Clock Conundrum

Firstly, note that the first time the hands cross after 00:00:00 will come between 1 'o' clock and 2 'o' clock. Hence, we can take 1 'o' clock as a starting point in any calculations of distance travelled by the hands.

The hour hand moves at a rate of $5x/hr$ , where $1x$ is an increment of 1 minute of the clock. Meanwhile, the minute hand moves at a rate of $60x/hr$ . However, since we already know that the point of overlap comes after 1 'o' clock, we can simplify the problem by immediately giving the hour hand it's 5x headstart: see the picture attached if this isn't clear.

$v=d/t$ , so using these previously calculated "rates", we can formulate expressions for the distance travelled by each hand. The hour hand's distance (i.e. number of increments) from the 1 'o' clock position is given by $5t +5$ whereas the minute hand's position is given by $60t$ . These can be solved simultaneously: $60t=5t+5$ $55t=5$ $t=\frac{5}{55}=\frac{1}{11}$ So the hands overlap at $\frac{1}{11}$ of a minute - 27 seconds - after 01:05:00. So the solution is $\boxed{01:05:27}$ .