Sum of digits of $1000^{20}-20$ in decimal notation

Question

Find the sum of the digits of $$1000^{20}-20$$ written in decimal notation.

What exactly do they mean by decimal notation? How will the answer be any different if it is not in decimal form?

Answer 1

$(10^n)^m =10^{nm}$ and this is, in decimal, a one followed by $nm$ zeroes, so the sum of the digits is one.

Answer 2

Note that in base $10$,

$$1000^{20}-20=10^{60}-20=999\ldots9980$$

so the trick in base $10$ is simply to figure out how many nines there are before the $80$. This is easiest to do by first considering a few smaller exponents:

$$\begin{align} 10^3-20&=980\\10^4-20&=9980\\10^5-20&=99980 \end{align}$$

which suggests there are $60-2=58$ nines in $10^{60}-20$, so that the sum of the digits in $1000^{20}-20$ is $58\cdot9+8+0=530$.

My guess is that whoever wrote the problem added "written in decimal form" just to emphasize they were talking about the usual way we work with numbers. If you want to pose the problem for some other base, you need to somehow specify whether the expression $1000^{20}-20$ is already written in the other base (which presumably could not be base $2$, since it uses the three digits $0$, $1$, and $2$) or if it needs to be converted from decimal to the other base.

Answer 3

The result will have $60$ digits.
$\underbrace{99\dots9}_{58 \text { times}}80$.
$\text{Sum of digits}= (9\times58)+8+0= 530$.
$\text{In decimal notation}= 5.3\times10^2$.