Sharky's Conjecture

Hint: $n\equiv \text{digit sum of }n\pmod 3$

Furthermore, you can do this as well: $144\equiv (14+4)\equiv (1+44)\equiv (1+4+4)\pmod 3$

What do we notice about the given conditions? i.e. that $2, 5, 8, 11 \equiv 2 \pmod{3}$.

Lemma 1

All numbers in base $n$ such that $n \equiv 1 \pmod{3}$ which have a digit sum divisible by three will themselves be divisible by three.

Proof

All numbers are in the form $\overline{abcd\dots}$. In decimal representation, we take this to mean $10^n(a)+10^{n-1}(b)+10^{n-2} \dots$. Before going further, note that $10 \equiv 1 \pmod{3}$. Thus any power of $10$, say $100$ will equal $1 \times 1=1 \pmod{3}$, and by multiplying by 10 again, the number is still $1 \pmod{3}$. For example, $10^5-1=99999=3 \times 33333$. Anyways, using this property, we can split this given polynomial form for decimal numbers into two parts,

$(10^n-1)a+a+(10^{n-1}-1)b+b+(10^{n-2}-1)c+c \dots$

Which, when grouping, becomes

$(10^n-1)a+(10^{n-1}-1)b+(10^{n-2}-1)c \dots+(a+b+c \dots)$

From what we've derived about powers of $10$ minus $1$, the left part of the expression will have all of it's coefficients divisible by $3$, since they're all powers of $10$ minus $1$, which will make that whole part divisible by $3$ regardless of what the actual numbers are. Now, as long as the right hand part ($(a+b+c\dots)$) is divisible by $3$, the WHOLE expression will be divisible by three. This is simply the digit sum! We can further extend this to encompass numbers in all bases that are $1 \pmod{3}$ because as shown above, they will be able to be split up into a clearly divisible-by-three part, and a digit sum part. Thus we have proven the lemma.

Lemma 2

If $n \equiv a \pmod{3}$, then $n+b$ will be equivalent to $a+b \pmod{3}$.

Proof

This is easily proven by recalling Lemma 1. Because a number $n$ that is $\overline{abcd\dots}$ can be split up into a part that is clearly divisible by $3$, and then added to $a+b+c\dots$. Thus, by adding some arbitrary $a$ to $n$, you will add to the digit sum which will cause it to change by $a \pmod{3}$ and we're done.

With these little lemmas in hand, we can begin to approach this problem. Anyways, we'll consider three different cases for $n$:

Case 1

$n \equiv 0 \pmod{3}$. This will obviously work for all cases, since the first part which is simply $n$ will have a digit sum divisible by $3$ and so will $2n$, $5n$, $8n$, and $11n$.

Case 2

$n \equiv 1 \pmod{3}$. In this case, $n$ will have a digit sum of $1 \pmod{3}$. So for any given $2n$, the digit sum will for sure be equivalent to $2 \pmod{3}$ from Lemma 2 and thus by adding the digit sum of $2n$, $5n$, $8n$, or $11n$ to the digit sum of $n$, the result will be equivalent to $1+2 \pmod{3}$ and we're finished with this case.

Case 3

$n \equiv 2 \pmod{3}$. Similarly, we can construct that the digit sum of $n$ plus the digit sum of $2n$, $5n$, $8n$, or $11n$ will equal $2+2^2 \pmod{3}$ which is $0$ and we're done!

Absolutely amazing proof problem @Sharky Kesa! I loved writing this obnoxiously long proof. @Mursalin Habib how you like me now? :D

What does $\overline{n2n}$ mean?

@Finn Hulse I expect you to have a proof.

This is quite simple.

The digit sum n + 2n = 3n. Therefore, it is divisible by three. Similarly, note that 5, 8 and 11 are all one less than a multiple of three. Hence, similiar logic applies.

hey! .... it may seem simple but i think that it has some interesting outcomes. You can tease your friend by giving him a very large number and ask him if it is prime or not....... you could give him 396943659 which is obtained when 3969 and 3969*11 are concatenated. :p