Only Number Twice Sum of Digits is 18

Theorem

There exists only one (strictly) positive integer that is exactly twice the sum of its digits.

Proof

Let $n$ be equal to twice the sum of its digits.

Suppose $n$ has $1$ digit.

Then $n$ equals the sum of its digits.

Thus $n$ does not have only $1$ digit.

Suppose $n$ has $3$ digits.

Then $n > 99$.

But the highest number that can be formed by the sum of $3$ digits is $9 + 9 + 9 = 27$.

Similarly for if $n$ has more than $3$ digits.

Hence, if there exists such a number, it must have exactly $2$ digits.

Let the $2$ digits of $n$ be $x$ and $y$.

Thus:

Since $x$ and $y$ are both single digits, it follows that $x = 1$ and $y = 8$.

Hence the result.

$\blacksquare$

Sources

• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $18$