Project Euler Problem 27 Solution

Project Euler Problem 27 Solution

Quadratic primes

by {BetaProjects} | Project Euler & HackerRank

Project Euler Problem 27 Statement

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.

Considering quadratics of the form:
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.


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, for a quadratic expression, n2 + an + b, which 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 a simple iteration of a and b over their range will produce a result for this specific problem, but takes too long for the more complex inquiry from HackerRank.

There 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 expression 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.

The 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 be evaluated to an odd number to yield a possible prime. We know b will always be an odd prime so 1+b will be even. We can now infer that a must be odd and |a| < b.

HackerRank's Project Euler 27 adds the constraint that the coefficients a and b are 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 as the original problem required.

HackerRank version

HackerRank simply changes the 1000 limit to any N, where 42 ≤ N ≤ 2000.

Python Source Code

from Euler import prime_sieve, is_prime
L = int(input())
nmax = 0
for b in prime_sieve(L+1)[1:]:	#start with prime number 3
    for a in range(-b+2, 0, 2):
        n = 1
        while is_prime(n*n + a*n + b): n += 1
        if n>nmax: nmax, p = n, (a,b)
print (p[0],p[1])	

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