The question —

"Find all 2 digit good numbers. A natural number is said to be good if and only if it is divisible by the product of its digits."

My work —

Let $\overline{ab}$ be a good number.

$\Longrightarrow ab|\overline{ab}$

$\Longrightarrow \frac{10a+b}{ab} \in \mathbb{Z}$

$\Longrightarrow (\frac{10}{b} + \frac1a) \in \mathbb{Z}$

$\Longrightarrow a = 1$ and $b = 1,2,5$

$\Longrightarrow \overline{ab} = 11, 12, 15$

But I don't think that this is all. I have missed out on some solutions. $24$ was one I could think of but not get. Is there a mistake in my solution? Can you please help me?

  • $\begingroup$ One must remember that two fractions can sum to an integer. Your case of $24$ corresponds to $\frac{10}{4} + \frac{1}{2} = 3 \in \mathbb{Z}$. $\endgroup$ – Hyperion May 11 '19 at 4:46
  • $\begingroup$ Hmm...True. But how am I gonna find all solutions that I missed out on? $\endgroup$ – PranavGupta53535 May 11 '19 at 4:49
  • $\begingroup$ @JeanMarie that question was related to the sum of digits,not the product. $\endgroup$ – PranavGupta53535 May 11 '19 at 5:22
  • $\begingroup$ @JeanMarie the last question was related to the sum of the digits, and this one is related to the product of the digits. Can you please tell me how the title of my last question was misleadin. $\endgroup$ – PranavGupta53535 May 11 '19 at 5:34
  • $\begingroup$ All right. I have been mislead by the fact that you use the same term "good number" for two different definitions. I erase my previous comment. $\endgroup$ – Jean Marie May 11 '19 at 5:35

If $ab|10a+b$, then in particular, you have $b|10a+b$ and $a|10a+b$. So, $b|10a$ and $a|b$. I think from here is easier to check all the possible cases.

You can check the cases when $a=1,2,3,4$ by hand. And if $a\geq 5$, deduce that $b=a$, so $a^2|11a$, so $a|11$, which gives you no solutions.

  • $\begingroup$ You're welcome. Just edited it, to show that you actually only need to check cases $a=1,2,3,4$. $\endgroup$ – Julian Mejia May 11 '19 at 5:05

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