Question

I just got this question on a SE position interview, and I’m not quite sure how to answer it, other than brute force:

Given a natural number N, find two numbers, A and P, such that:

N = A + (A+1) + (A+2) + … + (A+P-1)

P should be the maximum possible.

Ex: For N=14, A = 2 and P = 4

N = 2 + (2+1) + (2+2) + (4+2-1)
N = 2 + 3 + 4 + 5

Any ideas?

Answer 1

If N is even/odd, we need an even/odd number of odd numbers in the sum. This already halfes the number of possible solutions. E.g. for N=14, there is no point in checking any combinations where P is odd.

Rewriting the formula given, we get:

N = A + (A+1) + (A+2) + ... + (A+P-1)
    = P*A + 1 + 2 + ... + (P-1)
    = P*A + (P-1)P/2 *
    = P*(A + (P-1)/2)
    = P/2*(2*A + P-1)

The last line means that N must be divisible by P/2, this also rules out a number of possibilities. E.g. 14 only has these divisors: 1, 2, 7, 14. So possible values for P would be 2, 4, 14 and 28. 14 and 28 are ruled our for obvious reasons (in fact, any P above N/2 can be ignored).

This should be a lot faster than the brute-force approach.

(* The sum of the first n natural numbers is n(n+1)/2)