Question

Ten ants are on the real line. At time $$t = 0$$ the $${k^{th}}$$ ant start as the point $${k^2}$$ and travelling at uniform speed, reaches the point $${\left( {11 - k} \right)^2}$$ at time $$t = 1$$. The number of distinct times at which at least two ants are at the same location is?

Answer 1

Hint:
If we read the problem carefully we can see that velocity and position of $${k^{th}}$$ term are given, from these conditions we have to form an equation after solving the equation we can find the answer.

Complete step by step solution:
Velocity of $${k^{th}}$$ ant $$ = {\left( {11 - k} \right)^2} - {k^2} = 121 - 22k$$
Position of $${k^{th}}$$ ant at time $$t:{x_k} = {k^2} + t\left( {121 - 22k} \right)$$

Let $${p^{th}}$$ ant and $${k^{th}}$$ ant (k and p are distinct) be at the same position at time t.
$$\matrix
   { \Rightarrow {k^2} + t\left( {121 - 22k} \right) = {p^2} + t\left( {121 - 22p} \right)} \\
   { \Rightarrow \left( {{k^2} - {p^2}} \right) - 22t\left( {k - p} \right) = 0} \\
   { \Rightarrow k + p = 22t} \\
 $$
$$k$$ and $$p$$ both range from $$1$$ to $$10$$ and are distinct, so $$\left( {k + p} \right)\;$$ attains each value in the set $$\left\{ {3,4,5...18,19} \right\}$$

Hence, there are $$17$$ distinct times at which at least two ants are at the same location.

Note:
Here it was given that At time $$t = 0$$ the $${k^{th}}$$ ant start as the point $${k^2}$$ and travelling at uniform speed, reaches the point $${\left( {11 - k} \right)^2}$$ at time $$t = 1$$. All we did is form conditions from the equation to solve those to find the answer.