A recent meme showed the following "algebra":

$$\frac{777}{4773} = \frac{7(77)}{4*(77)*3}=\frac{7}{43}$$

which is, amazingly, true!

Clearly, there are infinitely many natural number pairs that satisfy this property (by adding on a zero to numerator and denominator, hence multiplying by 10), but we ignore those solutions as trivial.

We define $\text{value}(a_M,a_{M-1},...,a_1,a_0) = \sum_{m=0}^M a_m 10^m$

let $A, B \in \mathbb{N}$ be given. We then say that $A = \text{value}(a_M,a_{M-1},...,a_1,a_0)$ and $B= \text{value}(b_N,b_{N-1},...,b_1,b_0)$ are dodgy-reducible if


where the $k_i$s and the $l_i$s are any selection of the numbers $\{0,1,...,M\}$ and $\{0,1,...,N\}$ respectively.

The first handful are:

64  16
65  26
95  19
98  49
121 22
132 33
136 34
143 44
154 55
165 66
176 77
184 138
185 148
187 88
192 96
194 97
195 39
196 49
196 98
198 99

I have tried (and failed) to make any analytic headway, but my friend wrote a program to naïvely iterate through all integer pairs below a given threshold, yielding the following distribution:

0 - 300 (excluding trivial solutions) 0-300 0-~4800 (excluding trivial solutions) 0-4800

The question: Is this a special case of a known sequence in number theory, and if not, what can be determined about its distribution?

  • 1
    $\begingroup$ A common way to write $\text{value}(a_M,a_{M-1},\cdots,a_1,a_0)$ is just the variables concatenated with a line over them: $$\overline{a_0a_1\cdots a_{M-1}a_M}$$ This is just how we would represent the number if we knew what the variables are: if $a=2$ and $b=3$, then $\overline{ab}=23$. However, since you don't state that the $a_i$ are single-digit integers ($0$ through $9$), this may not be applicable to this situation $\endgroup$ May 15, 2017 at 14:02
  • $\begingroup$ Also, it seemed like you'd made some typo's; I edited your post, please have a look and make sure I didn't change it to something you didn't mean $\endgroup$ May 15, 2017 at 14:09
  • $\begingroup$ I don't think your definition of dodge-reducible is complete. You haven't said anything about digits being the same. That is, not all subsequences are acceptable: you only want those subsequences the leave out identical subsequences from the original. $\endgroup$ May 15, 2017 at 20:37
  • 1
    $\begingroup$ In case that wasn't clear, an example: 13/26 = 3/6 where we cancel the 1 and the 2 is not what you want, but it is allowed by your definition. $\endgroup$ May 15, 2017 at 22:17
  • 1
    $\begingroup$ see also: MathWorld which also gives the more dull name of Anomalous Cancellation. $\endgroup$
    – mdave16
    May 18, 2017 at 12:53


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