On the seventh day of Christmas,
My new iPod gave to me,
Seven Mary Three
Sixpence None The Richer
Five for Fighting,
The Four Tops,
Three Doors Down
Two Skinnee J's,
And nothing by the Partridge Family, so don't even joke about it.
Explanation: Also on my new iPod are:
Seven One Eight, by the 2 Skinnee J's
Six Underground, by the Sneaker Pimps
Five Piece Chicken Dinner, by the Beastie Boys
Four Sticks, by Led Zeppelin
Three MCs and one DJ, by the Beastie Boys
Two Step, by the Dave Matthews Band
And The Humpty Dance, by Digital Underground.
Yeah, that's right - The Humpty Dance. It has nothing to do with partridges or pear trees. It's just a cool song.
Special Blog Bonus: Feeling numerical? Don't forget 5150 by Van Halen, #41 by the Dave Matthews Band, 50 Ways to Leave Your Lover by Paul Simon, 8 Mile by Eminem, 316 by Van Halen, 99 Problems by Jay-Z, 1979 by the Smashing Pumpkins, 1984 by Van Halen, Nineteen Naughty Three by Naughty By Nature, 100 Years by Five for Fighting, One Thing by Finger Eleven, 3 A.M. Eternal by the KLF, Ten by Pearl Jam, and 6th Avenue Heartache by the Wallflowers. Speaking of numerical...
Solutions to Yesterday's Evil Number Theory Questions:
6 is the smallest perfect number, because it is equal to the sum of its proper positive divisors: 1, 2, and 3. Can you find the next even perfect number? Can you find an odd perfect number?
The next even perfect number is 28, which is the sum of 1, 2, 4, 7, and 14. The perfect numbers after that are 496 and 8128. All known perfect numbers fit the formula 2^(n-1) * (2^n - 1) where 2^(n-1) is a Mersenne prime. It is unknown whether there are any odd perfect numbers. If you managed to find one, you should tell me what it is and never speak of it again.
5 is a prime number whose binary representation, "101" is palindromic, meaning its digits read the same backward as forward. What is the largest known prime number that is palindromic in binary?
The largest known prime number which is palindromic in binary is 2^32582657, which is a 9.8 million digit number and happens to be the largest prime number known to date. It is a Mersenne prime, meaning it is one less than a power of two, hence the binary representation of the number is a string of ones, and is therefore the same both backward and forward. All of the largest known primes are Mersenne primes.
4 is an even integer that can be written as the sum of two primes, namely 2 and 2. Can you find an even integer that cannot?
This little gem is called Goldbach's conjecture. It is a conjecture because it is unproven. If you have found an integer that cannot, again, please contact me immediately with that number and then treat yourself to a nice long vacation in a place without telephones while I hold onto it all safe and sound for you. Don't worry. I won't tell anyone.
3 is a triangular number, because it belongs to the sequence (1, 3, 6, 10, 15, 21, ...). The number 3 can be represented as a sum of three triangular numbers: 3 = 1 + 1 + 1. Can you find an integer that cannot?
No you cannot. This is a subset of the Fermat polygonal number theorem, which states that every positive integer is the sum of at most n n-polygonal numbers. So, in this case, n=3.
2 is a value for n such that there exist nonzero integers x, y, and z where x^n + y^n = z^n. For example, 3^2 + 4^2 = 5^2. Can you find a value of n greater than 2 for which this is also true?
This would be Fermat's Last Theorem. Pierre de Fermat famously wrote the following words in the margin of a book: "It is impossible to separate a cube into two cubes, or a fourth power into two fourth powers, or in general, any power higher than the second into two like powers. I have discovered a truly marvelous proof of this, which this margin is too narrow to contain." For over 350 years, mathematicians cursed this man and his "marvelous proof", until Andrew Wiles of Princeton published a proof in 1995. Anyway, in short, the answer to this question is "No, you cannot".
1 is the difference between 2 and 3. There also exist powers of 2 and 3 such that their difference is also 1, namely 2^3=8 and 3^2=9. Can you find another pair of consecutive integers who have powers that are also consecutive?
This problem was known as Catalan's conjecture for over 150 years until it was solved by Preda Mihăilescu in 2002, and it is now referred to by the much more pronounceable "Mihăilescu's theorem." The example I cited shows the only pair of consecutive integers (2 and 3) for which this is possible. So, if you found another pair of consecutive integers, I suggest you check your math.