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## Project Euler 27 Solution ## Project Euler 27: Find a quadratic formula that produces the maximum number of primes for consecutive values of n

#### Project Euler 27 Problem Description

Project Euler 27: Euler published the remarkable quadratic formula:

n² + n + 41

It turns out that the formula will produce 40 primes for the consecutive values n = 0 to 39. However, when n = 40, 402 + 40 + 41 = 40(40 + 1) + 41 is divisible by 41, and certainly when n = 41, 41² + 41 + 41 is clearly divisible by 41.

Using computers, the incredible formula  n² − 79n + 1601 was discovered, which produces 80 primes for the consecutive values n = 0 to 79. The product of the coefficients, −79 and 1601, is −126479.

n² + an + b, where |a| < 1000 and |b| < 1000, where |n| is the modulus/absolute value of n
e.g. |11| = 11 and |−4| = 4

Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n = 0.

#### Analysis

Note: The quadratic formula needs only to generate prime numbers, not necessarily unique nor consecutive prime numbers.

This problem is wanting us to find two coefficients, a and b, to a quadratic expression, n² + an + b, that will generate the most prime numbers when evaluated with n starting from zero and incremented by one until a non-prime is evaluated. The coefficients a and b are constrained as -1000 < a < 1000 and -1000 < b < 1000.

Using simple iteration of a and b over their range will produce a result for this specific problem, but take too long for the more complex inquiry from HackerRank.

The are a few observations that will help improve performance.

### b has to be prime

The problem requires our n to start at 0, and since we are looking for prime numbers with consecutive values of n, our first result from a quadratic equation would be 02 – 0 + b which evaluates simply to b. So b has to be positive, prime and less than 1000 (for this problem) which helps reduce the search space significantly.

Additionally, we can now start `n` at 1 since we know the first outcome is guaranteed prime. Our sieve, `prime_sieve(L)`, generates primes less than `L`. Adding 1 to our limit, guarantees `L` gets included should it be a prime number.

Prime number 2 is excluded from the possible starting primes as it evaluates to an even number after 1 or 2 iterations.

### 1-a+b must be odd and possibly prime

If you further consider when n=1, then the expression becomes 1-a+b or 1+a+b, which must always evaluate to an odd number to yield a possible prime. We know b+1 will be even so a must be odd. We can now infer that a must be odd and |a| < b which supports our thinking.

HackerRank Project Euler 27 adds the constraint that the coefficients a and b in the range [-2000,-42] U [42,2000]. The positive constraint on a and b can be ignored as we previously established; effectively halving the search range. We are asked to print the two coefficients a and b instead of the product of a and b. This program and method
solves all test cases for
Project Euler 27 on HackerRank

#### Project Euler 27 Solution

Runs < 0.070 seconds in Python 2.7. Use this link to get the Project Euler 27 Solution Python 2.7 source.

#### Afterthoughts

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