As this year comes to a close, people around the world will be counting down the last few seconds of 2012. But, how come we never count up the first few seconds of the new year? What is it about the last few seconds of the year that makes them so special? Maybe it has to do with surviving a Mayan apocalypse, but that was just this year. I guess it always comes down to letting go of the good and of the difficult moments in our past. The champagne helps. Still, I would like to take a moment to pay tribute to the first few seconds of 2013 (the future) with a simple problem and a twist…
The question is simple enough:
What is the longest sequence of consecutive numbers, such that each element of the sequence is a power of a prime number?
You will arrive at the answer soon after asking yourself the question: How often is a number divisible by both 2 and 3? The answer is every 6 numbers (since the number in question has to be divisible by $latex 2cdot3$). So, the best one can do is to count the numbers from one to five. That is, if you count 1 as the power of a prime number, then the first five numbers have the incredible property of being the only such sequence of numbers (right?).
Now, here comes the twist…
Challenge: Is there a sequence of 10 consecutive numbers such that none of them is a power of a prime number?
To get the ball rolling, here are the first such sequences with one, two and three elements, respectively: , [14,15], [20,21,22].
Super Challenge: Can you find a sequence of 2013 consecutive numbers, such that none of them is a power of a prime number?
Impossible Challenge: Can you solve the first challenge, giving the sequence containing the smallest numbers satisfying the conditions of the problem?
Good luck and enjoy the rest of 2012! Who knows, maybe some genius will solve the above challenges before 2013 rolls around…