Problem 12 - What is the value of the first triangle number to have over five hundred divisors?

링크

The sequence of triangle numbers is generated by adding the natural numbers. So the 7^th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

Let us list the factors of the first seven triangle numbers:

 1: 1
 3: 1,3
 6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28

We can see that 28 is the first triangle number to have over five divisors.

What is the value of the first triangle number to have over five hundred divisors?

python - slow

def numFactor(n): res = 0 for x in range(1, int(n**0.5) + 1): if n % x == 0: res = res + 2 if x * x == n: res = res - 1 return res n = 1 step = 1 limit = 500 while numFactor(n) <= limit: step = step + 1 n = n + step print n

python - fast

An integer n can be expressed as follows

n = p1^a1 * p2^a2 * p3^a3 * ...

The number of divisors D(n) of an integer n is

D(n) = (a1 + 1) * (a2 + 1) * (a3 + 1) * ...

And, n-th triangle number t is 

t = n * (n+1) / 2

We can calculate D(t) as follows

D(t) = D(n/2) * D(n+1)  if n is even, or

D(t) = D(n) * D((n+1)/2) if n+1 is even

prime = [True]*1001 prime[1] = False for i in range(1, int(round(1001**0.5)) + 1): if prime[i] == True: prime[2*i:1001:i] = [False] * (1000 / i - 1) primes = [] for i in range(1, len(prime)): if prime[i] == True: primes.append(i) n = 1 dn = 1 cnt = 0 while cnt <= 500: n = n+1 nn = n dnn = 1 if nn % 2 == 0: nn = nn / 2 for i in primes: exp = 1 while nn % i == 0: exp = exp + 1 nn = nn / i dnn = dnn * exp if nn == 1: break; cnt = dn * dnn dn = dnn print (n-1)*n / 2