I was crossing over the fact that some three digit (0-9) numbers can be expressed with the equation:

$(n+2m)*10^2 + (n+m)*10^1 + n*10^0 = 3(70m+37n)$

But how would you determine, how many percents of the all three digit numbers from 000 to 999 can be expressed with the equation? As a follow up, I'd like to know if there is a single equation to express rest of the three digit numbers in a similar manner?

  • $\begingroup$ Are negative values for $m$ and $n$ allowed? $\endgroup$ – Christian Blatter Jan 2 '18 at 15:56
  • $\begingroup$ No, just positive 0-9. $\endgroup$ – MarkokraM Jan 2 '18 at 15:56
  • $\begingroup$ $0 \le n \le 9$ and $0 \le m \le 4$ is all I have understood so far. $\endgroup$ – Mohammad Zuhair Khan Jan 20 '18 at 14:58
  • $\begingroup$ And that the difference between the numbers has to be constant. So all numbers of the form $n*10^2+n*10+n$ (there are $10$ such numbers) are allowed. $\endgroup$ – Mohammad Zuhair Khan Jan 20 '18 at 15:03
  • $\begingroup$ BTW do you consider $012$ as a three digit number? $\endgroup$ – Mohammad Zuhair Khan Jan 20 '18 at 15:03

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