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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

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3^4 = 81: I feel it in my heart. –  Benjamin Dickman Oct 30 '12 at 5:37
1  
@B.: Heart is right on. –  André Nicolas Oct 30 '12 at 5:54

3 Answers 3

up vote 2 down vote accepted

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 $3^4$.

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In Maple:

padic:-ordp(19^93 - 13^99, 3);

$4$

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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 $(2).$

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