Comments on: Project Euler Problem 132 Solution

By: Mike

Mike — Wed, 17 Mar 2010 23:36:37 +0000

Let me explain my variables and why I chose them. 'n' is an odd integer variable that starts from 7. I only want 'n' that is prime so I used the is_prime() to select them. This was just a lazy invention and could be made much more efficient. 'q' is just a constant 10⁹ which is needed in our power mod function (pow() function with third parameter specified). Here's how this solution works. Any R(n) x 9 + 1 = 10ⁿ. This allows us to use a power mod function, which Python supports natively, to check each prime with a remainder of 1 when 10 is raised to q (10⁹). This solution is extensible, however, and can solve problems with similar R(n). Also, instead of summing the first 40 primes I used a set so I could learn which primes were used. Also, it's never a bother, just have little time to respond quickly sometimes.

By: Vijay

Vijay — Thu, 11 Mar 2010 09:17:14 +0000

Thank you for the reply. I still don’t get the exact idea behind the solution. Can you throw more light on this ?
Sorry to bother you.

By: Mike

Mike — Thu, 11 Mar 2010 01:41:14 +0000

This is a good question.

Firstly, I should rename the variable ‘n’ to ‘p’ for prime. That may make things a bit clearer.

Secondly, 7 is the first possible prime that could be considered since 2 and 5 cannot factor a number ending in 1. Also 3 doesn’t divide 10⁹ (summing the digits of R(10⁹)).

Lastly, since we start at ‘n’ = 7 we never consider ‘n’ = 3.

Thanks for asking. Does this answer your question?

By: Vijay

Vijay — Wed, 10 Mar 2010 12:04:50 +0000

Hi,
I have a doubt in the solution above.
Why does ‘n’ start with 7 ?
When ‘n’ = 3, pow(10, q, n) == 1 but 3 is not a prime factor of R(10^9). Anyone kindly explain the idea behind this solution. Thank You.