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

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


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

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


Let base $b$ is given

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



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

  • $\begingroup$ 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. $\endgroup$ – IMOPUTFIE Sep 14 '19 at 15:20
  • $\begingroup$ @IMOPUTFIE Okay thanks, you are right, I change $n_0\ne 0$ $\endgroup$ – Pruthviraj Sep 14 '19 at 15:36
  • $\begingroup$ 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: $T(q)>T(p)$ $\endgroup$ – IMOPUTFIE Sep 14 '19 at 15:40
  • $\begingroup$ @IMOPUTFIE Okay if change $T_b(q)\geq T_b(p)$ then is it satisfied for problem? $\endgroup$ – Pruthviraj Sep 14 '19 at 15:45
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    $\begingroup$ ... and the "$\forall p\ \forall b\ \exists q\ \ldots$" case is an easy consequence of this. $\endgroup$ – 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.


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