Nins_Leprechaun

Hopefully this should answer the second question

Well in any event I enjoyed solving it. So duplicate or not thank you .

I think I have it...

And another one (was trying to develop a system to easily find them and gave up)

Another example

I must be missing something then. If the clock starts at exactly 6 then in approx 30 seconds the second hand would overlap the hour hand. In that time the ant will travel 6 inches, the length of the hour hand. Yes there will be ever so slightly more than 30 seconds for this overlap to occur, but that fractional increase is irrelevant. If you could clarify the problem with my reasoning here it would be appreciated. Edit allow me to explain my interpretation of your description so you may correct any mistakes. A clock has a 12 inch second hand, 9 inch minute hand and 6 inch hour hand. The time is exactly 6 (hour hand at 6, second and minute hands at 12). The ant is on the tip of the hour hand. The ant must get to the tip of the second hand when it is pointed exactly at 12. The ant moves 12 inches a minutes up and down a clocks hand, turning around when it reaches the tip or pivot of the hand it is on. The ant may switch hands any time they overlap with no change to his current direction or speed of travel.

Wording made some of the rules sort of shaky.

Nice call, I forgot the first click

oops mispost

Love all the problems you post Bmad

the first and biggest problem is that neither realizes that their wife will kill them if they give away their christmas present also...