Good Luck! (some of you will need it, lol)
This game is based on the “Petals Around the Rose” dice game. Both games are easy in the sense that once you know the “secret”, you can easily determine the answer in seconds. After hearing of this game, I was able to figure out the secret immediately. Just remember, as in “Petals Around the Rose”, the name of the game is important. Note, however, you do not actually need to know anything about chess to figure out the secret.
The net has a great amount of interactive cool maths games. Below are a few sites worth taking a look at. Most of the math games are fun for all ages and they are all absolutely free.
1. http://www.mathplayground.com/games.html
2. http://www.coolmath-games.com/
3. http://resources.kaboose.com/games/math2.html
The following is a tutorial about how to make the Yoshimoto cube using paper (though it takes a very long time!). To make it you’ll need paper, scissors, some glue, and some adhesive tape.
To make one yourself you can print out the following Yoshimoto sheet and try to assemble it yourself 🙂
Five pirates (of different ages) have a treasure of 100 gold coins. On the ship, they decide to split the coins using the following scheme:
Assuming
that all five pirates are intelligent, rational, greedy, and do not wish
to die, what should the oldest pirate propose to a) survive and b) maximize his profit?
Using prime numbers, you can amaze your friends with a prime prediction…
1. Ask your friends to pick any prime number greater than 3.
2. Square it.
3. Add 14.
4. Divide by 12.
Without knowing which prime number your friends picked, you can still tell them:
There will be a remainder of 3.
But HOW does it work?
Let’s do an example:
13 is a prime number, squaring it gives 169, adding 14 gives 183 which has a remainder of 3 upon division by 12.
This works for every prime number greater than 3, but how exactly does it work?
The mathematics behind this is rather simple.
1. Let p be a prime number, p > 3.
2. Squaring gives:
p^2.
3. Adding 14 gives:
p^2 + 14
4. Taking it modulo 12 gives:
(p^2 + 14) mod 12
We want to show that:
(p^2 + 14) mod 12 = 3
This is equivalent to:
p^2 – 1 is divisible by 12.
That is:
(p-1)(p+1) is divisible by 12.
For a number to be divisible by twelve, it has to be divisible both by 3 and by 4. We know that, out of p-1, p and p+1, one of them must be
divisible by 3; and it can’t be p, because p is prime and greater than 3. Thus, either p-1 or p+1 is divisible by 3, and so their product is also:
(p-1)(p+1) is divisible by 3.
Now, since p is a prime greater than 3, we know that it is odd. Therefore, both p-1 and p+1 are even numbers. The product of two even numbers is divisible by 4, so:
(p-1)(p+1) is divisible by 4.
Combining this with the above, we get that:
(p-1)(p+1) is divisible by 12.
And hence:
(p^2 + 14) mod 12 = 3
This is one of my favourite maths dingbats (quite an easy one!) It’s one of over 50 maths dingbats on the website – some easy, some pretty tough.
Are you ready for another brain teaser? You’re in for a treat.
The gentle Pardoner […] produced the accompanying plan, and said that it represented sixty-four towns through which he had to pass during some of his pilgrimages, and the lines connecting them were roads. He explained that the puzzle was to start from the large black town and visit all the other towns once, and once only, in fifteen straight pilgrimages. Try to trace the route in fifteen straight lines with your pencil. You may end where you like, but note that the omission of a little road at the bottom is intentional, as it seems that it was impossible to go that way.
Check the original puzzle here.
No. Color the squares as shown. Now each tile must cover a blue, a yellow, and a red square. If we’re to cover the figure completely, then it must contain an equal number of squares of each color. But it contains 17 blue squares, 18 yellow squares, and 19 red squares. So the task is impossible.
