Ten ants are on the real number line. At time $t=0$, the $k-th$ ant starts at the point $k^2$ and travelling at uniform speed, reaches the point $(11-k)^2$ at time $t=1$. The number of distinct times at which at least two ants are at the same location is:
My method:- i found out the velovity of the k th particle and it turned out to be $11(11-2k)$
so the first ant move 99 m second moves 11 m and so on so the first ant crosses 9 other ants second crosses 8 and so on therefore the answer is $9+8+7...+1=45$
but the given answer is $17$
Note:-this question was asked in the exam and i want to challenge its key.So can anybody tell what the correct answer is?