What is the largest power of $3$ that divides $999\dots 999$ ($300$ $9$'s)?
I have looked at the pattern of $9$, $99$, $999$, $9999$, $99999\dots 999999999$ and found the pattern that the largest power of $3$ that divides is: $2, 2, 3, 2, 2, 3, 2, 2, 4$
How would I work with prime factorization of this large power of $3$ that divides $999\dots 999$ ($300$ $9$'s)?
On
The number is $N=10^{300}-1$. The question is what is the largest $k$ with $3^k\mid N$, i.e., $10^{300}\equiv1\pmod{3^k}$. Now \begin{align} 10^{300}&=(1+9)^{300}=1+300\times 9+\binom{300}29^2+\binom{300}39^3+ \cdots\\ &=1+2700+81M \end{align} for some $M\in\Bbb N$. Then $N\equiv0\pmod{27}$ but $N\not\equiv0 \pmod{81}$.
On
$$ 10^{300}-1=(10^{75}-1)(10^{75}+1)(10^{150}+1) $$
now analyzing $10^{75}-1 = (9+1)^{75}-1\;\;$ which is divisible by $3^3 = 27$
Concluding
10^75 + 1 // FactorInteger
{{7, 1}, {11, 1}, {13, 1}, {211, 1}, {241, 1}, {251, 1}, {2161, 1}, {5051, 1}, {9091, 1}, {78875943472201, 1}, {10000099999999989999899999000000000100001, 1}}
10^150 + 1 // FactorInteger
{{61, 1}, {101, 1}, {601, 1}, {3541, 1}, {9901, 1}, {27961, 1}, {60101, 1}, {261301, 1}, {3903901, 1}, {4188901, 1}, {7019801, 1}, {39526741, 1}, {168290119201, 1}, {14103673319201, 1}, {1680588011350901, 1}, {25074091038628125301, 1}, {38654658795718156456729958859629701, 1}}
so finally we can conclude that the maximum is $3^3$
On
Your number $N$ can be expressed as $ N = 10^{300}-1 $
In the general formulation we have for numbers of such a form (by applying "Little" Fermat and Euler's totient and a bit more, see here)
$$ \{10^n -1 ,3\} = [n:1](2 + \{n,3\})$$
(where the braced expression $\{a,b\}$ means the highest power to which $b$ occurs in $a$, often called "valuation" or $v_b(a)$ and$[a:b]$ means $1$ if $b$ divides $a$ and $0$ if not)
this gives
$$ \{10^{300} -1 ,3\} = [300:1](2 + \{300,3\})=1\cdot(2+1)=3$$
Hints: