Prove that $10^{340} < \frac{5^{496}}{1985}$

$$2^{340}\cdot 5^{340} \lt \dfrac{5^{496}}{1985}$$ is equivalent to $$2^{340}\lt \frac{5^{496-340}}{5\times 397}=\frac{5^{155}}{397}$$ which is equivalent to $$397\cdot 2^{340}\lt 5^{155}$$

Since $397\lt 1024=2^{10}$, it is sufficient to prove that $$2^{350}\lt 5^{155}$$ which is equivalent to $$2^{70}\lt 5\cdot 5^{30}$$ which is equivalent to $$5\gt \left(\frac{2^7}{5^3}\right)^{10}=(1.024)^{10}$$

Now, $(1.024)^{10}\lt (1.1)^{10}=(1.21)^5\lt 1.3\times (1.3)^4\lt 1.3\times (1.7)^2=3.757\lt 5$.

$625=5^4>2^9=512$

$496-340 = 156 = 4\times 39 $

So $5^{156}>2^{351}$ and $5^{496}>10^{340}\cdot2^{11}>10^{340}\cdot 1985$

So $10^{340} < \dfrac{5^{496}}{1985}$ as required.