Number with largest number of divisors

We know from the Fundamental Theorem of Arithmetic that any composite number can be constructed from a unique key of primes. The amount of ways these primes can be combined in gives the amount of divisors for the number they generate. Let's explore some initial keys:

• The longest prime key containing duplicates is $2^{10}=1024$ . It is composed from 10 primes and has 11 divisors total: $\{1, 2, \ldots, 1024\}$ .
• The longest prime key that doesn't contain duplicates consists of 4 prime numbers. The smallest one possible is $2*3*5*7=210$ and it has 15 divisors.

I didn't find a general pattern for finding the combination of primes that gives the largest amount of divisors so I just narrowed down the possibilities by altering the initial $2^{10}$ expression and counting the combinations:

• $2^9*3^1=1536$ gives 21 combinations. Trading $2^2$ for 3 the key becomes:
• $2^7*3^2=1152$ gives 25 combinations. Trading 2 for 3:
• $2^6*3^3=1728$ gives 29 combinations. Trading $2^2$ for 3:
• $2^4*3^4=1296$ gives 26 combinations. Divisor count won't get any higher than 29 for the key consisting of multiples of two primes. Adding a new prime:
• $2^3*3^3*5=1080$ gives 33 combinations. Adding a new prime:
• $2^4*3*5*7=1680$ gives 41 combinations.

At this point new primes cannot be added without removing all instances of any other prime already present in the key. This guarantees both that this is the key that gives the most divisors and that it is also the largest integer in the set.

The correct answer is then 1680.

Completely computational approach gives $D = 40$ divisors for the number $N = \boxed{1680}$ . Here is the code:

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for n = 1:2018
    k  = 1:n ;
    s0(n) = size( k( rem(n,k)==0 ), 2 ) ;
end
[D,N] = max(s0)