Coin of the ream
Problem 180 — Where there’s a will, there’s a weigh, as this month’s problem by Knute Hancock of Kinnelon, N.J., demonstrates.
Paparazzi, tabloid magazine reporters, and hordes of curious spectators all jostled each other for a view inside the offices of Luhpohl, Fuddle, and Dodge, Attorneys at Law. Their client, the famous old multimillionaire recluse, Thaddeus Wadd IV, had died of mysterious causes earlier in the week, and two people had a vested interest in the testamentary depositions.
Inside the offices, Wadd’s former business partner, who had just been released from prison, and his 21-year-old wife of one month craned forward eagerly. Luhpohl cleared his throat and began to read the last will and testament.
Wadd had converted his property into Krugerrands — gold coins that weigh 1 troy oz — and divided the coins evenly into two large bags. Into another bag he had put and equal number of fake gold coins that weighed 1.1 troy oz each. The bags were labeled #1, #2, and #3.
Under terms of the will, Maurice Flote would choose two bags for himself and Lulu “Gold” Digha Wadd would get the remaining bag. Wadd had also ordered that a direct-reading scale (not a balancetype), accurate to 0.1 oz, be in the room. The scale could be used only once. How can Flote determine which bag has the fake coins, and thus run no risk of choosing it?
Technical consultant, Jack Couillard, Menasha, Wis.
Solution to last month’s problem 179 — Your cup doesn’t runneth over, if you answered 2 hr, 26 min.. Here’s how Augustus Pfizz’s speech was all wet.
First compute the volume of the inside of the cup. The height is given as 7 ft, and the equation for the slope of the sides is given as:
The equation is continuous on [1, 8], and so we can compute the volume by revolving the region bounded by ƒ(y) = x3 around the y axis. By definition of a solid of revolution, the equation for volume becomes:
437 gal ÷ 3 gpm is approximately 146 min, or 2 hr, 26 min.
Scrubbers at the d’Poo will be working lots of overtime.
