Web Puzzles

A Farmer's Good Fortune Puzzle Solution

Here's one combination:

 94 sheep =  $470
  1 pig   =   $30
  5 cows  =  $500
-----------------
100 heads = $1000

Are there anymore?

A Special Old Guy Puzzle Solution

Firstly, we can say that the date of birth (DOB) for the special old guy (SOG) is:
Equation {a}:
DOB(SOG) = x^2 – x

Similarly, for the son of the special old guy (or SOSOG for short):

Equation {b}:
DOB(SOSOG) = y^3 – y

SOG's age when SOSOG was born was:
Equation {c}:
Age(SOG) = DOB(SOSOG) – DOB(SOG) = (y^3 – y) – (x^2 – x)

There are unlimited solutions to equation {c} so we need some assumptions and limits.

We know that SOG must have been between, say 10 years old and 100 years old when SOSOG was born:

Equation {d}:
10 < Age(SOG) < 100

We also know that both SOG and SOSOG were born some time since the year 500 AD:
Equation {e}:
500 < DOB(SOSOG) < 2004.

Based on equation {b} and {e} we can clearly see that there are only 5 solutions for y. They are y = 8, 9, 10, 11 or 12. Any other solutions for y are in breach of equation {e}.
For each of these possible solutions for y there is only a limited number of solutions for x that comply with {d}. They are:

 x     y
--------
21     8
22     8
26     9
27     9
31    10
36    11
41    12

Any other solutions for x and y are in breach of equation {d}.
SOG's age when his son was born can be calculated for each of these possible solutions by using equation {c} as follows:

 x      y     Age(SOG)
----------------------
21      8           84
22      8           42
26      9           70
27      9           18
31     10           60
36     11           60
41     12           76

So there are 7 different solutions, which is why Buzz said he couldn't solve for x and y.
Now comes the lateral part of the puzzle: Alvin informs Buzz that "...he was your age when his son was born". Of course, Buzz knows his own age. He should therefore be able to pick the correct solution from the list of 7 possible solutions shown above. However, he can't. That means that Buzz (and the special old guy) must be 60 because there are multiple solutions for an age of 60, whereas any other age would yield a unique solution. Any other age and Buzz would be able to solve.
A 60 year old father – quite a special old guy.

The Diagram Puzzle Solution

Joe counted how many times each of the 7 LEDs lit up for all 10 digits...

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Mountaineer Puzzle Solution

Of course there is. To make sure, imagine two mountaineers: one is in the village, and the other one is at the refuge. They'll both leave at eight o'clock, travel along the same path as our mountaineer, and at his same speed. At some point along the path they'll obviously meet.

Napoleon's Star Puzzle Solution

To be able to place nine coins, it is necessary to make the 3rd point of each step equal to the start point of the previous step. For example:

a-g; i-a; c-i; f-c; e-f; h-e; b-h; j-b; d-j.


 With such a simple solution, it's hard to believe that Napoleon stayed in charge for so long.

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!