Sum of the digits of $2001^{10}$

Question

I saw this question in internet and I could not solve it. I saw that many people are trying to solve these kind of questions by Python. Then I decided that this question is a computer-science kind of question. If a program is solving my problem, where is the fun and pride on this?

My try: Since $2001^{10}\equiv0 \pmod{9}$, the sum of the digits $SD$ is a multiple of $9$. Also, this number contains $34$ digits. So, $SD\leq9\times34=306$. I thought about the binomial expansion $(2000+1)^{10}$ but it is complicated.

The answer is $108$. How can I find it without calculator? Is there a method? Thanks in advance.

Answer 1

Not really an answer;

On rewriting it as $(2000+1)^{10}$, we can apply binomial expansion on this as follows,

$$(2000+1)^{10}=\sum_{r=0}^{10} \binom nr (2000)^r$$

$$\sum_{r=0}^{10} \binom nr (2000)^r=\binom{10}{0}+\binom{10}{1}(2000)+\binom{10}{2}(2000^2)+\binom{10}{3}(2000)^3+\binom{10}{4}(2000)^4+\binom{10}{5}(2000)^5+\binom{10}{6}(2000)^6+\binom{10}{7}(2000)^7+\binom{10}{8}(2000)^8+\binom{10}{9}(2000)^9+\binom{10}{10}(2000)^{10}$$

$$=1+10(2000)+45(2000)^2+120(2000)^3+210(2000)^4+252(2000)^5+\cdots$$

$$=1029131535373448067360960180020001$$

The sum of digits = $1+2+9+1+3+1+5+3+5+3+7+3+4+4+8+6+7+3+6+9+6+1+8+2+1 = 108$

Extra; for $T=(2000)^n$, for any natural $n>0$, the sum of digits in $T$ is essentially the sum of digits in $2^n$

Answer 2

The easiest and quickest is pen and paper, write down 2001, then add 2000x nine times, then count the digits.

Or start with x = 4,004,001, then add 4,000x and 4,000,000x four times, and count the digits.

Pari/GP? Seriously?