Showing posts with label logical games online. Show all posts

Saturday, June 27, 2015

Four Hats

No comments :
Four men have been buried all the way to the neck, only their heads stick out. They cannot turn their heads, so they can see only in front of them. A wall has been placed between A and B, so that A cannot see the other 3 (B, C, D), and viceversa. All of them know in which position the others have been buried. So, for example, B knows that C and D can see him, even though he can't see them.

A hat has been placed on top of each man's head. All of them know that there are two black hats and two white hats, but no one is told the colour of the hat he's wearing.

four hats
They will all be saved if at least one of them can safely say what colour is the hat he's wearing. Otherwise they'll all be decapitated.

Which one of them saved the day? And, most importantly, how?

Four Hats Puzzle Solution

C saves the day.

D clearly has the most information at his hands, but seeing one white and one black hat doesn't give him any certainty about his own hat's colour. Would B and C both have been wearing the same colour, D would have been able to provide the answer.

But C is one clever guy and he knows that if D doesn't answer, it means that B is wearing a different colour than him. Because B is wearing white, C knows he's wearing black.

Note that A is redundant: the puzzle could have included only B, C, D. That way, the hats would have been three, with two hats of the same unspecified colour, and one other hat of the opposite colour.

Friday, June 12, 2015

Carpet Layer

No comments :
Walter Wall is a carpet layer. He and his two apprentices are asked by a nightclub owner to give a quote on laying carpet.

The owner indicates an oblong dance floor (figure on the right) and tells them that he wants a square of carpet adjacent each of the sides and running its entire length, making four squares in all (figure below left).

Walter asks both apprentices how many measurements must be made to calculate the total area of carpet needed in order to give a quote.
 
Sam, the slower of the two, replies that eight measurements are needed: two sides of each square.

Walter reprimands him, reminding him that these are squares and therefore have all sides the same length, and that they are in identical pairs, "So we only need to take two measurements - one side of one of the large squares and one side of one of the smaller squares".

Brian, the bright apprentice, points out that they can give the quote after taking only one measurement.


 How can the total area (that is, the sum of the areas of the four red squares) be calculated by taking just one measurement?

Carpet Layer Puzzle Solution

The only measurement that needs to be taken is the distance between opposite angles of the rectangular dance floor (figure on the left). That distance then will be squared (figure on the right) and doubled to get the sum of the areas of the 4 squares.

Friday, April 3, 2015

Trick Mules

No comments :
trick mules
Start with the three puzzle pieces arranged so that the two jockeys are correctly riding the two weary-looking mules.

Rearrange the pieces so that the mules miraculously break into a frenzied gallop!

 

Notes:
The mules do not overlap. There is a clever, but not deceptive solution to this puzzle. 

Trick Mules Puzzle Solution

This ambiguous "Trick Mules Puzzle" is solved by the realisation that the mule can have two different orientations. Here the same lines and contours have two interpretations, one horizontal and one vertical.
trick mules solution

Wednesday, March 4, 2015

Poker Results

No comments :
poker game riddle
Alice, Barbara, Claire, Daniel and Edward are discussing the result of the card game of the previous night.

Person #1 (woman): "Claire is single. The sisters and brothers all together totalled a loss of £9."

Person #2 (man): "My wife and I have lost a total of £1."

Person #3 (woman): "My sisters-in-law, all together, have lost £2."

Person #4 (man): "My brother-in-law and I have managed to lose £12 all together."

Person #5 (woman): "My result combined with that of Alice - who is Daniel's wife and an only child - is overall positive."

How much has each of them won or lost?

Poker Results Puzzle Solution

The 5 people and their wins/losses are:

Person #1: Barbara (+ £4).
Person #2: Edward (- £5).
Person #3: Alice (+ £14).
Person #4: Daniel (- £7).
Person #5: Claire (- £6).

