Web Puzzles

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.

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.

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.

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.

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!

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?