Show that there always exist prime $q$ such that $T_b(q)>T_b(p)$ for any prime $p$

Question

Let $n$ be a positive integer with some base $b$. Then $n$ can be represent as

$$n=( n_1 ... n_{l-1} n_l)_b$$

Let $T$ be the function defined as

$$T_b(n)= \sum_{i=1}^{l}n_i$$

Example

Let $n= (3596)_{10} = 3596$

So $$T_{10}(3596) = 3+5+9+6=23$$

Question

Let base $b$ is given

Show that for any prime $p$ there exist prime $q$ such that

$$T_b(q)>T_b(p)$$

Example

Let base $b = 2$ and $p=17=(10001)_2$ then $T_2(17)=2$

So we can choose any prime $q\in \{7,11,13,19,...\}$ for $T_b(q)>T_b(p)$.

Answer 1

Following the comment I posted yesterday, we can take the prime $p$ and convert it to base $b$, $(p)_{10}=(n_1n_2...n_{t_p})_b$. Then, there exists a prime $q$ which starts with the sequence of digits $n_1,n_2,...,n_{t_p}$, i.e. $(q)_{10}=(n_1n_2...n_{t_p}...n_{t_q})_b$ and $n_{t_q}\ne0$ (otherwise $b\mid q$, but $q$ is prime). Then $$T_b(q)=n_1+n_2+...+n_{t_p}+...+n_{t_q}>n_1+n_2+...+n_{t_p}=T_b(p)$$

This is a direct result of the fact that for any sequence of digits in base $b$, there is a prime number starting with that sequence, in that base.