Question

Help!! ignore the answers the only right one is yes.

Answer 1

wait 33 minutes

he will have 23 minutes

Explanation:

laser 1: 3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60

laser2: 2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40...,60

laser 3: 5,10,15,20,25,30,35,40,45,50,55,60,

laser 4: 4,8,12,16,20,24,28,32,36,40,44,48,52,56,60

laser 5: 1,2,3,4,5,6,...35,36,37,38,39,40,...58,59,60

from 33min to 37 min is enouigh time to escape

he will have 23 minutes because they will all be off at 60 minutes and he escaped at 37 minutes so 60-37 = 23minutes

Hope this helps! :)

Answer 2

wait 33 minutes, pass lasers 1-5 at minutes 33, 34, 35, 36, 37; have 23 minutes before the alarm sounds

Explanation:

If it takes 1 minute between laser passes, we can only minimize the time if the lasers are off at consecutive minutes. We notice the first three are off at minutes 3, 4, and 5, but the fourth laser isn't off again until minute 8--too late.

That same sequence of {divisible by 3, divisible by 2, divisible by 5} will repeat every 3×2×5 = 30 minutes. That is the first three lasers are off at minutes 33, 34, 35. The fourth laser will be off at minute 36 (a multiple of 4), and the last laser will be off at the next minute, 37.

The prisoner can escape by waiting until minute 33, then passing one laser each minute to achieve freedom at minute 37. There will be 60-37 = 23 minutes after that before the alarm sounds.

The prisoner must wait 33 minutes. He will have 23 minutes before the alarm sounds.