51st Ukrainian National Mathematical Olympiad, 3rd Round

WLOG, $a_5=5$. If $b_5\ne 5$, then there will be two numbers that are divisible by $5$. Let $b_5\ne 5$ and suppose that the statement of the problem is false. Then the numbers $a_1{b}_1,a_2{b}_2,a_3{b}_3,a_4{b}_4$ have remainders $1,2,3$ and $4$ modulo $5$ (in some order). Then from one hand the product has remainder $4$.

From the other, since the numbers $a_1,a_2,a_3,a_4$ and $b_1,b_2,b_3,b_4$ are permutations of $1,2,3$ and $4$, the product is $a_1{b}_1\cdot a_2{b}_2\cdot a_3{b}_3\cdot a_4{b}_4=24\cdot 24=576$ has remainder $1$ modulo $5$. This contradiction finishes the proof.