Number pairs such that their fraction $\frac{AB}{CD}$ (in decimal representation) is the same as the fraction of sums $\frac{A+B}{C+D}$

Question

I'm looking for all number pairs $(M,N)$ where $M,N$ have an even amount and the same amount of digits $2k$. Let's say $M=m_1m_2...m_{2k},N=n_1n_2...n_{2k}$ in decimal representation with their digits $m_i,n_i$. I am curious, for which number pairs we get the equality $$ \frac{m_1m_2...m_{2k}}{n_1n_2..n_{2k}} = \frac{m_1m_2...m_k + m_{k+1}...m_{2k}}{n_1n_2...n_k + n_{k+1}...n_{2k}} $$

It would be interesting how this works out in other number systems (besides decimal) as well.

I've already found a simple example, if the last $r \geq k$ digits are zeros, we get e.g.: $$ \frac{230000}{189000} = \frac{230 + 0}{189 + 0} = \frac{230}{189} $$

But I'm interested in more non-trivial examples.

As we already have an answer. What about applying this multiple times? How can we find a number, such that using this procedure iteratively we still get the same fraction?

Example: $$ \frac{4228}{5436} = \frac{42+28}{54+36} = \frac{70}{90} = \frac{7+0}{9+0} = \frac{7}{9} $$

Answer 1

There is a fact explaining that the solutions are somewhat, how to say, deceiving.

Consider the case of 6 digits numbers, like $230189$. It can be written:

$$230000+189=230 \times \underbrace{1000}_N + 189$$

With this notation, your condition can be written:

$$\frac{Na+b}{Nc+d}=\frac{a+b}{c+d}=\color{red}{\frac{(N-1)a}{(N-1)c}=\frac{a}{c}}\tag{1}$$

(see remark below) which is possible iff considering the extreme fractions in (1)

$$(Na+b)c=(Nc+d)a \ \ \iff \ \ ad=bc$$

i.e, if and only if

• (case 1) $2$ of them at least are zero (you have given such an example in your question).

• (case 2) none of them is zero, we have this proportionality ratio:

$$\frac{b}{d}=\frac{a}{c}$$

which (for case 2!) is a necessary and sufficient condition for a solution.

Remark: The red fractions in (1) have been obtained by using the rule

$$\frac{P}{Q}=\frac{R}{S}=\frac{uP+vR}{uQ+vS}$$

valid for any factors $u,v$, under the condition $uQ+vS \ne 0$ of course.

Answer 2

More complex answer than @Jean Marie:

Let's calculate both fractions and look which conditions are sufficient to make these fractions equal: \begin{align} \frac{m_1m_2...m_{2k}}{n_1n_2...n_{2k}} &= \frac{m_1m_2...m_k + m_{k+1}...m_{2k}}{n_1n_2...n_k + n_{k+1}...n_{2k}} \\ \frac{\sum_{i=1}^{2k} m_i\cdot 10^{2k-i}}{\sum_{i=1}^{2k} n_i\cdot 10^{2k-i}}&=\frac{\sum_{i=1}^{k} m_i\cdot 10^{k-i}+\sum_{i=1}^k m_{i+k}\cdot 10^{k-i}}{\sum_{i=1}^{k} n_i\cdot 10^{k-i}+\sum_{i=1}^k n_{i+k}\cdot 10^{k-i}} \\ \left(\sum_{i=1}^{2k} m_i\cdot 10^{2k-i}\right)\left(\sum_{i=1}^{k} n_i\cdot 10^{k-i}+\sum_{i=1}^k n_{i+k}\cdot 10^{k-i}\right)&=\left(\sum_{i=1}^{k} m_i\cdot 10^{k-i}+\sum_{i=1}^k m_{i+k}\cdot 10^{k-i}\right)\left(\sum_{i=1}^{2k} n_i\cdot 10^{2k-i}\right) \end{align}

