47th Known Mersenne Prime Found!

On April 12^th, the 47^th known Mersenne prime, 2^42,643,801-1, a 12,837,064 digit number was found by Odd Magnar Strindmo from Melhus, Norway! This prime is the second largest known prime number, a "mere" 141,125 digits smaller than the Mersenne prime found last August.

Click here for more information.

Best Visual Illusion of the Year Contest

The Best Visual illusion of the Year Contest is a celebration of the ingenuity and creativity of the world’s premier visual illusion research community. Contestants from all around the world have submitted novel visual illusions (unpublished, or published no earlier than 2008), and an international panel of judges has rated them and narrowed them to the TOP TEN.

Free Mathematics Books

Found this website with links for free mathematical e-books.

Quite Basic -- Sieve of Eratosthenes

A BASIC programming website that uses the Sieve of Eratosthenes as an example.

Gallery of Illustrations

An interesting website that has plenty of mathematical illustrations.

Number Puzzles

Found some great number puzzles on the Internet.

Bridges
Dominosa
Light-up
Loop
Nonograms
Nurikabe
Shikaku
Sudoku

Drunk Men Work Here

This is a very interesting website. Problems 170 and 129 are a couple of great logic games.

Security Codes

I was entering the four-digit security code to our house, and I realized that it doesn't end with an ENTER (E) key to accept it. The problem with that is you can keep pressing numbers until the right combination is found. Since most keypads use all ten digits (0-9), the total number of combinations should be $10^n$. However, it is the ENTER key that makes it so difficult.

Let D equal the number of digits for the security code. Let K equal the number of digits on the keypad.

For a binary keypad (0, 1) and a three-digit code, our set consists of (000E, 001E, 010E, 011E, 100E, 101E, 110E, 111E). Thus, we have at most $2^3 \times (3+1)$ possible keys to enter, where the plus one is for the ENTER key. Generally speaking, we'd have $K^D \times (D+1)$ possible keys to enter.

Without the ENTER key, we could keep pressing the keypad until all combinations have formed. The minimum number of keys required would be $K^D + D$.

So for a binary keypad, we'd have:

D	String	Keys (wo/ E)	Keys (w/ E)
1	01	2	4
2	00110	5	12
3	0001110100	10	32
4	0000111100101101000	19	80
5	000001111100010010101110110011010000	36	192

Thus, one only needs to know the string concatenations to be able to guess the right combination if no ENTER or code-stopper key is required.