I'm working on a number theory problem for the Regional Mathematical Olympiad, Stage 2 of the Indian Olympiad Programme, and it's been challenging to solve.
The problem is as follows.
A function $\delta$ is defined such that if $m$ is a natural number, $\delta(m)$ denotes the repeated sum of digits of $m$. For example, $\delta(36) = 3 + 6 = 9$, but $\delta(429) = \delta(4 + 2 + 9) = 1 + 5 = 6$. Prove that $\delta(n^k) = \delta(\delta(n)^k)$ for all natural numbers $n$ and $k$.
I've verified this for multiple examples, and it seems to hold:
$$\delta(25^3) = \delta(15625) = 1$$ $$\delta(7^3) = \delta(343) = 1$$
...
$$\delta(33^2) = \delta(1089) = 9$$ $$\delta(6^2) = \delta(36) = 9$$
etc.
I have tried using double induction with the multinomial theorem, but it didn't seem to work out. I'd appreciate any guidance or insights on how to prove this.
Thank You.
EDIT: One comment mentioned that $\delta(n)$ is essentially equivalent to $n\pmod9$, which might be helpful in solving this problem, but I'm still struggling to find a suitable approach.
Hello fellow RMO aspirant, I am also preparing for it! I have come up with something over here. (my first answer)
Proof:
Let us take $a_1,a_2,a_3,...,a_n$
Claim: $$\delta(\sum_{i=1}^{n} a_i) = \delta(\sum_{i=1}^{n} \delta(a_i))$$
The claim can be easily proved by the decimal expansion.
Now, $$\delta(n^k) = \delta(n\times n^{k-1}) = \delta(\sum_{i=1}^{n^{k-1}} n)$$
$$\implies \delta(n^k) = \delta(\sum_{i=1}^{n^{k-1}} \delta(n)) = \delta(\delta(n)\times n^{k-1})$$
Thus, $$\delta(n^k) = \delta(\delta(n)\times n^{k-1})$$
Repeating the process,
$$\delta(n^k) = \delta(\delta(n)\times n^{k-1}) = \delta(\delta(n)^2\times n^{k-2}) = ... = \delta(\delta(n)^k)$$
Hence, $$\delta(n^k) = \delta(\delta(n)^k)$$
QED.