The original incarnation of today’s puzzle talks about winding a clock but who, under the age of 40, knows what that means anymore? Here’s a slightly more modern formulationâ€¦

While you’re asleep one night, there’s a power outage. When you wake up, your alarm clock is flashing (but running), alerting you to the fact that you lost power for some period of time but you have no idea how long the outage lasted, and therefore no idea what time it is. You would like to reset your alarm clock to the current time but you don’t own any other clocks or watches or any communication devices whatsoever – no TV, no radio, no telephone, no cell phone, no computer, etc. And you live all alone in a very isolated place. Your only hope is a friend who has an accurate clock, however, he lives a long walk away from your house. So you walk to your friend’s house, stay overnight, then walk back home (assume the same amount of travel time in both directions). When you return home, you are able to accurately set your clock. How do you do it?

The answer will be posted tomorrow, along with the names of any readers who guess the solution.

** Solution:** Hats off to Al P. & Kimberly C., who cleverly worked out the correct solution. The trick is to record the times shown on your clock when you leave and return home, as well as the times shown on your friend’s clock when you arrive at, and leave, his house. Here’s what you do with that information:

- When you return home, subtract the time shown on your clock from the time shown when you left home – this is the total amount of time you’ve been gone.
- Subtract the time shown on your friend’s clock when you left from the time shown when you arrived – this is how long you stayed at your friend’s house.
- Subtract the amount of time spent at your friend’s house from your total time away from home, this is the two-way travel time to and from your friend’s house.
- Divide the travel time by two, to get the one-way travel time.
- Add the one-way travel time to the time shown on your friend’s clock when you left his house. That result is the current time.

That might seem a bit complicated but it’s not really that bad – let’s work through a specific example. Suppose I wake up to a flashing alarm clock displaying 9:00 am. I immediately leave for my friend’s house and, as soon as I arrive, I note that his clock displays 11:24 am. I stay at my friend’s house exactly 24 hours and leave the next morning at 11:24 am (per his clock). When I return home, I note that my clock displays 9:48 am. Now we simply follow the steps above:

- Subtracting the time shown on my clock when I returned home from the time shown when I left tells me I’ve been gone for 1 day and 48 minutes (9:48 – 9:00 on the previous day).
- I know that I spent exactly 1 day at my friend’s house.
- Subtracting the time spent at my friend’s house (1 day) from the total time away from home (1 day, 48 mins) gives me the total travel time of 48 minutes.
- Dividing the total travel time by two gives me a one-way travel time of 24 minutes.
- Therefore, I left my friend’s house 24 minutes ago, when his (accurate) clock showed the time to be 11:24 am so I can confidently set my clock to 11:48.

Now my alarm clock is back in synch with the current time and I can rest easy – until the next power outage!

This seems like an algebra problem with 2 equations and 2 unknowns, with the following variables:

A = time on my clock when I leave home for my friends’ house (expressed using 24-hour system)

x = amount of time my clock is behind the true time

y = amount of time to get to my friend’s house

B = time on my friend’s clock when I arrive

So: A + x + y = B

Next, on my return trip, I have the following variables:

C = time on my friend’s clock when I leave his/her house

y = amount of time to get home

D = time on my clock when I arrive (which is still x hours behind true time)

Now I have: C + y = D + x

If I rewrite these in terms of y, I get:

y = B – A – x

y = D + x – C

Adding these two statements together, I get:

2y = B – A + D – C, which can be rewritten as: y = (B – A + D – C)/2

I can solve for y since I know A, B, C and D. So the amount of time my clock is behind actual time is: x = (C + y) – D

Now I can set my clock accurately, by moving it ahead by x.

Use your “bad” clock to note the time you leave, this is not the accurate time but

you don’t care.

From your friends clock, you know how much time you spend overnight, say 8 hours.

Note the time you leave your friend’s using the “good” clock. Say 8am.

When you arrive home, your “bad” clock will tell you how long you’ve been gone. Say 10 hours.

Total time away (10) – overnight(8) = 2hours walking time. 1 hour each way.

Since you left you’re friend at 8am, you have arrived home at 9am and can set your clock.

Keep ‘em coming Marc.