I agree that the sum of the base ten digits seems rather peripheral. However: given a positive integer $M;$ let $S_M$ be the set of primes $p$ so that the base $10$ digit sum of $Mp$ is a prime.

Q: is $S_M$ infinite as long as the digit sum of $M$ is not a multiple of $3?$

I would guess that one could find encouraging evidence of YES (I haven’t checked) although no proof. You only care about $M$ with prime digit sum but that would follow from the more general question I gave.

Flimsy reasoning for why I suspect yes: Suppose $M$ has $t$ digits and consider primes with $s$ digits. Then there will be a very large number of products $Mp$ and the average digit sum should be roughly $5(s+t)$. There seems no reason to doubt that these digit sums behave as if they were randomly chosen (among non-multiples of $3$) in the appropriate range. Estimating the number of sums and density of primes in that range should suggest a healthy number of successes.

EXTREMELY LIMITED CALCULATIONS

I just looked at $M=314159$ and the $1228$ primes $p \neq 3$ less than $10^5.$ The digit sums for $Mp$ range from $13$ to $70.$ The only missing sums from that little experiment (among non-multiples of $3$ from $13$ to $70$ are $14, 16.$ Here are the counts in the middle of that range

$[28, 34], [29, 25], [31, 39], [32, 44], [34, 54], [35, 74], [37, 67], [38, 82], [40, 75], [41, 75], $

$[43, 81], [44, 69], [46, 69], [47, 62], [49, 58], [50, 60], [52, 44], [53, 33], [55, 29], [56, 29]$

One might expect from this that in general among the $Mp$ are every possible digit sum (prime and otherwise, but not multiples of $3$) multiple times (with perhaps a small number of small exceptions.)

I'm not sure I would have predicted it but the comment by მამუკა ჯიბლაძე suggests that there are infinitely many primes such that $3p$ has digit sum $3$

There are $k-1$ integers $x$ with digit sum $3$ and $k$ digits. The $k-1$ values $\frac{x}{3}$ are all relatively prime to $6$ so one might expect the number of primes there to be roughly $\frac3{\ln(10^k/3)}(k-1) \sim 1.3$. Up to $k=100$ there are on average about $1.6$ in each range. I wonder if something related is true for $M$ with digit sum a multiple of $3.$

I agree that the sum of the base ten digits seems rather peripheral. However: given a positive integer $M;$ let $S_M$ be the set of primes $p$ so that the base $10$ digit sum of $Mp$ is a prime.

Q: is $S_M$ infinite as long as the digit sum of $M$ is not a multiple of $3?$

I would guess that one could find encouraging evidence of YES (I haven’t checked) although no proof. You only care about $M$ with prime digit sum but that would follow from the more general question I gave.

Flimsy reasoning for why I suspect yes: Suppose $M$ has $t$ digits and consider primes with $s$ digits. Then there will be a very large number of products $Mp$ and the average digit sum should be roughly $5(s+t)$. There seems no reason to doubt that these digit sums behave as if they were randomly chosen (among non-multiples of $3$) in the appropriate range. Estimating the number of sums and density of primes in that range should suggest a healthy number of successes.

EXTREMELY LIMITED CALCULATIONS

I just looked at $M=314159$ and the $1228$ primes $p \neq 3$ less than $10^5.$ The digit sums for $Mp$ range from $13$ to $70.$ The only missing sums from that little experiment (among non-multiples of $3$ from $13$ to $70$ are $14, 16.$ Here are the counts in the middle of that range

$[28, 34], [29, 25], [31, 39], [32, 44], [34, 54], [35, 74], [37, 67], [38, 82], [40, 75], [41, 75], $

$[43, 81], [44, 69], [46, 69], [47, 62], [49, 58], [50, 60], [52, 44], [53, 33], [55, 29], [56, 29]$

One might expect from this that in general among the $Mp$ are every possible digit sum (prime and otherwise, but not multiples of $3$) multiple times (with perhaps a small number of small exceptions.)

I'm not sure I would have predicted it but the comment by მამუკა ჯიბლაძე suggests that there are infinitely many primes such that $3p$ has digit sum $3$

There are $k-1$ odd integers $x$ with digit sum $3$ and $k$ digits. The $k-1$ values $\frac{x}{3}$ are all relatively prime to $6$ so one might expect the number of primes there to be roughly $\frac3{\ln(10^k/3)}(k-1) \sim 1.3$. Up to $k=100$ there are on average about $1.6$ in each range. I wonder if something related is true for $M$ with digit sum a multiple of $3.$

I agree that the sum of the base ten digits seems rather peripheral. However: given a positive integer $M;$ let $S_M$ be the set of primes $p$ so that the base $10$ digit sum of $Mp$ is a prime.

Q: is $S_M$ infinite as long as the digit sum of $M$ is not a multiple of $3?$

I would guess that one could find encouraging evidence of YES (I haven’t checked) although no proof. You only care about $M$ with prime digit sum but that would follow from the more general question I gave.

Flimsy reasoning for why I suspect yes: Suppose $M$ has $t$ digits and consider primes with $s$ digits. Then there will be a very large number of products $Mp$ and the average digit sum should be roughly $5(s+t)$. There seems no reason to doubt that these digit sums behave as if they were randomly chosen (among non-multiples of $3$) in the appropriate range. Estimating the number of sums and density of primes in that range should suggest a healthy number of successes.

I agree that the sum of the base ten digits seems rather peripheral. However: given a positive integer $M;$ let $S_M$ be the set of primes $p$ so that the base $10$ digit sum of $Mp$ is a prime.

Q: is $S_M$ infinite as long as the digit sum of $M$ is not a multiple of $3?$

I would guess that one could find encouraging evidence of YES (I haven’t checked) although no proof. You only care about $M$ with prime digit sum but that would follow from the more general question I gave.

Flimsy reasoning for why I suspect yes: Suppose $M$ has $t$ digits and consider primes with $s$ digits. Then there will be a very large number of products $Mp$ and the average digit sum should be roughly $5(s+t)$. There seems no reason to doubt that these digit sums behave as if they were randomly chosen (among non-multiples of $3$) in the appropriate range. Estimating the number of sums and density of primes in that range should suggest a healthy number of successes.

EXTREMELY LIMITED CALCULATIONS

I just looked at $M=314159$ and the $1228$ primes $p \neq 3$ less than $10^5.$ The digit sums for $Mp$ range from $13$ to $70.$ The only missing sums from that little experiment (among non-multiples of $3$ from $13$ to $70$ are $14, 16.$ Here are the counts in the middle of that range

$[28, 34], [29, 25], [31, 39], [32, 44], [34, 54], [35, 74], [37, 67], [38, 82], [40, 75], [41, 75], $

$[43, 81], [44, 69], [46, 69], [47, 62], [49, 58], [50, 60], [52, 44], [53, 33], [55, 29], [56, 29]$

One might expect from this that in general among the $Mp$ are every possible digit sum (prime and otherwise, but not multiples of $3$) multiple times (with perhaps a small number of small exceptions.)

I'm not sure I would have predicted it but the comment by მამუკა ჯიბლაძე suggests that there are infinitely many primes such that $3p$ has digit sum $3$

There are $k-1$ integers $x$ with digit sum $3$ and $k$ digits. The $k-1$ values $\frac{x}{3}$ are all relatively prime to $6$ so one might expect the number of primes there to be roughly $\frac3{\ln(10^k/3)}(k-1) \sim 1.3$. Up to $k=100$ there are on average about $1.6$ in each range. I wonder if something related is true for $M$ with digit sum a multiple of $3.$