# Web Puzzles

Showing posts with label mathematical riddles. Show all posts

## A Farmer's Good Fortune

A farmer from a small community is out of money. After a mysterious desease spread among and killed his lifestock, he now needs to quickly make up for the lost animals. He needs a whole grand.
Knowing that the bank won't lend him any money, he pays a visit to the local loan shark. The outlaw, who's known to have a bit of an obsession with puzzles, proposes a deal.

With the \$1,000 he gets, the farmer has to be able to buy a combination of cows, pigs, and sheep, to total exactly 100 heads of lifestock. The combination has to include at least one cow (\$100 each), one pig (\$30 each), and one sheep (\$5 each). The total amount of money spent for the 100 animals has to equal exactly \$1,000. If the farmer manages to accomplish the task, he'll have to return the money with a "friendly" interest rate. Otherwise, he'll get the normal rate, and the threat of a broken pinkie...

How many of each kind of livestock did the farmer buy?

### A Farmer's Good Fortune Puzzle Solution

Here's one combination:

``` 94 sheep =  \$470
1 pig   =   \$30
5 cows  =  \$500
-----------------
```

Are there anymore?

## A Special Old Guy Alvin and Buzz are nerds and like doing nerdy things. So Alvin called Buzz one day...

"Buzz, I've finished tracing my family tree back from the year 500 AD, and I found one quite special guy".

"Well, he was x years old in the year x^2 (x squared) and he had a son who was y years old in the year y^3 (y cubed)".

Buzz looked perplexed "Sorry Alvin, but I can't solve for x or y".

"Well, he was your age when his son was born." said Alvin.

"You're right" said Buzz "He was a special old guy! But I still can't solve for x or y".

How old was the old guy when his son was born?

Notes:
Assume that the nerds have the conversation this year, ie 2004 AD.

### 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

Bob looked up from his book and noticed that Joe was staring at the VCR clock, holding a pencil and a pad of paper. He knew the clock was set correctly, having set it himself. But he noticed that Joe would occasionally write something down.

Finally Bob's curiosity got the best of him: "What are you doing, Joe?"

"Just another minute" came the reply.

"How long are you going to stare at the clock like that?" As Bob finished he noticed the clock advance a minute.

At this, Joe started scribbling some more until he produced this diagram:
 8 6 8 7 4 9 7
"Why, just ten minutes, of course", Joe beamed as he showed the diagram to Bob.

What was Joe doing, and what does his diagram mean?

### The Diagram Puzzle Solution

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

## Mountaineer

An Austrian mountaineer left Zurglatt, his village, at eight o'clock in the morning, and started his climb towards the refuge Tirpitz, on Gross Glossen mountain. He walked at a steady pace, without stopping, and his increase in heart pulse rate was negligible. He reached the refuge at three in the afternoon, i.e. seven hours since he left the village. At the refuge he rested, admired the view, scribbled some notes on his diary, sang three lieder, ate two sausages and drank a litre of beer. He then slipped into his sleeping bag and fell asleep. The next morning, at eight o'clock, he started his descent, again with a steady pace, but faster, since he was travelling downhill. He reached Zurglatt at one in the afternoon, after walking for five hours.

Could there be a point along the path where the mountaineer walked, on the outbound and the return journey, exactly at the same time of day?

### 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

Napoleon had an obsession: a star. His star. He would talk about it to everyone, and whoever would listen to him out of respect, would point at the star in the sky. Napoleon even talked about the star during the Russian campaign, while his troops were receding.

It seems like Talleyrand sent him the game - Napoleon's Star - on the evening of 17th June, 1815, the day before the Battle of Waterloo. It has been said that the great general spent the entire night and the following day, until sunset, trying to solve the game, without hearing the noise of the battle and without listening to his officers pleading for help. When he came out of his tent to breath some fresh air, looking tired and unshaven, but with the solution in his grasp, Waterloo had already been won by the English, and his troops were fleeing with no order or hope. Here's the game: start from any of the ten points, marked with a letter, and follow - in a staight line - to the third point from the starting position (eg from a to g); place a coin on this third point. Then pick another point unoccupied by any coin, and again go to a third unoccupied point in a straight line (a coin on the second point doesn't matter), and place a coin on it. Repeat the procedure until you've placed nine coins.

### 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

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!