Two brothers, Jack and Joe, were born in May. One was born in 1932, and the other, a year later. Each brother owned a watch, which was previously possessed by their father. Considering their age, the watches worked pretty well.
Nonetheless, one watch gained ten seconds per hour, while the other lost ten seconds per hour.

On a day of January, they set their watches at 12:00 PM. Joe said: "Hey, Jack, listen. The next time that our watches will both show exactly the same time, it will be on your 47th birthday!" Jack was surprised at first, but nodded.

Who is older, and what year is it?

  • $\begingroup$ Did they get the watches at exactly the same time or depending upon their ages? $\endgroup$
    – Sid
    Commented Dec 22, 2016 at 16:04

3 Answers 3


It is currently

January 31, 1979

and the older brother is


How I figured this out:

If the watches' times are moving in opposite directions (one is gaining and one is losing), then they will have the same time when each has gained/lost 6 hours, since that will make for a gap of 12 hours between their times.
Since they gain/lose time at a rate of 10 seconds/hour, it will take $60 \div 10 = 6$ hours to change by a minute, $6 \times 60 = 360$ hours to change by one hour, and $360 \times 6 = 2160$ hours to change by the required 6 hours.
At this point, one will have gained 6 hours, and one will have lost 6 hours, so they will be showing the same time (assuming a 12-hour clock).

$2160 \div 24 = 90$, so it will be Jack's birthday in exactly 90 days.

If it is currently January, and Jack's birthday is in May, then the only 90-day gap that works is if it is January 31, and Jack's birthday is May 1st (90 days later, assuming no leap year).

Since the brothers were born in 1932 and 1933, their 47th birthdays will be in 1979 and 1980. But if it is currently 1980 (a leap year), then Joe's math doesn't work out (there are 91 days from January 31 to May 1), so it is currently 1979, and Jack is the older brother (Joe's 47th birthday will be next year, in 1980).

  • 1
    $\begingroup$ damn it...got distracted with a meeting and didn't get to finish my answer in time. good one. $\endgroup$
    – Marius
    Commented Dec 22, 2016 at 16:26

Assuming they ignore daylight savings time

The year in question is either 1979 or 1980.
one clock loses 10s/hour. This means in 360h it loses 1h. So in 15 days, one clock loses 1h and the other one gains 1h.
in 90 days one clock loses 6h and the other one gains 6h.
This would mean that both of the watches should show the same time. 6 o'clock (there were not digital clock back then).


90 days from january puts you in april or in may only if the january day is 31st. This will make it 90 days later May 1st if the year is not a leap year. Otherwise it's not possible. 1979 was not a leap year. so Jack is born on May 1st 1932 and Joe was born somewhere in May 1933. and they both set their watches on January 31st 1979


If one watch gains 10 seconds and the other loses 10 per hour, there'll be a difference of 20 seconds per hour, and every $60*60*12/20=2160$ hours (90 days), both watches will show the same time. Since Jack's birthday must still be in May despite coming 90 days after a day in January, the 47th one can only be May 1, 90 days after Jan 31 in a leapless year. This day can't be on 1933+47=1980 (a leap year), which means he was born on May 1, 1932, and the day they set their watches was Jan 31, 1979.


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