How to find the sum of digits of a number whose prime factorisation is given For example, if $n = 15^2 \times 5^{18}$ in base $10$. Is there a way to find the upper and lower bounds for the sum of digits of this number? The answer is given to be $6\leq s \leq 140$.
Also, is there a way to calculate the exact sum?
Also, how do i find the number of digits of this number?
 A: For an upper bound, without logarithms or calculators:
$$\begin{array}{ccc}15^2\cdot5^{18}&<&16^2\cdot 15^{18}\\
&=&16^2\cdot 5^3\cdot5^{15}\\
&<&16^2\cdot 2^7\cdot5^{15}\\
&=&2^{15}\cdot 5^{15}\\
&=&10^{15}
\end{array}$$
So the number has at most $15$ digits, one of which is a $5$.  So the digit sum can be upper bounded by $9\cdot 14+5=131$.
Whatever method the given answer used seems to have allowed for up to $16$ digits including one $5$.
A: Since the number ends with $5$ , the sum of the digits at the minimum would be $6$ (Since at least one of the other terms is $\ge 1$).
As for the upper bound , As pointed out in the comments , since the number has $15$ digits , I believe the maximum sum should be $9*14 +5 = 131$
Hence the limits should be $6\le s\le 131.$
A: Well, if you want any upper bound then $n =15^2*5^{18} < 10^{100}$ so it has fewer than a hundred digits each at most $9$ so an upper bound is $900$
And a lower bound is $0$ as the sum is surely positive.
So $0 \le s < 900$
I'm being facetious but whoever did this didn't do much more than that.
I figure s/he figured being an odd multiple of $5$ then last digit was $5$ and having more than one digit the sum $s\ge 6$.  
That is bare minimum of work.  As $n = 15^2*5^{18}=9*5^{20}$ the number must be a multiple of $9$.  So $s \ge 9$.
And the number of digits is $\lfloor \log 9*5^20\rfloor + 1=15$ so the upper bound is $15*9 = 135$.  Don't know how the upper bound of $140$ was figured.  We know the last digit if $5$; not $9$ so $131$ is an upper bound. 
[Credit where credit is due.  Even though I knew the last digit it didn't occur to me to consider it an exception to digits being $9$ and thus $131$ is legitimate upper bound. Credit to the Demonix Hermit for that basic observation.]
We can probably figure the first digit of $n=9*5^{20} = 9*10^{20}\frac 1{2^{20}} = 9*\frac 1{1024*1024}10^{20}\approx 9*\frac 1{1000000 + 50000 + 25^2}*10^{20}$ has $8$ and not $9$ as a first digit so $18\le s \le 130$.
Now of course we can actually do the problem.  $n =858306884765625$ and $s = 81$.
I'm really unsure what degree of accuracy is expected. 
A: Another estimation for the number of digits is that, since $2^{10}> 10^3$, you have that $\log_{10} 2 >0.3$ so $\log_{10} 5 =1-\log_{10} 2 <0.7$ so $\log_{10} 5^{20}=20\log_{10} 5< 14.$
So $5^{20}$ is no more than $14$ digits, and thus $15^2\cdot 5^{18}=9\cdot 5^{20}$ is no more than $15$ digits.
