Solve Enjoyable Enigmas with Mr.E's Challenge

  • Thread starter Enigman
  • Start date
In summary: Four princes approach the king vying for the hand of the princess. In order to choose the best among the four suitors the king and the princess arrange a test: the suitors are brought to a large rectangular hall. The floor is covered with a carpet all over except at the corners- where there are four squares of bare floor and the suitors are told to stand at these corner. Each suitor takes a corner and stands there while the princess stands at the center of the hall. The king then proclaims the prince who without leaving their respective squares shall put a ring on the princess's hand will be declared to be the bridegroom of his daughter and the heir to Enigmania. No ropes or rods are
  • #1,016
Dunno, maybe they had 69 barrels and 22 barrels respectively before paying duty?
 
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  • #1,017
zoki85 said:
Dunno, maybe they had 69 barrels and 22 barrels respectively before paying duty?
No, this isn't it. The number of barrels given is exactly what they arrived at the gates of Paris with.

I'll let this sit a while and see if anyone else hits on the alternative answer.
 
  • #1,018
Ok, my last try on it. If the barrels they are paying with are considered money and not goods, they are actually paying duty for 59 barrels and 18 barrels respectively. So the equations would be:
59x=5y+40
18x=2y-40
?
 
  • #1,019
zoki85 said:
Ok, my last try on it. If the barrels they are paying with are considered money and not goods, they are actually paying duty for 59 barrels and 18 barrels respectively. So the equations would be:
59x=5y+40
18x=2y-40
?
So, looking at this new way, what is the value of each barrel of wine, and what is the duty payable on it?
 
  • #1,020
zoobyshoe said:
So, looking at this new way, what is the value of each barrel of wine, and what is the duty payable on it?
That doesn't interest me anymore now if you confirm this way that was the catch;)
 
  • #1,021
zoki85 said:
That doesn't interest me anymore now if you confirm this way that was the catch;)
There never was a catch. There is an ambiguity. You could assume, as you did, that they were required to pay customs duty on the whole amount they originally wanted to take into Paris, (even though they wouldn't actually be taking that amount in because they were using part of it to pay the duty) or, you could assume they would only be required to pay duty on the amount they actually ended up taking in, 59 and 18 barrels, respectively.

The problem is that either way, the amount they end up paying, and the number of barrels they end up taking into Paris, are exactly the same, despite the fact the price per barrel and the duty amount charged per barrel are different in the two different scenarios. The puzzle, as its stated (and I copied it verbatim from the book) has no indication of which of these two possibilities it is looking for. It makes a difference because they specifically ask for the price per barrel and the amount of duty.

I didn't see any way around it but to calculate both possibilities and then check to see which the book wanted.
 
  • #1,022
Using ten coins, construct a triangle like this:

puzzle1.jpg


Now, completely invert the triangle by moving only three of the coins.
 
  • #1,023
puzzle1.jpg
 
  • #1,024
zoki85 said:
puzzle1.jpg
Correct! And nice graphic work!
 
  • #1,025
There are 5 sacks, of which Nos. 1 and 2 weigh a total of 12 lbs; Nos. 2 and 3, 13 1/2 lbs; Nos. 3 and 4, 11 1/2 lbs; Nos. 4 and 5, 8 lbs; Nos. 1, 3, and 5, 16 lbs.

Find the weight of each sack.
 
  • #1,026
5.5, 6.5, 7, 4.5, 3.5
 
  • #1,027
Enigman said:
5.5, 6.5, 7, 4.5, 3.5
Yes.

Could you describe how you found the answer? The way I did it (which got me the right answer) seems remote from the method described in the book, and I can't even understand the book's procedure.
 
  • #1,029
Enigman said:
I did it using Matrix method of solving equations.
http://www.mathsisfun.com/algebra/systems-linear-equations-matrices.html
Thanks. I'll have to study that a while to see if I can make sense out of it.

Here's what the book said:

"The sum of all the weighings, 61 lbs, includes sack 3 thrice and each of the others twice. Deducting twice the sum of the first and fourth weighings, i.e. 21 lbs for thrice 3 - i.e. 7 lbs for sack 3. The rest follows."

