Does there exist a Benny number? For positive integers $x$, let $S(x)$ denote the sum of the  digits of $x$, and let $L(x)$ denote the number of digits of $x$. It can be shown that there are infinitely many numbers that cannot be expressed as $x+S(x)$ or $x+L(x)$ or $x+S(x)+L(x)$ individually or any method of those three i.e. $x+S(x)$, $x+L(x)$, and $x+S(x)+L(x)$ 


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*[Edit note: the question this is based on, Natural numbers not expressible as $x+s(x)$ nor $x+s(x)+l(x)$,  does not include $x+L(x)$ among the allowed formation methods].


And now a Benny number or Naughty number is a natural number greater than one that cannot be expressed as $x+S(x)$, $x-S(x)$, $x+L(x)$, $x-L(x)$, $x+S(x)+L(x)$, $x+S(x)-L(x)$, $x-S(x)+L(x)$, nor $x-S(x)-L(x)$. 
I've verified that there are no Benny numbers up to $10^{20}$. My question is: Does there exist a Benny number?

For all $n$, we either have $L(n)=L(n+L(n))$ or $L(n)=L(n-L(n))$ (or both).  If $L(n)=L(n+L(n))$, let $x=n+L(n)$.  Then 
$$x-L(x)=(n+L(n))-L(n+L(n))=(n+L(n))-L(n)=n$$
Likewise, if $L(n)=L(n-L(n))$, let $x=n-L(n)$, in which case
$$x+L(x)=(n-L(n))+L(n-L(n))=(n-L(n))+L(n)=n$$
