find all two digit numbers with the property that if i sum the digits and add this sum to the product of the digits, i get the number.

I noticed that every number that ends with $9$ works but i would like to know the reason behind this.

for example:

$19 \rightarrow 1+9=10 \rightarrow 10+(9 \times 1)=19$

Also works for $29,39,\ldots,89,99$

Any reason why this works?

  • $\begingroup$ It is not true that $1+9 = 10 + (9\times 1)$. I understand what you meant, but that is certainly not standard use of notation and in many contexts it can lead to confusion. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 9 '15 at 4:45
  • $\begingroup$ you are right, lemme edit. $\endgroup$ – Caddy Heron Oct 9 '15 at 5:03

$x+y + xy=10x + y\implies 9x=xy$

If $x\ne 0$ then $y=9$

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  • $\begingroup$ Made so much sense, thanks! $\endgroup$ – Caddy Heron Oct 10 '15 at 18:41

Well to answer your second question about why this works:

We can write a two-digit number as $10t+u$ where $t \leq 9$ is the integer in the tens place and $u \leq 9$ is the integer in the ones place. So your question leads to:

$10t+u = t+u+tu$

$9t = tu$


So it's clear that the unit digit must be a 9 and that works for all integers $t \in \left\lbrace 1,2, \cdots, 9 \right\rbrace$

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