THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

The last digit of $3^{2017}$ is 3, so the last digit of the numerator is 5. $$(a_1+a_2)+(a_2+a_3)+\cdots +(a_{2016}+a_{2017})+(a_{2017}+a_1)=2(a_1+a_2+\cdots +a_{2017}).$$ The sum of these 2017 nonnegative integers being even, one of the summands, and hence their product, is even, so the last digit of the number $$9^{(a_1+a_2)(a_2+a_3)\dots (a_{2016}+a_{2017})(a_{2017}+a_1)}$$ is 1, and the denominator is divisible by 10. Hence, the fraction can be simplified by 5.