2-Digit Positive Integer equals Product plus Sum of Digits iff ends in 9

Theorem

Let $n$ be a $2$-digit positive integer.

Then:

$n$ equals the sum added to the product of its digits

if and only if:

the last digit of $n$ is $9$.

Proof

Let $n = 10 x + y$ where $0 < x \le 9, 0 \le y \le 9$.

Then:

\(\ds \paren {x + y} + \paren {x y}\)

\(=\)

\(\ds 10 x + y\)

\(\ds \leadstoandfrom \ \ \)

\(\ds x y - 9 x\)

\(=\)

\(\ds 0\)

\(\ds \leadstoandfrom \ \ \)

\(\ds x \paren {y - 9}\)

\(=\)

\(\ds 0\)

\(\ds \leadstoandfrom \ \ \)

\(\ds y\)

\(=\)

\(\ds 9\)

as $x \ne 0$

$\blacksquare$

Sources

• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $39$