BRILLIAthon Day $2$ , Problem $1$ of $2$

For integers $0<n<m$ , $n \bmod m = n$ . Therefore,

$\begin{aligned} x \bmod (x+1) + x & = 20 \\ x + x & = 20 \\ 2x & = 20 \\ \implies x & = 10 \end{aligned}$

The answer is $x+9 = \boxed{19}$ .

$x \pmod {(x + 1)}$ stands for the remainder of $x$ when divided by $x + 1$ , which is just $x$ for any positive $x$ , since $x \geq \frac{1}{2}(x + 1)$ .

Therefore, the equation becomes $x + x = 20$ , which solves to $x = 10$ , and $10 + 9 = \boxed{19}$ .

@Yajat Shamji - Hope I'm not late, I was sleeping, timezone differences and all :)

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$x \mod y$ Just equals to the value of the remainder, when $x$ is divided by $y$ . Here's a fun GIF to explain the Modulo Operator visually -

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Now, Let's look at our problem -

$\begin{aligned} x \mod (x+1) + x &= 20 \end{aligned}$

$x \mod (x + 1)$ must be equal to $x$ as for all integers $a \Rightarrow$ $0 < a < b, a \mod b = a$

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Simplifying the problem -

$\begin{aligned} x \mod (x+1) + x &= 20 \\ x + x &= 20 \\ 2x &= 20 \\ \Rightarrow x &= 10\end{aligned}$

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As the answer is $x+9 \Rightarrow$ $10 + 9 =$ $\textbf{\Huge \boxed{19}}$