Great Internet Mersenne Prime Search (GIMPS) group has reported an all new Mersenne Prime Number (a prime number of type $2^P-1$) which is, now officially the largest prime number ever discovered. This number is valued to a whopping $2^{74207281}-1$ and contains 22,338,618 digits. It is quoted as M747207281 and is almost 5 million digits longer than the previous record holding prime number M57885161.

It took non-stop calculations of an i7 computer for 31 days to prove the primality of the number. A press release about this discovery is available at GIMPS’ official website.

Feel free to ask questions, send feedback and even point out mistakes. Great conversations start with just a single word. How to write better comments?

This site uses Akismet to reduce spam. Learn how your comment data is processed.

## Fox – Rabbit Chase Problems

Part I: A fox chases a rabbit. Both run at the same speed $v$ . At all times, the fox runs directly toward the instantaneous position of the rabbit , and the rabbit runs at an angle $\alpha$ relative to the direction directly away from the fox. The initial separation between the fox and the rabbit is…

## The Collatz Conjecture : Unsolved but Useless

The Collatz Conjecture is one of the Unsolved problems in mathematics, especially in Number Theory. The Collatz Conjecture is also termed as 3n+1 conjecture, Ulam Conjecture, Kakutani’s Problem, Thwaites Conjecture, Hasse’s Algorithm, Syracuse Problem. Statement: Start with any positive integer. • Halve it, if it is even. Or • triple it and add 1, if it is odd. If you…

## Chess Problems

In how many ways can two queens, two rooks, one white bishop, one black bishop, and a knight be placed on a standard $8 \times 8$ chessboard so that every position on the board is under attack by at least one piece? Note: The color of a bishop refers to the color of the square on which it sits,…
This mathematical fallacy is due to a simple assumption, that $-1=\dfrac{-1}{1}=\dfrac{1}{-1}$ . Proceeding with $\dfrac{-1}{1}=\dfrac{1}{-1}$ and taking square-roots of both sides, we get: $\dfrac{\sqrt{-1}}{\sqrt{1}}=\dfrac{\sqrt{1}}{\sqrt{-1}}$ Now, as the Euler’s constant $i= \sqrt{-1}$ and $\sqrt{1}=1$ , we can have $\dfrac{i}{1}=\dfrac{1}{i} \ldots \{1 \}$ $\Rightarrow i^2=1 \ldots \{2 \}$ . This is complete contradiction to the…