Prime numbers are related by $q=2p+1$ Let primes $p$ and $q$ be related by $q=2p+1$.  Prove that there is a positive multiple of $q$ for which the sum of its digits does not exceed $3$.
My work so far:
$p,q -$ primes and $q=2p+1 \Rightarrow \exists n,k \in \mathbb N: n=qk$ and $S(n) \le 3$
$p=2 \Rightarrow q=5 \Rightarrow 5|10=n; S(10)=1 \le3$.
$p=3 \Rightarrow q=7 \Rightarrow 7|21=n; S(21)=3 \le3$.
$p=5 \Rightarrow q=11 \Rightarrow 1|11=n; S(11)=2 \le3$.
$p=7 \Rightarrow q=15=5 \cdot3 $.
$p=11 \Rightarrow q=23 \Rightarrow 23|n; S(n) \le3$.
 A: (ALMOST an answer) In order for $S(kq)\leq3$, $kq$ must be in one of the following forms:
(1): $$kq=2*10^n+1$$
(2): $$kq=10^n+1$$
(3): $$kq=10^n+10^m+1$$, where $n\neq m$
We will look into the cases one by one and show that there must exist some $n$ or $n,m$ that satisfy at least one of the above cases, following @Peter 's direction.

Case (1):
$2*10^n+1\equiv2p+1$ (mod $q$)
$10^n\equiv p$
Now let $g$ be a primitive root for mod $q$, such that $g^a\equiv10$ and $g^b\equiv p$ (mod $q$), where $1\leq a,b\leq 2p$
Then the question reduces to whether $g^{an}=g^b$ (mod q) is solvable or not. And since the primitive root has order $2p$,
$an\equiv b$ (mod $2p$)
Is only solvable iff
$b\equiv 0$ (mod $GCD(a,2p)$)
So if $a,2p$ are coprime, there must be an integer $0\geq n\geq 2p$ that satisfies $kq=2*10^n+1$

Case (2):
If the above case fails, that means $b\not\equiv 0$ (mod $GCD(a,2p)$)
Then $a$ and $2p$ must not be coprime, since $p$ is a prime, $a=p$ or $a=2$.
$2p\equiv -1\equiv 10^n$ (mod $q$)
Let $g^c\equiv 2$ (mod $q$), $1\leq c\leq 2p$
We know,
$g^p\equiv g^{b+c}\equiv -1$
Because $(g^p)^2\equiv 1$ (mod $q$) but $p\not\equiv 0$ (mod $2p$)
If $g^p\equiv g^{an}$ is solvable
Then,
$p\equiv an$ (mod $2p$) has a solution for $n$, and it has solution iff
$p\equiv 0$ (mod $GCD(a,2p)$)

Sub-case: $a=p$, 
$p\equiv 0$ (mod $a=p$) is trivial
So there must be solution for $n$ if $a=p$

Sub-case: $a=2$,  there must be solution for $n$ if $b,c$ have the same parity.
In fact, $1 \leq b,c \leq 2p$ implies that $2\leq b+c\leq 4p$, 
so $b+c=p$ or $b+c=3p$, 
because if $b+c=2p$ or $4p$, it would contradict that $g^{b+c}\equiv -1$ (mod $q$)
So $b+c\equiv 1$ (mod $a=2$), for the case that $a=2$, there will be no solution in the form of (1) or (2).

Case (3):
$kq=10^n+10^m+1$
$2p\equiv 10^n+10^m$ (mod $q$)
$g^{b+c}\equiv -1\equiv g^{2n}+g^{2m}$
I cant figure how to show that this type of equation has solution $n,m$. If you can prove that there exist solutions, then we have covered all possible cases. 
(Side note: it can be shown that if $a=p$, then there must be no solution in the form $10^n+10^m+1$ since
$g^{pm}+g^{pn}+1\not\equiv 0$ (mod $q$) since $g^{pm},g^{pn}\equiv \pm 1$)
