How would you solve this A level style question?

This is the puzzle where I will be exploring some nice pieces of maths I talked about before. I am discussing this because in a recent class this problem came up and the entire class independently tried it and when we looked at our solutions, none of them were the same, but there were many different correct methods. Please write in the comments how you solved or tried to solve this problem. Here it is:

Prove that n⁵-n is divisible by 30, where n is a member of the natural numbers (ℕ)

I hope you enjoy!

Breaking Maths News- Prime Numbers Aren't Random

I have mentioned before that primes are one of my favourite topics in maths, so when I heard recently about a great breakthrough in my favourite field I had to share it. Mathematicians Kannan Soundararajan and Robert Lemke Oliver of Stanford University in California, have discovered a hidden pattern in the primes. This is so amazing because the primes are believed to be random, because there is no way of knowing where along the number line the primes are located. A hidden pattern would suggest that the primes are not as random as we thought.

 The pattern is found when examining the last digit of prime numbers. The last digit of all primes, other than 2 and 5, ends in either a 1,3,7 or 9. Therefore, is we assume the primes are random the probability of a certain prime ending in each is 25% each. As a result,  if Primes were random an equal number of primes would end in either a 1,3,7 or 9, which they do. However, if the last prime ended in a 1, then the probability of each should still be 25%, because it is random. However, they found that in the first hundred million primes, a primes ending in a 1 is only followed by another 18.5% of the time. 
Then primes ending in a 3 or 7 each occur 30% of the time and 9 22% of the time.

Similar patterns were discovered for 3,7 and 9 also.  They also discovered the pattern in different bases which means the pattern is inherent to primes and not caused by our base 10 numbering system.

We already know why this pattern occurs. Littlewood and Hardy's k tuples conjecture, which estimates how often pairs triples and larger groups of primes will appear, places restrictions on what the last digit of primes can be, which explains the unequal distribution. 

Sadly, this result won't help us solve any long standing problems involving primes such as the Twin Prime Conjecture and the Riemann Hypothesis, but is has excited mathematicians because it reveals another of the primes' hidden secrets. I will leave you with the words of James Maynard, a world expert on primes from Oxford University, "Mathematicians go around assuming primes are random, and 99% of the time this is correct, but you need to remember the 1 per cent of the time it isn’t".

Please check regularly for more maths news when it arrives.

Palindromic Number Puzzle Solution

I hope you enjoyed that puzzle. I mentioned that there are many possible ways to prove that every 4 digit palindrome is divisible by 11.  However, I have decided to talk about these nice pieces of maths in the next puzzle, which is a better example of these techniques. Now the simple proof.

First call an eleven digit palindrome, abba. (That is why I like this puzzle)

abba can be factorise like this a x 1000 + b x 100 + b x 10 + a.
This can be rewritten as 1000a + 100b +10a +b
This is obviously equal to 1001a + 110b
This can be simplified further by taking a factor of 11 out
Which leaves   11(91a+ 10b)
The bracketed 91a + 10b must be an integer, so that it is palindromic. Any integer where 11 is a factor  must divide by 11. Therefore, all 4 digit palindromes are divisible by 11.

I hope you enjoyed that puzzle the next puzzle will be up shortly.

Palindromic Number Puzzle

Today I have another puzzle which you can tackle from many angles. This puzzle is deceptively simple. Prove that all  4 digit palindromic numbers  (a number which reads the same backwards and forwards) are divisible by 11.
Enjoy!