> Two friends, George and Harry, were born in May, one in 1964 and the other > a year later. Each has an antique 12-hour clock. One clock loses 10 > seconds an hour and the other gains 10 seconds per hour. On a day in > January the two friends set both clocks right at exactly 12 noon. "Do you > realize, " says George, "that the clocks will drift apart and won't be > together again until ... good grief, your 23rd birthday!" How long will > it take for the two clocks to come together again? Which friend is older, > George or Harry?
Your conundrum could also have asked for the date they set their clocks and the date of Harry's birthday - seemingly even more impossible from the given facts than divining which of them is older! Anyway, on with the solution... Since they are 12 hour clocks, they will first be in sync again when one has lost 6 hours and the other has gained 6 hours. At 10 seconds per hour, this will take 2160 hours, i.e. 90 days. The current date is somewhere in January and their birthdays are not until May. February, March and April add up to 89 days if it's not a leap year and 90 if it is. The latter case doesn't fit the facts given, because then the current date couldn't be in January with their birthdays in May, hence it isn't a leap year. So, the date they set their clocks is 31st January, and one of their birthdays is on 1st May. If George were born in 1964 then Harry was born in 1965. He will be 23 in May which would make this year 1968. But this is a leap year, which contradicts our findings above! If Harry were born in 1964 and was 23 in May, this year would be 1987, and George was born in 1965, making him 22. Thus Harry is older. Hope this hasn't spoilt your fun - but then if you didn't want to know the answer, you shouldn't have downloaded the message! Regards to all, David Higgon (never could resist a challenge!) London
