Formulated this question while on a break from a break (repeated digit sum and composite numbers related question)

Question

Suppose $a_1...a_m$ is some composite natural number with at least two different prime factors written in decimal notation.

Is there an infinite number of composites with at least two different prime factors $a_1...a_m=\prod_{k=1}^{r}p_k^{b_k}$ such that $\text{rds}_{10}(a_1...a_m)=\text{rds}_{10}(p_1^{b_1}+...+p_r^{b_r})$

Here, $\text{rds}_{10}$ is the repeated sum of digits in base $10$.

For example $\text{rds}_{10}(123456789)=1+2+3+4+5+6+7+8+9=45=4+5=9$

Answer 1

Yes, there are an infinite number of such composites. One example of a set of such numbers is

$$(3^2)(2^{b_2})(5^{b_3}) \tag{1}\label{eq1A}$$

where $b_2 = 6n_2, \; n_2 \in \mathbb{N}$ and $b_3 = 3 + 6n_3, \; n_3 \in \mathbb{N}$. This is because the product is a multiple of $9$, so the repeated sum of digits will always end up at $9$. Also, since

$$2^{b_2} \equiv 2^{6n_2} \equiv (2^6)^{n_2} \equiv (64)^{n_2} \equiv (1)^{n_2} \equiv 1 \pmod{9} \tag{2}\label{eq2A}$$

and

$$5^{b_3} \equiv 5^{3 + 6n_3} \equiv (125)(5^6)^{n_3} \equiv (8)(1)^{n_3} \equiv 8 \pmod{9} \tag{3}\label{eq3A}$$

you then have

$$9 + 2^{b_2} + 5^{b_3} \equiv 0 + 1 + 8 \equiv 0 \pmod{9} \tag{4}\label{eq4A}$$

so it's repeated sum of digits will also end being $9$.

Answer 2

Yes. It is well known that $\text{rds}_{10} (x) \equiv x \pmod{9}$ (It takes values $1$ to $9$). We claim that any composite of the form: $$n= \prod_{i=1}^{10} p_i^{6k}$$ works, where $\{p_i \mid 1 \leqslant i \leqslant 10 \}$ is the set of first $10$ prime numbers, namely primes from $2$ to $29$ and $k \in \mathbb{N}$

Clearly, $9 \mid n \implies \text{rds}_{10} (n) =9$. Now, we have:

$$\text{rds}_{10} \bigg( \sum_{i=1}^{10} p_i^{6k} \bigg) \equiv \sum_{i=1}^{10} p_i^{6k} \equiv 0+9 \cdot 1 \equiv 9 \pmod {9}$$

Above, the sum was congruent to $0+9 \cdot 1$ since the prime $p_2=3$ raised to the power is clearly $0 \pmod{9}$ while the rest of the primes raised to the power $6k$ are all $1 \pmod{9}$ by Euler's Totient Theorem ($\varphi(9)=6$).

This gives us: $$\text{rds}_{10} (n) = \text{rds}_{10} \bigg( \sum_{i=1}^{10} p_i^{6k} \bigg) = 9$$

Since $k$ has infinitely many options (all natural numbers), we are done!