A number theory problem by Josh Rowley

$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+........+ab^{n-2}+b^{n-1})$

Using this, $802^{11}-73^{11}=(802-73)(802^{10}+802^{9}73+........+73^{10})$

$=729(802^{10}+802^{9}73+........+73^{10})=3^6(802^{10}+802^{9}73+........+73^{10})$

So $N$ is at least $6$ . Now since we need to find out the number of factors of 3 in the second expression,we at first check if $3|(802^{10}+802^{9}73+........+73^{10})$ . Since $802\equiv 1\pmod 3$ $73\equiv 1\pmod 3$ we can conclude easily that $(802^{10}+802^{9}73+........+73^{10})\equiv 11\equiv 2\pmod 3$ and hence is not divisible by $3$ .

So we conclude that $N=\fbox{6}$

We use the Lifting the Exponent Lemma, and we get:

$\begin{aligned} \text{ord}_3 (802^{11}-73^{11}) &= \text{ord}_3(802-73) \\&= \text{ord}_3(729) \\&= \boxed {6}. \end{aligned}$

We know that $x^{n}$ - $y^{n}$ is divisible by x-y when n is odd.

We can prove this, but to reduce the step I had written directly.

Therefore $802^{11}$ - $73^{11}$ is divisible by 802-73=729

Since $3^6$ is 729

Therefore the value of N is $\boxed{6}$