If we set the clocks like this (hour hands omitted, since we don't care about them)
we get touching minute hands at every possible occasion, that is, whenever a minute hand points at another clock, there will always be a minute hand pointing back.
Since we cannot change how often a minute hand points at another clock, and there cannot be any touching when a ...
NOTE: This answer assumes we must deduce the restart times of the clocks without restarting them, and hence being unable to observe hands' behavior; I made this assumption so as to not trivialize parts of the puzzle.
How to fix the clocks:
The current time is
(deduced from the Pendulum clock)
To fix the Precision Clock,
Restart the precision clock at
One key observation is that
So if there are more Fridays and Saturdays than other days of the week in the year Marina was born then it must be that
Now you are not asked for her birth year but given that she is between 20-29 years old in 2012 we find that
The best answer is "time to get a new clock" but the two best times are 9:05:25.452 and 2:54:34.548, with angles HM=119.83°, HS=120.00° and MS=120.17°. Python code:
import numpy as np
def bestN(a, b, c, N=1):
e1 = np.abs(((b-a) % 60) -20)
e2 = np.abs(((c-b) % 60) -20)
e3 = np.abs(((a-c) % 60) -20)
As with the other question,
That's not actually quite true, because
But (confession: I failed to spot this despite attempting to check; thanks to Jaap Scherphuis in comments for being less stupid than me)
So in fact
An analog clock mechanism with only two hands has
meaning that every angle (counting from the minute hand to the hour hand in the same direction) repeats exactly that many times during a 12 hour period.
The question then becomes: can the second hand (which is the third hand, really) distinguish between those symmetrical cases?
If all the hands move by "...
Assuming they synchronise their watches when leaving the office, they’ll have a 10 min difference accumulating every hour.
Assuming a standard 5-day week and 8-hour day,
Tuesday through Friday, they’ll be separated for a nominal 16h (by each watch), so they will out by 160 min: one early by 80 min, the other late by 80 min;
Monday, they’ll be separated for ...