This problem was invented by Édouard Lucas, a French nineteenth-century mathematician.
“Every day at noon,” Lucas said, “a ship leaves Le Havre for New York and another ship leaves New York for Le Havre. The trip lasts 7 days and 7 nights. How many New York-Le Havre ships will the ship leaving Le Havre today meet during its journey to New York?”
Warning: Use of undefined constant bfa_comments - assumed 'bfa_comments' (this will throw an Error in a future version of PHP) in /home/customer/www/riddledude.com/public_html/wp-content/themes/atahualpa/comments.php on line 132
14 ships
7 that left during the 7 days prior to its departure
+
7 That will leave during the 7 days trip to New York
Nope. Think about it again.
6 ships will pass eachother
Nope! You’re getting colder (like the soup).
Try again.
Assuming you would also count the ship docked in New York and waiting to leave when the Le Havre ship arrives, the total would be Roolstar’s answer plus one – 15.
15 is correct!
At the moment you take off there will be 6 NY-LH ships already at sea, one docking at LH and one casting off at NY. At noon of each day of the voyage a new ship will be launched. You must encounter every ship that was already at sea (6) and each one that was launched during your voyage (including the one that launched at the same time you did, so 7 more). So you will pass 13 ships at sea and also encounter a ship that docked as you cast off and another that cast off as you docked.
You’re today’s winner.
15 if you count the ships that are just arriving as the other ship leaves.
13 at sea
The question in this classic problem is supposed to be “at sea” discounting the other 2 at quay
13