Weekly puzzle 14 – solution
Day Of The Week
The answer is 17 days and 3 hours later, which would have been a Wednesday. This is the only other time in the same month when the two would agree at all.
In 17 days the slow clock loses 17*24*7 minutes = 2856 minutes, or 47 hours and 36 minutes. In 3 hours more it loses 21 minutes, so it has lost a total of 47 hours and 57 minutes. Modulo 12 hours, it has *gained* 3 minutes so as to make up the 3 minutes it was slow on Sunday. It is now (fortnight plus 3 days) exactly accurate.
Since the clock was not adjusted since the last visit, it’s also possible that the radio time shifted by one hour due to a change to or from summer daylight saving time. However, it turns out that the only additional possibilities that need to be considered are those of 4 days 15 hours later, when the clock would have lost 12 hours 57 minutes, and 29 days 15 hours later, when the clock would have lost 3 days 10 hours 57 minutes. Without even considering the rules for when in the month the clock is changed, these possible solutions are ruled out because we know that both visits were in the evening (“I spent a second evening with him”). and they involve times in a different part of the day.
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Oops, I forgot to tell, 3h30m is assuming Tom’s cottage is in England.
To not waste this comment, I’ll add one more explanation to Andrew excellent solution. After the second event in which the radio time and the clock are exactly right, the next time that happens is when 7n is a multiple of 12*60, n being the number of hours until the coincidence happens again, in other words when 7n=12*60*m, m being an integer to know. This can be easily solved by anyone who knows how to do calculus with modules:
7n=12*60*m => (mod 7) 0=5*4*m=-m => m=7
Knowing m, we can calculate n as the number of hours until that happens again:
n=12*60*(7)/7 = 720h = 30d
In other words, this kind of coincidences only happens in fixed periods of 30 days if we don’t count with the summer hour change… The first event is the only one we have to worry about, because the next one is surely after that month.
If we count with the summer change, the equation to search is a little different:
7n=(12*m +/- 1)60 =>0= ( – m +/- 4) (mod 7) => m=4 or 3
n= (12*4 + 1)60/7 = 420h = 17.5d
n= (12*3 – 1)60/7 = 300h = 12.5d
Now, it could appear that we are in trouble because these results can combine with the 4days and 15 hours of Andrew to create a proper evening event in the same month (for instance, 4d15h+12d12h=17d3h). However, that can only happen if there are 2 changes of summer hour in the some month… which doesn’t happen.
And since both visits were between 6pm – midnight, and I know the event in the second one happened 3 hours later in the evening in relation to the first event, that means it happened at 10pm or 11pm. Add to it 3h30m and we’ll have the first hours of panj-šambe (Persa for Thursday), at Iran.
In the end, I cannot claim I solved the puzzle because, I didn’t had the courage to face the summer hour problem (I simply ignore it, although while searching for weird hours fuses, I tried to find summer hour country followers). Nice touch that one.
PS.: Weird, panj-šambe is very alike Perşembe… same linguistic root?