A number theory problem

Lemma:

Let $n$ be a positive integer and let $Q(n)$ denote the sum of digits of $n$ then if $n \equiv r \pmod{9}$ then $Q(n) \equiv r \pmod{9}$ .

Proof:

Let $n = \displaystyle \sum_{i=0}^k a_i \cdot {10}^i$ . Since ${10}^i \equiv 1 \pmod{9} \ \forall \ i \in \mathbb{N} \cup \{0\}$ , so if

$n \equiv r \pmod{9} \implies \displaystyle \sum_{i=0}^k a_i = Q(n) \equiv r \pmod{9}$ .

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Now, we have

$\begin{aligned} & n - 2 Q(n) = 2016 \\ \equiv & r - 2r = 0 \pmod{9} \\ \implies & r = 0 \end{aligned}$

Hence $n$ is a multiple of $9$ .

Now if $n$ is a multiple of $9$ then $Q(n) = 9$ and so we get

$n - 2 \cdot 9 = 2016 \implies n = \boxed{2034}$

n-2q(n)=2016 Let's assume that n=2ABC cause wer'e looking for the n to be minimal so Q(n)= 2+B+A+C pluging it in: 2000+100A+10B+C-2(2+B+A+C)=2016 After a bit of algebra: 98A-8B-C=20 let A be 0 so that: 8B-C=20 we can see that the only option is B=3 and C=4 So that: n=2034