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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

<p>Show that for any prime <span class="math-container">$p$</span>  there exist prime <span class="math-container">$q$</span> such that</p>

<p><span class="math-container">$$T_b(q)&gt;T_b(p)$$</span></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 $T_b(q)>T_b(p)$.

rtybase
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Pruthviraj
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  • 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
  • Note that $q<p$. Thus choosing $p=2$, it is not correct. – IMOPUTFIE Sep 14 '19 at 15:48
  • I suspect that "for every base..." is meant to be a part of the premise. Currently, the claim can be interpreted as "$\forall p\ \exists q\ \forall b\ \ldots$" which is clearly false. – metamorphy Sep 14 '19 at 19:22
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    ... and the "$\forall p\ \forall b\ \exists q\ \ldots$" case is an easy consequence of this. – metamorphy Sep 14 '19 at 19:31
  • @metamorphy Actually I'm not understand your conclusion, let suppose b=2 then what's wrong going in this problem, please explain – Pruthviraj Sep 14 '19 at 19:46
  • I mean, currently your claim can be understood as "for every prime $p$ there exists a prime $q>p$ such that for every base $b$ we have $T_b(q)>T_b(p)$", which is false (we can simply take $b=q$, when $T_q(q)=1$). – metamorphy Sep 15 '19 at 04:45
  • @metamorphy Now I fixed my statement please check. – Pruthviraj Sep 15 '19 at 06:47
  • The answer is yes, because ... see this. – rtybase Sep 15 '19 at 15:24

1 Answers1

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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.

rtybase
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