How to calculate smallest number with certain number of divisors?

Question

From Project Euler problem 500

The number of divisors of 120 is 16. In fact 120 is the smallest number having 16 divisors.

Find the smalle

Answer 1

The highest divisor any number has, other than itself, is the half of the number. For example, 120 has a max divisor of 60 other than itself. So, you can easily reduce the range from (n+1) to (n/2).

More over, for a number to have m divisors, the number must be atleast ((m-1) * 2) following the above logic (-1 because the m th number is itself). For example, a number with 4 divisors has to be atleast 6. So, your search for n has a smaller range now.

These two will reduce the runtime a little.

Answer 2

Not full answer some hints instead:

1. the max integer divisor of n is n/2

   so no need to check all divisors up to n

2. can use prime decomposition

   max prime divisor is sqrt(n) so no need to test up to n instead test up to sqrt(n) or up to number than has half of the n bits

   m=(2^(ceil(ceil(log2(n))/2)+1))-1
   

   that should speed up things a bit but you need to add the computation of non prime divisors

3. this looks like some kind of series (prime decomposition)

   divisors  | prime_divisors | non_prime_divisors              | LCM(all divisors)
   1         | 1               |                                 | 1
   2         | 1,2             |                                 | 2
   3         | 1,2             | 4                               | 4
   4         | 1,2,3           | 6                               | 6
   5         | 1,2             | 4,8,16                          | 16
   6         | 1,2,3           | 4,6,12                          | 12
   7         | 1,2             | 4,8,16,32,64                    | 64
   8         | 1,2,3           | 4,6,8,12,24                     | 24
   ...
   16        | 1,2,3,5        |4,6,8,10,12,15,20,24,30,40,60,120 | 120
   

   so try to find the equation that generates this order and then simply compute the n-th iteration in modulo arithmetics (simple PI operation ... modmul). I can see the even and odd elements have separate equation ...

[edit1] decomposition up to 16 divisors

  1:    1
  2:    1,   2
  3:    1,   2,   4
  4:    1,   2,   3,   6
  5:    1,   2,   4,   8,  16
  6:    1,   2,   3,   4,   6,  12
  7:    1,   2,   4,   8,  16,  32,  64
  8:    1,   2,   3,   4,   6,   8,  12,  24
  9:    1,   2,   3,   4,   6,   9,  12,  18,  36
 10:    1,   2,   3,   4,   6,   8,  12,  16,  24,  48
 11:    1,   2,   4,   8,  16,  32,  64, 128, 256, 512,1024
 12:    1,   2,   3,   4,   5,   6,  10,  12,  15,  20,  30,  60
 13:    1,   2,   4,   8,  16,  32,  64, 128, 256, 512,1024,2048,4096
 14:    1,   2,   3,   4,   6,   8,  12,  16,  24,  32,  48,  64,  96, 192
 15:    1,   2,   3,   4,   6,   8,   9,  12,  16,  18,  24,  36,  48,  72, 144
 16:    1,   2,   3,   4,   5,   6,   8,  10,  12,  15,  20,  24,  30,  40,  60, 120

Answer 3

Fully functional code.

def find_smallest_number(num):
    number2=0

    if(num%8==0):
        number2=min(2**((num/4)-1)*3**1*5**1 , 2**((num//2)-1)*3**1)
    elif(num%9==0):
        number2=2**((num/9)-1)*3**2*5**2   
    elif(num%2==0 and num%3==0):
        number2=2**((num/6)-1)*3**2*5**1   
    elif((num%4==0)):
        number2=2**((num/4)-1)*3**1*5**1
    elif(num%2==0):
        number2=2**((num/2)-1)*3**1
    else:
        number2=2**(num-1)         

    return number2   


num=32
print("The number of divisors :",num)
result=find_smallest_number(num)
print("The smallest number having",num," divisors:",result)