The divisors can be found by combining all the possible products of the prime factors. To show this, let's consider 120 = 2³ x 3¹ x 5¹ , which has 16 divisors: 1,2,3,4,5,6,10,12,15,20,24,30,40,60,120. We can write out these divisors as follows:

1	= 20 x 30 x 50
2	= 21 x 30 x 50
3	= 20 x 31 x 50
4	= 22 x 30 x 50
5	= 20 x 30 x 51
6	= 21 x 31 x 50
8	= 23 x 30 x 50
10	= 21 x 30 x 51
12	= 22 x 31 x 50
15	= 20 x 31 x 51
20	= 22 x 30 x 51
24	= 23 x 31 x 50
30	= 21 x 31 x 51
40	= 23 x 30 x 51
60	= 22 x 31 x 51
120	= 23 x 31 x 51

(To understand this table, you need to know that n⁰ = 1 for any value of n).

Do you see how all possible combinations appear? There are four possibilities for the power of 2 (i.e. 0,1,2, and 3), and two possibilities each for the power of 3 and 5 (i.e. 0 and 1). So there are 16 divisors in total, as 4 x 2 x 2 = 16.

Now let's consider a general number n with prime factorization

where p, q, r, etc. are prime numbers, and a, b, c etc. are integers (e.g. 0,1,2,3, ...). How many divisors does this number have? Following the argument above, it must have

divisors in total. This formula is the key to answering the problem.

Now, we need to find the prime factorization of 2013:

To find a number with 2013 divisors, we can set (a + 1) = 3, (b + 1) = 11, (c + 1) = 61, so that a = 2, b = 10, c = 60

These are the indices. To find the smallest number with 2013 divisors, we should build our number from the smallest possible primes factors, 5, 3, and 2, so

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Is this really the smallest possible number with 2013 divisors? Well, we could try using other (non-prime) factorizations of 2013, e.g., 33 x 61, 11 x 183, 3 x 671.  This leads to other solutions, e.g.

But these are all clearly bigger than the first solution we found.

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One aim of this Mini-Challenge was to show that there are some very "simple" mathematical questions which cannot be solved easily with a naive computer code. We still need to use our brains!