This result is obtained from the statements of the 5 people.
  • Since the sisters-in-law (#1 and #5) lost, wile #5 and Alice have won, then #3 is Alice.
  • Alice won £14, because £2 lost by the sister-in-law, and £12 lost by the men.
  • #1 is Barbara, as she talks about Claire.
  • Alice is the only child, therefore the sisters-in-law must be sisters of the husband (Daniel).
  • Daniel lost £7, as the bunch of brothers and sisters have lost a total of £9, of which only £2 was lost by the sisters.
  • Edward lost £5, as the two men lost a total of £12, of which £7 was lost by Daniel.
  • Barbara - Edward's wife because Claire is single - won £4, as the total lost by the couple Barbara-Edward is £1.
  • Claire lost £6, as the total loss of the two sisters is £2, but Barbara won £4.

Monday, December 29, 2014

100

No comments :
Find at least four ways of writing the number 100, each time using only one digit repeated five times.
For example: (999 / 9) - 9 = 102 (but you must get 100, not 102!!!)

Good luck!

100 Puzzle Solution

  • 111 - 11 = 100
  • (3 * 33) + (3 / 3) = 100
  • (5 * 5 * 5) - (5 * 5) = 100
  • (5 + 5 + 5 + 5) * 5 = 100
  • (11 - 1) ^ ( 1 + 1) = 100 [Thanks to Steven Renich for that one!]
  • ((2 * 2 * 2) + 2) ^ 2 = 100 [Thanks to David Cohen for this other one!]
  • ((99 * 9) + 9) / 9 = 100 [Thanks to Taylor Lowry for this other one!]
  • ((22 - 2) / 2) ^ 2 = 100 [Thanks to Karen D. Miller for this other one!]
  • 6! / 6 - 6! / (6 * 6) = 100 [Thanks to Karen D. Miller for this other one!]
  • 5! - 5 - 5 - 5 - 5 = 100 [Thanks to Karen D. Miller for this other one!]
  • 5! - (5 + 5 + 5 + 5) = 100 [Thanks to Jim St. Clair for this other one!]
  • 5 * 5 * (5 - 5 / 5) = 100 [Thanks to Rishi Mohan Sanwal for this one!]
  • 4! + 4! + 4! + 4! + 4 = 100 [Thanks Karen D. Miller and Saurabh Gupta!]
  • 99 + 9 ^ (9 - 9) = 100 [Thanks Gopalakrishnan Thirumurthy for this one!]
  • 5! - 5 * (5 - 5 / 5) = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
  • (4! + 4 ^ (4 - 4)) * 4 = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
  • 4! * 4 + 4 - 4 + 4 = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
  • (4! * 4) + (4 * 4 / 4) = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
  • 5 * 5 * (5 - (5 - 5)!) = 100 [Thanks Gopalakrishnan Thirumurthy for it!]
  • ((9 - 9) */ 9)! + 99 = 0! + 99 = 1 + 99 = 100 [Thanks to Bala Neerumalla!]
  • (3 - 3)! + 33 * 3 = 100 [Thanks Bala Neerumalla for this other one!]
  • (5 + 5) ^ ((5 + 5) / 5) = 100 [Thanks Bala Neerumalla for this one!]
  • ((2 ^ 2) * 2 + 2) ^ 2 = 100 [Thanks Bala Neerumalla for this one!]
  • (4! + ((4 - 4) */ 4)!) * 4 = 100 [Thanks Bala Neerumalla for this one!]
Forum member Bealzbob reminds us that, whenever we have a subtraction like A - B, we can rewrite it as an addition with a negative number, like A + (-B). In the case of the first solution, 111 - 11 = 111 + (-11).

