Selection and Training Session

Answer: 14 zeroes.

First solution. Derive the maximum power of 2 as a common factor: $$A=2^7\cdot (7^{14}+1+2\cdot 5^2\cdot 7^{11}+2^3\cdot 5^4\cdot 7^7+2^3\cdot 5^6\cdot 7^3).$$ Note that $50=7^2+1$ and represent 50 in this way in all terms: $$2^{-7}A=7^{14}+(7^2+1)\cdot 7^{11}+2\cdot (7^2+1{)}^2\cdot 7^7+(7^2+1{)}^3\cdot 7^3+1.$$ Expand the right-hand side in powers of 7: $$2^{-7}A=7^{14}+7^{13}+7^{11}+2\cdot 7^{11}+4\cdot 7^9+2\cdot 7^7+7^9+3\cdot 7^7+3\cdot 7^5+7^3+1.$$ Once again, use equality $7^2+1=50$, for which derive even degrees of 7 in all terms: $$2^{-7}A=7^{14}+7\cdot 7^{12}+7\cdot 7^{10}+14\cdot 7^{10}+28\cdot 7^8+14\cdot 7^6+7\cdot 7^8+21\cdot 7^6+21\cdot 7^4+7\cdot 7^2+1$$ and expand the right-hand side in even powers of 7: $$2^{-7}A=7^{14}+7\cdot 7^{12}+21\cdot 7^{10}+35\cdot 7^8+35\cdot 7^6+21\cdot 7^4+7\cdot 7^2+1.$$ Now it is clear that the right-hand side is the Newton binomial formula for $(7^2+1{)}^7$, hence the number $A=2^7\cdot 50^7=10^{14}$ ends with 14 zeroes.

Second solution For any $x_1$ and $x_2$ set $S_k=x_1^k+x_2^k$, $x_1+x_2=p$ and $x_1{x}_2=q$ (note that $S_0=2$, $S_1=p$). Using the obvious recurrence formula $S_{k+2}=p{S}_{k+1}-q{S}_k$, we successively get $S_2=p^2-2q$, $S_3=p^3-3pq$, $S_4=p^4-4p^{2}q+2q^2$, $S_5=p^5-5p^{3}q+5p{q}^3$, $S_6=p^6-6p^4+9p^2{q}^2-2q^3$ and, finally, $$S_7=p^7-7p^{5}q+14p^3{q}^2-7p{q}^3(1)$$ Now, take $x_1=100$, $x_2=-2$ then $p=2\cdot 7^2$, $q=-2\cdot 10^2$ and (1) takes the form $10^{14}-2^7=2^7\cdot 7^{14}+2^6\cdot 7^{11}\cdot 10^2+2^6\cdot 7^7\cdot 10^4+2^4\cdot 7^3\cdot 10^6$ which is equivalent to $A=10^{14}$.