n\hspace1em ^nC_r.a^{n-r}.b^r\).

So, all cases proved.

Application of Digit Math

Now time to see Digit Math in action. I will be using this to prove some empirical concepts.

Link to list of applications of Digit Math.