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

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

• Dear Pruthviraj, first note that your sum telescopes to $n_l-n_0$. Thus choosing say 19 ($T_10=8), you can't reach a higher number than 8(Contradicting the greater than). Besides, could you clarify your use of leading zeroes. I'd be very appreciative. – IMOPUTFIE Sep 14 '19 at 15:20 • @IMOPUTFIE Okay thanks, you are right, I change$n_0\ne 0$– Pruthviraj Sep 14 '19 at 15:36 • But when$n_0$cannot be$0$, then as already stated 8 is the maximum for$T_10$, hence there doesn't exist$q$such that:$T(q)>T(p)$– IMOPUTFIE Sep 14 '19 at 15:40 • @IMOPUTFIE Okay if change$T_b(q)\geq T_b(p)$then is it satisfied for problem? – Pruthviraj Sep 14 '19 at 15:45 • ... and the "$\forall p\ \forall b\ \exists q\ \ldots\$" case is an easy consequence of this. – metamorphy Sep 14 '19 at 19:31

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.