Only Number Twice Sum of Digits is 18
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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:
\(\ds n\) | \(=\) | \(\ds 10 x + y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \left({x + y}\right)\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 8 x\) | \(=\) | \(\ds y\) |
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): $18$