36th Hellenic Mathematical Olympiad

$${\alpha }_v=5^{v-1}\cdot \left[1+\frac{3}{5}+{\left(\frac{3}{5}\right)}^2+\cdots +{\left(\frac{3}{5}\right)}^{v-1}\right]=\frac{1}{2}(5^v-3^v),v=1,2,\dots$$ 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: $$5^{2v}\equiv 1(\mathrm{mod}4)\text{ and }{3}^{2v}\equiv 1(\mathrm{mod}4)\Rightarrow 5^{2v}+3^{2v}\equiv 2(\mathrm{mod}4),\text{ for all }v\ge 1.$$

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: $v_2(a_k)=2021$.