Hellenic Mathematical Olympiad

$$\alpha_v = 5^{v-1} \cdot \left[ 1 + \frac{3}{5} + \left(\frac{3}{5}\right)^2 + \dots + \left(\frac{3}{5}\right)^{v-1} \right] = \frac{1}{2}(5^v - 3^v),\ v = 1, 2, \dots$$
Now for $k=2^{2019}$, we have: $2a_k = 5^{2019} - 3^{2019} = 2 \cdot (5+3)(5^2+3^2) \dots (5^{2018} + 3^{2018})$, and hence:
$$a_k = (5+3)(5^2+3^2)\dots(5^{2018}+3^{2018}).$$
We observe that the first factor is divided by $8$ and all the others are divided by $2$ and are not divided by $4$. In fact, we have:
$$5^{2v} \equiv 1 \pmod{4} \text{ and } 3^{2v} \equiv 1 \pmod{4} \Rightarrow 5^{2v} + 3^{2v} \equiv 2 \pmod{4}, \text{ for all } v \ge 1.$$

The factors from $5^2 + 3^2$ to $(5^{2^{2018}} + 3^{2^{2018}})$, are totally $2018$, and therefore the greatest power of $2$ dividing $a_k$ is $2^{2021}$.

Alternatively, we can use a special form of the Lifting the Exponent Lemma concerning the greatest power of $2$ dividing a difference of powers of integers. We denote by $v_p(\alpha)$ the greatest exponent of power of a prime number $p$ which divide the integer $\alpha$, that is: $p^{v_p(\alpha)}|\alpha$ and $p^{v_p(\alpha)+1} \nmid \alpha$. We have the following:

Lemma: Let $\alpha, \beta$ two odd integers and $v$ an even positive integer. Then:
$$v_2(\alpha^v - \beta^v) = v_2(\alpha - \beta) + v_2(\alpha + \beta) + v_2(v) - 1.$$
By applying the lemma to the integer $2a_{2^{2019}} = 5^{2^{2019}} - 3^{2^{2019}}$ we find:
$$v_2(2a_{2^{2019}}) = v_2(5^{2^{2019}} - 3^{2^{2019}}) = v_2(5-3) + v_2(5+3) + v_2(2^{2019}) - 1 = 1 + 3 + 2019 - 1 = 2022,$$
and hence: $v_2(a_k) = 2021$.