This enigma, incidentally, was written by Lewis Carroll
 
  • #1,030
Cute little puzzle for people who like chess.
White has just made the last move and the game ended in a draw (stalemate-see diagram). Determine that move.
movediagram.php.gif
 
  • #1,031
zoki85 said:
Cute little puzzle for people who like chess.
White has just made the last move and the game ended in a draw (stalemate-see diagram). Determine that move.
movediagram.php.gif

King (at d1) takes knight at c1.

[Edit: that was a toughy! I was halfway convinced for awhile that this enigma did not have an answer. But there is one! :w]
 
Last edited:
  • #1,032
The black king must have moved all the way below the pawns at some point. As those did not move, A3/B3 were always blocked. It would be tempting to assume the knight moved in, but then the black king would have had to come from A1, which was not a legal position before. Therefore, the black king did not move last time, and the white king captured a black figure at C1. The king was at D1 before, this just allows a black knight to move in. The capture scenario is not possible with the white knight as there is no possible way the black figure could have moved to A1 before.
Summary: Black knight from somewhere moves to C1, white king from D1 captures it.
This was a bad decision, taking the rook would have been better.

Edit: Wow, took me way too long to write that answer. Well, with proof this is the only option.
 
  • #1,033
That's right, king takes knight on C1 :)
 
  • #1,034
zoki85 said:
That's right, king takes knight on C1 :)
This puzzle was beyond me, and I'm amazed mfb and collinsmark were able to figure out the last move involved taking a specific piece which is no longer even on the board. Are they PSYCHIC?
 
  • #1,035
They must be.oo)
 
  • #1,036
zoobyshoe said:
This puzzle was beyond me, and I'm amazed mfb and collinsmark were able to figure out the last move involved taking a specific piece which is no longer even on the board. Are they PSYCHIC?
drizzle said:
They must be.oo)
Haha, doing a little of retrograde analysis certainly helps more here
 
  • #1,037
If you like this sort of chess puzzles, I'll post one more:

movediagram.phpd2436.gif

White just played a move (diagram), but instantly noticed that wasn't the best move.
With allowance of Black, White takes back the move, and then executes a new move giving a checkmate with it.
Can you reconstruct the moves?
 
  • #1,038
Nice one
White took on black's bishop on e8 with his pawn promoting to a knight. Best was pawn to f8 knight promotion resulting in mate.
[\spoiler]
 
  • #1,039
zoki85 said:
White just played a move (diagram), but instantly noticed that wasn't the best move.
With allowance of Black, White takes back the move, and then executes a new move giving a checkmate with it.
Can you reconstruct the moves?
I'm confused. Does the image depict the situation just after white made the move it regretted, or does it depict the situation before white made the move it regretted?
 
  • #1,040
consciousness said:
Nice one
White took on black's bishop on e8 with his pawn promoting to a knight. Best was pawn to f8 knight promotion resulting in mate.
[\spoiler]
Correct:)
 
  • #1,041
zoki85 said:
White just played a move (diagram), but instantly noticed that wasn't the best move.
zoobyshoe said:
I'm confused. Does the image depict the situation just after white made the move it regretted, or does it depict the situation before white made the move it regretted?

I think the image is, "just after white made the move it regretted." :( That's based on zoki85 indicating, "White just played a move (diagram)."
- TD
 
  • #1,042
It is known that a quadratic equation has either 0, 1, or 2 unique real solutions. Well, look at this equation:
$$\frac{(x-a)(x-b)}{(c-a)(c-b)}+\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}=1$$
Without loss of generality, assume a < b < c. Now note that x=a, x=b, and x=c are all unique solutions! How can this equation have 3 solutions?!
 
  • Like
Likes collinsmark and Jonathan Scott
  • #1,043
1=1 has more than 2 solutions.

Very nice puzzle.
 
  • #1,044
Correct.
If a quadratic has more than two solutions then it will be an identity and therefore have infinite solutions.
 

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