Question on Clocks

Question

Three watches are set together. The first gains $5$ minutes a week, the second gains $8$ minutes a week, whilst the third loses $4$ minutes a week. When will they again indicate the same time?

Please give me some hint on how to tackle this problem.

1 second later; the time haven't change. (Unless your clocks are very specific...) — , Jun 18 '14 at 06:17
even in 1 sec all the three clocks will gain or loose a fraction of second that's why your assumption may be wrong. — panav2000k, Jun 18 '14 at 06:21
Are these 12 hour watches or 24 hour watches, i.e. does AM/PM matter? — JimmyK4542, Jun 18 '14 at 06:25

Eric Towers · Accepted Answer · 2014-06-18T06:34:45.427

1

Assumptions:

12-hour watches. So two watches that differ by $12 \times 60\ $ accumulated minutes are indicating the same time. So not one of these.
"The same time" means "have hands pointing at the same labels" (no a.m. vs. p.m. tracking, so not one of these)
Time is (miraculously) indicated with infinite precision. (The accuracy is somewhat lacking, though.)

The first two differ by 3 minutes per week, so at least $\frac{12 \times 60}{3} = 240$ weeks must pass before the first two watches indicate the same time. This will recur every 240 weeks. The first and last differ by 9 minutes per week, so $\frac{12 \times 60}{9} = 80$ weeks is the time between matches of the first and last watch. Conveniently, this divides 240 weeks, so all three watches will agree every 240 weeks.

edited Jun 18 '14 at 06:34

answered Jun 18 '14 at 06:29

Eric Towers

67,037

Pardon my ignorance, but why 12.60? – panav2000k Jun 18 '14 at 06:31
12 hours times 60 minutes -- the number of minutes between identical readings. – Eric Towers Jun 18 '14 at 06:34
Thank you Sir. It was helpful. – panav2000k Jun 18 '14 at 06:37

score 0 · Answer 2 · answered Jun 18 '14 at 06:29

0

There are $720$ minutes in $12$ hours. So, the time on the watches is modulo $720$ minutes.

Thus, the clocks show the same time after $n$ weeks iff $5n \equiv 8n \equiv -4n \pmod{720}$.

This condition is met iff $3n \equiv 0 \pmod{720}$, i.e. $n$ is a multiple of $240$.

So, the watches will show the same time after $240$ weeks.

answered Jun 18 '14 at 06:29

JimmyK4542

54,331

How would you explain it to a 9th grade student? – panav2000k Jun 18 '14 at 06:33
Replace "$5n \equiv 8n \equiv -4n \pmod{720}$" with "$8n-5n$ and $5n-(-4n)$ are multiples of $720$", and replace "$3n \equiv 0 \pmod{720}$" with "$3n$ is a multiple of $720$". – JimmyK4542 Jun 18 '14 at 06:37

Question on Clocks

2 Answers2