Sam Loyd
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Sam Loyd's Cyclopedia of 5000 Puzzles
Cyclopedia of 5000 Puzzles
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Sam Loyd's puzzles

Samuel Loyd (January 30, 1841 – April 10, 1911), born in Philadelphia and raised in New York, was an American chess player, chess composer, puzzle author, and recreational mathematician.
As a chess composer, he authored a number of chess problems, often with interesting themes. At his peak, Loyd was one of the best chess players in the US, and was ranked 15th in the world, according to chessmetrics.com.
He played in the strong Paris 1867 chess tournament (won by Ignatz von Kolisch) with little success, placing near the bottom of the field.
Following his death, his book Cyclopedia of 5000 Puzzles was published (1914) by his son. His son, named after his father, dropped the "Jr" from his name and started publishing reprints of his father's puzzles. Loyd (senior) was inducted into the US Chess Hall of Fame. (Wikipedia: Sam Loyd)


Sam Loyd's donkey puzzle: Trick Donkeys problem.


The solver must cut out the three rectangles along the dotted lines and rearrange the three pieces so that the two jockeys are riding the two donkeys.
Trick Donkeys problem


   Answer   Answer

 

Solution: First, Place the donkey pieces back-to-back.
Sam Loyd's donkey puzzle solution
Then place the jockeys piece on the middle.
Sam Loyd's donkey puzzle solution


Pythagoras' classical problem.


Pythagoras' classical problem

Proposition--Take a piece of paper of the dimensions of the two squares, as shown in the picture, and cut it into three pieces which will fit together and make a perfect square.

Pythagoras' classical problem


  

Answer

Pythagoras' classical problem Pythagoras' classical problem
Answer

Pythagoras' classical problem Pythagoras' classical problem Pythagoras' classical problem Pythagoras' classical problem

  

Question:
Pythagoras' classical problem
Proposition--Take a piece of paper of the dimensions of the two squares, as shown in the picture, and cut it into three pieces which will fit together and make a perfect square.


 


 


.... Sam Loyd's Mathematic puzzles ....

Puzzling scales.


Puzzling scales



  

Sam Loyd's answer:

Puzzling scales

Solve the puzzle by using algebra:

Let x= the weight of a top, y= the weight of a cube and z= the weight of a marble
In the first scale, x + 3y = 12z
In the second scale, x = y + 8z
If we add 3 cubes to each side of the second scale, we get x + 3y = 4y + 8z
Since x + 3y = 12z and x + 3y = 4y + 8z, therefore 12z = 4y + 8z
when we combine like terms, we get 12z - 8z = 4y
Therefore z = y
In other word, a marble weighs as much as a cube.


The lost cent.


The lost cent



It is told that 2 huckster ladies were selling apples at the market, when Mrs. Smith was called away and asked Mrs.Jones, the other apple lady, to dispose of her stock for her.

Now, it appears that each had an equal number of apples, but Mrs. Jones had larger fruit and was selling hers at the rate of two for a penny, while Mrs.Smith sold three of hers for a penny.

Upon accepting the responsibility of disposing her friend's stock, Mrs. Jones, wishing to be very impartial, mixed them all together and sold them off at the rate of five apples for two pence.

When Mrs. Smith returned the next day, the apples had all been disposed of but when they came to divide the proceeds they found that they were just seven pence short, and it is this shortage of the apple or financial market which has disturb the mathematical equilibrium for such a long period.

Supposing that they divided the money equally, each taking one-half, the problem is to tell just how much money Mrs.Jones lost by the unfortunate partnership.



  

Sam Loyd's answer:

The lost centThe lost cent


The lost cent - Sam Loyd's answer:
To tackle the problem from a somewhat new standpoint it can readily be shown that the apples,
if sold at 1-3 of a penny and 1-2 of a penny, would average 5-6 for two, or 25-60 of a penny for each apple,
but as they were closed out at the rate of five apples for two pence, which is the same as 2-5 or 24-60 of a penny per apple,
then 1-60 of a penny was lost on every apple.
As it was stated that seven pence was lost, we will multiply that 60 by 7, which shows there must have been 420 apples,
of which they each had one-half.
As Mrs. Jones had 210, for which she would have received 105 pence, but only got one-half of the proceeds of the entire sale at the rate of 5 for 2 pence, viz.: 84 pence, she lost twenty-one pence,
while Mrs. Smith, who should have received but seventy pence for her three-for-a-penny fruit, actually gets eighty-five.

The mysterious discrepancy occurs at the end of the seventieth combination sale.
Mrs. Smith's cheap fruit becomes exhausted on the seventieth sale, which takes 210 of three-fors, and 140 of the two-fors,
and at that stage of the game Mrs. Smith was entitled to half of the proceeds, and should have withdrawn with her seventy pence.
As there were now just seventy of the better class stock left, every sale now involves the giving of three apples for a penny, which should sell for two for a penny, Mrs. Jones' stock is sacrificed.



Similar puzzle can also be seen at www.jwstelly.org: The lost cent

 


 

Hill puzzle (Hipity-Hop, the lame peddler).


Hill puzzle

  

Sam Loyd's answer

Hill puzzle


Hill puzzle - Sam Loyd's answer:
Hipity-Hop could go one mile up the hill in 40 minutes, and could come down a mile in 13 1/3 minutes.
Therefore he would average a mile up and down in 53 1/3 minutes.
Since the particular hill which he tells about required six hours to climb and descend,
we may determine its height by dividing six hours by 53 1/3 minutes.
Thus we learn that hill must have been six and three-quarter miles high.



Similar puzzle can also be seen at www.jwstelly.org: Hill puzzle

The weight of a brick.


The weight of a brick

  

Sam Loyd's answer

The weight of a brick


The weight of a brick - Sam Loyd's answer:
Algebra teaches us that the balance is not affected by removing similar quantities from both sides of an equation,
so, in this puzzling little proposition we remove three-quarters of the whole brick and cancel off the three-quarter bat.
This leaves the weight balancing with one-quarter of a brick;
therefore if one quarter of a brick weighs three-quarters of a pound,
a whole brick weighs three pounds.