Merchant's Gold
Math brain teasers require computations to solve.
A rich merchant had collected many gold coins. He did not want anybody to know about them. One day, his wife asked, "How many gold coins do we have?"
After pausing a moment, he replied, "Well! If I divide the coins into two unequal numbers, then 32 times the difference between the two numbers equals the difference between the squares of the two numbers."
The wife looked puzzled. Can you help the merchant's wife by finding out how many gold coins they have?
Answer
The merchant has 32 gold coins.
It is easy to check this... Let's divide the 32 coins into two unequal numbers, say, 27 and 5. Then,
32 (27 - 5) = (27 x 27) - (5 x 5).
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Comments
majd 
Aug 19, 2002
| That was a nice one, I managed to solve it in a relatively short time |
Sir_Col  
Aug 19, 2002
| You can solve this algebraically. If x and y represent the number of coins in each pile, then 32(x-y)=x^2-y^2, but the difference of two square, x^2-y^2, can be written, (x+y)(x-y). Hence 32(x-y)=(x+y)(x-y). Divide by (x-y) and we get x+y=32. |
cs526columbia
Aug 19, 2002
| very cool |
mbunap
Jul 22, 2007
| I like Sir Col's technique. I tend to set up problems algebraically but had forgotten the (x+y)(x-y) manipulation. The number 32 is not unique. We can plug any number into this teaser and get a correct result. |
4demo
Aug 10, 2007
| Although it took me a couple of minutes to solve it algebraically, I actually found it very easy to actually solve it. Good teaser!  |
javaguru 
Jan 08, 2009
| While Sir Col's approach is simple, even simpler is to realize that if one of the unequal numbers is zero, then 32^2 = X^2, so X must be 32.
 |
javaguru
Jan 08, 2009
| Actually I didn't write what I meant last post. If one of the unequal numbers is zero then 32X = X^2 so X = 32. |
Jimbo
May 13, 2009
| I like Sir Col's approach but what I didn't realise from the wording until after I had solved it was that the 32 coins could be divided in any way and the result would still be true. I had assumed that the pile could only be divided one way in order to achieve the stated result. Javaguru's method seems a good one with hindsight but you have to know that you can divide the number of coins into any 2 random unequal piles.
Fun puzzle. |
opqpop
Sep 10, 2010
| This is too easy. Difference of squares and you're done. I do like the ingenuity however. Well done. |
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