Glen Parnell points out that, in any solution in which the digit 2 is used, it could be replaced by 2!, whenever the 2 is not only a digit but also a number, as 2 = 2! (but of course, 22 is not equal to 22!).
Bala Neerumalla has suggested some alternative solutions using trigonometric functions Sin() and Cos():
  • Sin(99 - 9) + 99 = Sin(90) + 99 = 1 + 99 = 100
  • Cos((9 - 9) */ 9) + 99 = Cos(0) + 99 = 1 + 99 = 100
  • Cos(3 - 3) + 33 * 3 = 100
  • (4! + (Sin(4-4)*4)!)*4 = 100
Gopalakrishnan Thirumurthy has intelligently played around with cubic root functions. If we were to represent this function with a name like, say, CR(), the following alternative solutions could be achieved:
  • 88 + 8 + CR(8) + CR(8) = 100
  • (8 + CR(8)) ^ (CR(8) * (8 / 8)) = 100
  • ((CR(8) * CR(8) * CR(8)) + CR(8)) ^ CR(8) = 100
  • ((8 * (8 / 8)) + CR(8)) ^ CR(8) = 100
However! If the cubic root function is represented as it should be, ie ³Ö, then there is the extra "3" which invalidates the solution. These alternative solutions using CR(), have therefore been given as possible examples that will not be accepted in the future. Unless, of course, the "3" in ³Ö is actually one of the five digits used in a solution. Same goes with any other root or power. This also applies to any usage of the symbol "%", as it really involves the number 100.
Bala Neerumalla smartly noticed that he can produce number 100 by using a different number base than the normal base 10 (b10). The following example is in base 5 (b5):
  • 444 - 44 = (4 * 5^2 + 4 * 5^1 + 4 * 5^0) - (4 * 5^1 + 4 * 5^0) = 100
However! The above equation mixes up two different number bases: the left hand side is b5, while the right hand side is b10 (100 b5 is equal to 25 b10). This is the reason why we will not be posting anymore solutions that mix different number bases. We will, however, accept solutions that use number bases different from 10, as long as they are consistent on both sides of the equation. Therefore, we will accept:
  • 111 - 11 = 100 [Wolfgang Solfrank pointed out that this works in any number base, thanks!]
Here are some more examples of equations using different number bases:
  • b16 4 ^ 4 +- 4 * (4 - 4) = 100 [Thanks to Glen Parnell for this one!]
  • b16 (2 * 2 * 2 * 2) ^ 2 = 100 [Thanks to Glen Parnell for this one!]
  • b8 4 * 4 * 4 +- (4 - 4) = 100 [Thanks to Glen Parnell for this one!]
  • b8 (4 ^ 4) / 4 +- (4 - 4) = 100 [Thanks to Glen Parnell for this one!]
  • b8 4 * 4 * 4 * (4 / 4) = 100 [Thanks to Glen Parnell for this one!]
  • b8 (2 ^ (2 * 2)) * 2 * 2 = 100 [Thanks to Glen Parnell for this one!]
  • b2 11 - 1 + 1 + 1 = 100 [Thanks to Glen Parnell for this one!]
Furthermore, Glen Parnell noticed that there exists a general rule that applies to equations using any number base:
  • bX+1 ((X - X) */ X)! + XX = 100
Here are some examples of it:
  • b16 ((F - F) */ F)! + FF = 100
  • b12 ((B - B) */ B)! + BB = 100
  • b10 ((9 - 9) */ 9)! + 99 = 100 [As already shown by Bala Neerumalla]
  • b8 ((7 - 7) */ 7)! + 77 = 100
  • b4 ((3 - 3) */ 3)! + 33 = 100
  • b2 ((1 - 1) */ 1)! + 11 = 100
Jeff Shall cleverly discovered that he can produce 100 using the Roman numeral 'L', which corresponds to number 50:

  • ((L / L) + (L / L)) * L = ((50 / 50) + (50 / 50)) * 50 = 100
Unashamedly based on the above solution by Jeff Shall, we found that we can also use the Roman numeral 'C', which corresponds to number 100:

  • (C / C) * (C / C) * C = (100 / 100) * (100 / 100) * 100 = 100
  • (C / C) - (C / C) + C = (100 / 100) - (100 / 100) + 100 = 100
  • CCC - CC = 300 - 200 = 100
  • CC - CC + C = 200 - 200 + 100 = 100
Gopalakrishnan Thirumurthy has discovered a lot of new combinations with Roman numerals, including numeral 'X', which is number 10:

