Prove that if 3 prime numbers greater than are in AP then common difference is divisible by 6.
2305A prime number is always of the form 6k+1 or 6k+5
(Since 6k+2 and 6k+4 are divisible by 2 and 6k+3 is divisible by 3).
Now to be in AP,
2b = a+c.
From here it is easy to see that this is possible only when a,b & c are 6m+1,6n+1 and 6p+1 or 6m+5,6n+5 and 6p+5.
Therefore the common difference is divisible by 6.
79This problem looked difficult first.But its not so........Try it.Its a good one.
36It has to be divisible by 2 so that following two numbers are even. It has to be divisible by three so that following two numbers are not divisible by 3( if divisible by any other odd number, one of the following two numbers will be divisible by three as that number is not)
So, it has to be divisible by 6