$a_1,a_2,\ldots,a_{11}$ and $b_1,b_2,\ldots,b_{11}$ are $2$ permutations of $1,2,\ldots,11$. Show that atleast...

Question

Let $a_1,a_2,\ldots,a_{11}$ and $b_1,b_2,\ldots,b_{11}$ be 2 permutations of $1,2,\ldots,11$. Show that atleast 2 of $a_1b_1,a_2b_2,\ldots,a_{11}b_{11}$ will have same remainder $\mod 11$

My attempt:
When $11=a_i=b_j [i\ne j]$, then it is obvious. When $11=a_i=b_i$, then $a_ib_i$ can be excluded from the list as no other number leaves 0 remainder. So, we have a permuation of $1,2,\ldots,10$. Now, we have to prove the given statement, which I cannot do any further. Please help. Thank you.

Answer 1

Suppose it is not true and let's $a_1,\ldots,a_{10}$ and $b_1,\ldots,b_{10}$ are permutations of $1,\ldots,10$. We have $$ a_1b_1=c_1(mod11) $$ $$ ... $$ $$ a_{10}b_{10}=c_{10}(mod11), $$ where $c_1,\ldots,c_{10}$ is also permutation of $1,\ldots,10$. By multiplying all these equations together we get $$10!(10!-1)=0(mod11),$$ which is contradiction.

Answer 2

We complete your proof: Without loss of generality let $a_{11}=b_{11}=11$. Now assume that $a_ib_i\ne a_jb_j\pmod{11}$ for all $i\ne j$ then $a_ib_i$ gives all values from $1$ to $10$ modulo $11$ hence by the Wilson's theorem we get

$$\prod_{i=1}^{10}a_ib_i=(10!)^2\equiv(-1)^2=1\equiv10!\equiv -1\pmod{11}$$ which is a contradiction.