Let's evaluate the LHS and RHS after each other: \begin{align} &\left(\sum_{i=1}^{2k} m_i\cdot 10^{2k-i}\right)\left(\sum_{i=1}^{k} n_i\cdot 10^{k-i}+\sum_{i=1}^k n_{i+k}\cdot 10^{k-i}\right) \\ &=\left(\sum_{i=1}^{k} m_i\cdot 10^{2k-i} + \sum_{i=1}^km_{i+k}\cdot 10^{k-i}\right)\left(\sum_{i=1}^{k} n_i\cdot 10^{k-i}+\sum_{i=1}^k n_{i+k}\cdot 10^{k-i}\right) \\ &=\left(\sum_{i=1}^{k} m_i10^{2k-i}\right)\left(\sum_{i=1}^{k} n_i10^{k-i}\right) + \left(\sum_{i=1}^{k} m_i10^{2k-i}\right)\left(\sum_{i=1}^k n_{i+k}10^{k-i}\right) + \left(\sum_{i=1}^km_{i+k}10^{k-i}\right)\left(\sum_{i=1}^{k} n_i10^{k-i}\right) +\left(\sum_{i=1}^km_{i+k}10^{k-i}\right)\left(\sum_{i=1}^k n_{i+k}10^{k-i}\right) \\ &=:S_{11} + S_{12} + S_{21}+ S_{22} \end{align} Analogously for the RHS we get: \begin{align} &\left(\sum_{i=1}^{k} m_i\cdot 10^{k-i}+\sum_{i=1}^k m_{i+k}\cdot 10^{k-i}\right)\left(\sum_{i=1}^{2k} n_i\cdot 10^{2k-i}\right) \\ &=\left(\sum_{i=1}^{k} m_i\cdot 10^{k-i} + \sum_{i=1}^km_{i+k}\cdot 10^{k-i}\right)\left(\sum_{i=1}^{k} n_i\cdot 10^{2k-i}+\sum_{i=1}^k n_{i+k}\cdot 10^{k-i}\right) \\ &=\left(\sum_{i=1}^{k} m_i10^{k-i}\right)\left(\sum_{i=1}^{k} n_i10^{2k-i}\right) + \left(\sum_{i=1}^{k} m_i10^{k-i}\right)\left(\sum_{i=1}^k n_{i+k}10^{k-i}\right) + \left(\sum_{i=1}^km_{i+k}10^{k-i}\right)\left(\sum_{i=1}^{k} n_i10^{2k-i}\right) +\left(\sum_{i=1}^km_{i+k}10^{k-i}\right)\left(\sum_{i=1}^k n_{i+k}10^{k-i}\right) \\ &=:T_{11} + T_{12} + T_{21}+ T_{22} \end{align}

Comparing those equations we get \begin{align} S_{11} &= T_{11} \\ S_{12} &= 10^k T_{12} \\ 10^k S_{21} &= T_{21} \\ S_{22} &= T_{22} \end{align}

This means the fractions are the same if \begin{align} T_{11}+ 10^k T_{12} + 10^{-k}T_{21} + T_{22} &= T_{11} + T_{12} + T_{21} + T_{22} \\ \Leftrightarrow (10^k-1) T_{12} &= (1- 10^{-k}) T_{21} \\ \Leftrightarrow 10^k(10^k-1)T_{12} &= (10^k-1)T_{21} \\ \Leftrightarrow 10^k T_{12} &= T_{21} \\ \Leftrightarrow 10^k\left(\sum_{i=1}^{k} m_i10^{k-i}\right)\left(\sum_{i=1}^k n_{i+k}10^{k-i}\right) &= 10^k\left(\sum_{i=1}^km_{i+k}10^{k-i}\right)\left(\sum_{i=1}^{k} n_i10^{k-i}\right) \\ \Leftrightarrow m_1...m_k \cdot n_{k+1}...n_{2k} &= m_{k+1}...m_{2k}\cdot n_1n_2...n_k \end{align}

To find those numbers, one might take a number with more than $4$ prime divisors (counted with multiplicity) and then take each of them for being one part of above equation, but all numbers have to have the same amount of digits!

Example: Take $2^3\cdot 3^2 \cdot 7^2 \cdot 11=38808$ with the four numbers $11,7\cdot 2=14,7\cdot 3=21,2^2\cdot 3 = 12$, multiplying each two pairs results in $154\cdot 252=132\cdot 294$ and thus we get:

$$ \frac{154294}{132252} = \frac{154+294}{132+252} $$