Highest power of $3$ in $19^{93}-13^{99}$
Was able to figure out that $9$ will suffice but how to know about more
Is there any other way than Binomial
Highest power of $3$ in $19^{93}-13^{99}$
Was able to figure out that $9$ will suffice but how to know about more
Is there any other way than Binomial
We can compute by hand, because the power is luckily not very high.
By the Binomial Theorem, we have $19^{93}=(1+18)^{93}=1+(18)(93)+\cdots,$ where each term in the part left out is divisible by at least $3^5$. Similarly, $13^{99}=(1+12)^{93}=1+(12)(99)+(12^2)(99)(98)/2+ (12^3)(99)(98)(97)/6+\cdots,$ where again the remaining terms are divisible by at least $3^5$. Subtract.
Note that $(18)(93)-(12)(99)=3^3(62-44)$, which is divisible by $3^5$. Now look at
$(12^2)(99)(98)/2+ (12^3)(99)(98)(97)/6.\tag{$1$}$ Dividing by $3^4$, we get $(4^2)(11)(49)+(4^3)(11)(49)(97).$ Modulo $3$, the first term is congruent to $-1$, and so is the second term. It follows that the highest power of $3$ that divides $(1)$ is $3^4$.
Once we know, the highest power to be $4,$ there can be several ways to establish it. Following is one of them.
$19^{93}-13^{99}=(19^{31})^{3}-(13^{33})^{3}=(19^{31}-13^{33})\{(19^{31})^2+ 19^{31} \cdot 13^{33}+(13^{33})^2\}$
$(19^{31})^2+ 19^{31} \cdot 13^{33}+(13^{33})^2$
$\equiv 1+(4)^{33}+\{(4)^{33}\}^2 \pmod 9$
$=1+(2^6)^{11}+(2^6)^{22} \mod 9\equiv 3\pmod 9$ as $\phi(9)=6\implies 2^6\equiv 1\pmod 9$
So, $3^1\mid\mid ((19^{31})^2+ 19^{31} \cdot 13^{33}+(13^{33})^2)--->(1)$
Now, $19^{31}=(1+18)^{31}=1+18\cdot 31+($ higher powers of $18)\equiv1+18\cdot 31\pmod{81}\equiv -8$ as $2\cdot 31\equiv -1$
$13^{33}=13\cdot (13^2)^8$ $\equiv 13\cdot (7)^{16}\pmod{81}$ $\equiv 13\cdot 7 \cdot(7^3)^5$ $\equiv 91\cdot(1+38\cdot 9)^5$
$\equiv 10(1+5\cdot 38\cdot 9+$ higher powers of $9)$ $\equiv10+ (10\cdot 9)(5\cdot 38)$ $\equiv10+9\cdot28=262\equiv 19\pmod{81}$
So, $19^{31}-13^{33}\equiv -8-19\pmod{81}\equiv 54$
So, $3^3\mid\mid (19^{31}-13^{33})--->(2)$
So, $3^4\mid\mid (19^{93}-13^{99})$ from $(1)$ and $(2).$
In Maple:
padic:-ordp(19^93 - 13^99, 3);
$4$