While working on a sermon the pastor heard a knock at his office door. "Come in," he invited. A sad-looking man in threadbare clothes came in, pulling a large pig on a rope. "Can I talk to you for a minute?" asked the... Read more of Quite In Church at Free Jokes.caInformational Site Network Informational
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THE SPANISH MISER.

(Money Puzzles)
There once lived in a small town in New Castile a noted miser named Don
Manuel Rodriguez. His love of money was only equalled by a strong
passion for arithmetical problems. These puzzles usually dealt in some
way or other with his accumulated treasure, and were propounded by him
solely in order that he might have the pleasure of solving them himself.
Unfortunately very few of them have survived, and when travelling
through Spain, collecting material for a proposed work on "The Spanish
Onion as a Cause of National Decadence," I only discovered a very few.
One of these concerns the three boxes that appear in the accompanying
authentic portrait.
Each box contained a different number of golden doubloons. The
difference between the number of doubloons in the upper box and the
number in the middle box was the same as the difference between the
number in the middle box and the number in the bottom box. And if the
contents of any two of the boxes were united they would form a square
number. What is the smallest number of doubloons that there could have
been in any one of the boxes?


Answer:

There must have been 386 doubloons in one box, 8,450 in another, and
16,514 in the third, because 386 is the smallest number that can occur.
If I had asked for the smallest aggregate number of coins, the answer
would have been 482, 3,362, and 6,242. It will be found in either case
that if the contents of any two of the three boxes be combined, they
form a square number of coins. It is a curious coincidence (nothing
more, for it will not always happen) that in the first solution the
digits of the three numbers add to 17 in every case, and in the second
solution to 14. It should be noted that the middle one of the three
numbers will always be half a square.










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