  • (L + L) * L ^ (L - L) = (50 + 50) * 50 ^ (50 - 50) = 100
  • (L + L) / L ^ (L - L) = (50 + 50) / 50 ^ (50 - 50) = 100
  • XX * X - X * X = 20 * 10 - 10 * 10 = 100
  • (X + (X / X)) * X - X = (10 + (10 / 10)) * 10 - 10= 100
  • X * X * (X ^ (X - X)) = 10 * 10 * (10 ^ (10 - 10) = 100
  • X * X / (X ^ (X - X)) = 10 * 10 / (10 ^ (10 - 10) = 100
  • C * C * C / C * C = 100 * 100 * 100 / 100 * 100 = 100
  • (C - C) + (C - C) + C = (100 - 100) + (100 - 100) + 100 = 100
  • L * (L - L) + L + L = 50 * (50 - 50) + 50 + 50 = 100
  • C * (CC / C) - C = 100 * (200 / 100) - 100 = 100
  • C * (C / C) + (C - C) = 100 * (100 / 100) + (100 - 100) = 100
Gopalakrishnan Thirumurthy has also noticed that all the solutions involving five times the number 5, could be rewritten by substituting it with Roman numeral 'V'. This is also true if, instead of number 5 and numeral 'V', we were talking about number 1 and Roman numeral 'I'.
Now, how about some solution using Roman numerals 'D' (500), or 'M' (1000)? If you find any alternative solutions, get in touch with us by email!

Easy!

Sunday, October 5, 2014

Faulty Batches

No comments :
"This time," said the Treasury Minister, "I ditched those dodgy Europeans, and I have assigned the manufacture of our gold coins to five American companies. Look, they are all shining and beautiful, and they are all exactly the same!"

The secretary looked at the coins, weighed some of them, and cleared his throat. "Ahem, Sir. I would like to point out that here we have at least three different kinds of coin; they all look the same, but their weight is different. Would you please come close to the scale? This coin weights 10 grams, as it should, but this other one is 11 grams, while this one is only 9 grams. Obviously two of our manufacturing companies haven't done a good job."

Sad as he could have been, for having been tricked agin by other dodgy companies, the Minister managed to raise his head. "Well.. it's just a matter of finding the fauly ones using the trick that you've showed me, by using the scale only once..."

"Sir. Actually, this is a different problem altogether, we need to find two sources of errors, rather than just one. One batch is heavier, another is lighter. The method I used before will not be sufficient this time. But we can nevertheless find the two offending batches by using the scale once."

How did they manage to use the scale only once?



Notes:
  • You may assume that each batch is made of a large amount of coins (thousands, millions, up to you! :)
  • All coins of the same batch weight the same amount.
  • The storyline in this puzzle follows from the story in Faulty Batch. It is however NOT necessary to have previously read/solved that puzzle in order to solve this one, even though it may be preferable.

Faulty Batches Puzzle Solution

They had to weigh 1 coin from the 1st batch, 2 from the 2nd, 4 from the 3rd, 8 from the 4th, and 16 from the 5th one.

If all coins weighed 10 grams as they should, the scale would display 310 grams ((1 + 2 + 4 + 8 + 16) * 10). However, since one batch has 9 grams coins, and another 11 grams coins, then the total weight of this combination of coins will be:
Total Weight Number of
9g coins
Number of
11g coins
311 1 2
313 1 4
317 1 8
325 1 16
312 2 4
316 2 8
324 2 16
314 4 8
322 4 16
318 8 16
309 2 1
307 4 1
303 8 1
295 16 1
308 4 2
304 8 2
296 16 2
306 8 4
298 16 4
302 16 8

After seeing the solution to this puzzle, it is clear that it would be a lot easier to simply use the scales up to 5 times rather than go through all this, but where is the fun in that?