Recently, I have found this problem:
Let $Q(n)$ denote the sum of the digits of a positive integer $n$. Prove that: $$Q(Q(Q(2005^{2005})))=7$$
This topic is related to other two question I have asked: Sum of number's digits and Simplifying an expression with modulus.
I have tried to solve this problem in many different ways, for example using the formula: $$Q(n)= n - 9 \sum_{k=1}^{\lfloor\log_{10} n\rfloor}\left\lfloor\frac{n}{10^k}\right\rfloor$$ but I'm stuck when I have to compute: $$Q(2005^{2005} - 9 \sum_{k=1}^{\lfloor\log_{10} 2005^{2005}\rfloor}\left\lfloor\frac{2005^{2005}}{10^k}\right\rfloor)= 2005^{2005} - 9 \sum_{k=1}^{\lfloor\log_{10} 2005^{2005}\rfloor}\left\lfloor\frac{2005^{2005}}{10^k}\right\rfloor- 9 \sum_{k=1}^{\lfloor\log_{10} \left ( 2005^{2005} - 9 \sum_{k=1}^{\lfloor\log_{10} 2005^{2005}\rfloor}\left\lfloor\frac{2005^{2005}}{10^k}\right\rfloor \right )\rfloor}\left\lfloor\frac{2005^{2005} - 9 \sum_{k=1}^{\lfloor\log_{10} 2005^{2005}\rfloor}\left\lfloor\frac{2005^{2005}}{10^k}\right\rfloor}{10^k}\right\rfloor$$
Or in other words, letting: $\Upsilon =2005^{2005} - 9 \sum_{k=1}^{\lfloor\log_{10} 2005^{2005}\rfloor}\left\lfloor\frac{2005^{2005}}{10^k}\right\rfloor $: $$Q(\Upsilon)=\Upsilon-9 \sum_{k=1}^{\lfloor\log_{10} \Upsilon\rfloor}\left\lfloor\frac{\Upsilon}{10^k}\right\rfloor$$
I have tried also using infinite sum, like complex Fourier series (see also this post Expression for finding sum of number's digits), but I don't know how to pass from an infinite series to the effective result.
