Will that be a small triangle?

$Lemma:$ Suppose we have two positive integers $x,y$ such that

$f \left(\frac{3^x}{10^4}\right) = f \left(\frac{3^y}{10^4}\right)$ . I say that this implies $x \equiv y (mod . 500)$ .

$Proof:$ Let $m,n$ be the remainders when $3^x, 3^y$ are divided by $10^4$ respectively. Then we can write $3^x = q_1 10^4 + m$ and $3^y = q_2 10^4 + n$ for some positive integers $q_1, q_2$ , such that $0 \le m,n < 10^4$ . Therefore,

$f \left(\frac{3^x}{10^4}\right) = f \left(q_1 + \frac{m}{10^4}\right) = \left(q_1 + \frac{m}{10^4}\right) - q_1 = \frac{m}{10^4}$

Similarly,

$f \left(\frac{3^y}{10^4} \right) = \frac{n}{10^4}$

Since those two quantities were assumed to be equal, then we must have $m = n$ . Thus, $3^x \equiv 3^y (mod. 10^4)$ .

Now it remains to find the order of 3 in the group of integers relatively prime to $10^4$ . By Euler's Theorem we know that $3^{4000} \equiv 1 (mod. 10^4)$ . With a little experimenting, you can find that $3^{500} \equiv 1 (mod. 10^4)$ , but $3^{250} \equiv 6249 (mod 10^4$ and $3^{100} \equiv 2001 (mod. 10^4)$ . Hence, the order of 3 is 500.

Therefore, it must be that $x \equiv y (mod. 500)$ . $\Box$

Now back to the problem. By the Lemma, we must have $a \equiv b \equiv c (mod. 500)$ . Since $a > b > c$ , this means there exist positive integers $p, q$ such that $b = c + 500p$ and $a = c + 500(p+q)$ .

Also, by the Triangle Inequality, we must have $b + c > a$ . Therefore,

$(c + 500p) + c > c + 500(p+q)$

$c > 500q$

Now since $q$ is positive, this means $c \ge 501$ . Then $b \ge 1001$ and $a \ge 1501$ .

Hence, $a + b + c \ge 3003$ .