Tag Archives: Eureka magazine

Railway Crossing Problem

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This is an interesting problem from the 1966 Eureka magazine.

“A railway and a road run together for seven miles from P to Q. Two miles from P there is a level crossing, which is closed one minute before, and opened one minute after, a train passes.

A train passes a Stationary car at P and travels on to Q at 60 m.p.h., and, forgetting to slow down, crashes at Q; the car passes the train as it crashes. Assuming that stopping for an instant from full speed loses the car one minute, of what speed must it be capable?”

Answer

See the Railway Crossing Problem for a solution.

The Umbrella Problem

This is a rather mind-boggling problem from the 1947 Eureka magazine.

“Six men, A, B, C, D, E, F, of negligible honesty, met on a perfectly rough day, each carrying a light inextensible umbrella. Each man brought his own umbrella, and took away—let us say “borrowed”—another’s. The umbrella borrowed by A belonged to the borrower of B’s umbrella. The owner of the umbrella borrowed by C borrowed the umbrella belonging to the borrower of D’s umbrella. If the borrower of E’s umbrella was not the owner of that borrowed by F, who borrowed A’s umbrella?”

Answer.

See the Umbrella Problem for solutions.

Pillar Wrapping Problem

This is a fun problem from the 1949 Eureka magazine.

“The following problems were set at the Archimedeans’ 1949 Problems Drive. Competitors were allowed five minutes for each question.  [This is problem #9.]

A pillar is in the form of a truncated right circular cone. The diameter at the top is 1 ft., at the bottom it is 2 ft. The slant height is 15 ft. A streamer is wound exactly five times round the pillar starting at the top and ending at the bottom. What is the shortest length the streamer can have?”

Answer.

See the Pillar Wrapping Problem for solution.

“Fermat’s Last Theorem” Puzzle

Here is a mind-numbing logic puzzle from Futility Closet.

“A puzzle by H.A. Thurston, from the April 1947 issue of Eureka, the journal of recreational mathematics published at Cambridge University:

Five people make the following statements:—

Which of these statements are true and which false?  It will be found on trial that there is only one possibility.  Thus, prove or disprove Fermat’s last theorem.”

Normally I would forgo something this complicated, but I thought I would give it a try.  I was surprised that I was able to solve it, though it took some tedious work.  (Hint: truth tables.  See the “Pointing Fingers” post regarding truth tables.)

One important note.  The author is a bit cavalier about the use of “Either …, or …”.  In common parlance this means “either P is true or Q is true, but not both” (exclusive “or”: XOR), whereas in logic “or” means “either P is true or Q is true, or possibly both” (inclusive “or”: OR).  I assumed all “Either …, or …” and “or” expressions were the logical inclusive “or”, which turned out to be the case.

See the Fermat’s Last Theorem Puzzle

Hard Time Conundrum

This problem comes from the “Problems Drive” section of the Eureka magazine published in 1955 by the Archimedeans at Cambridge University, England.  (“The problems drive is a competition conducted annually by the Archimedeans. Competitors work in pairs and are allowed five minutes per question ….”)

“There are ten times as many seconds remaining in the hour as there are minutes remaining in the day. There are half as many minutes remaining in the day as there will be hours remaining in the week at the end of the day.  What time is it on what day?”

One of the hardest parts of the problem is just being able to translate the statements into mathematical terms.  Solvable in 5 minutes?!!!

Answer.

See the Hard Time Conundrum for a solution.