# Give the smallest number that has N divisors

### APL, 2524 23 characters

f←{({+/⍵=⍵∧⍳⍵}¨⍳2*⍵)⍳⍵}


Defines a function f which can then be used to calculate the numbers:

> f 13
4096

> f 14
192


The solution utilizes the fact that LCM(n,x)==n iff x divides n. Thus, the block {+/⍵=⍵∧⍳⍵} simply calculates the number of divisors. This function is applied to all numbers from 1 to 2^d ¨⍳2*⍵. The resulting list is then searched for d itself (⍳⍵) which is the desired function f(d).

### GolfScript, 29 28 characters

{.{\.,{)1$\%},,-=}+2@?,?}:f;  Edit: A single char can be saved if we restrict the search to <2^n, thanks to Peter Taylor for this idea. Previous Version: {.{\)..,{)1$\%},,-@=!}+do}:f;


An attempt in GolfScript, run online.

Examples:

13 f p  # => 4096
14 f p  # => 192
15 f p  # => 144


The code contains essentially three blocks which are explained in detail in the following lines.

# Calculate numbers of divisors
#         .,{)1$\%},,- # Input stack: n # After application: D(n) ., # push array [0 .. n-1] to stack { # filter array by function ) # take array element and increase by one 1$\%      #   test division of n (\$1) by this value
},          # -> List of numbers x where n is NOT divisible by x+1
,           # count these numbers. Stack now is n xd(n)
-           # subtracting from n yields the result

# Test if number of divisors D(n) is equal to d
#         {\D=}+   , for D see above
# Input stack: n d
# After application: D(n)==d

{
\         # swap stack -> d n
D         # calculate D(n) -> d D(n)
=         # compare
}+          # consumes d from stack and prepends it to code block

# Search for the first number which D(n) is equal to d
#         .T2@?,?    , for T see above
# Input stack: d
# After application: f(d)

.           # duplicate -> d d
T           # push code block (!) for T(n,d) -> d T(n,d)
2@?         # swap and calculate 2^d -> T(n,d) 2^d
,           # make array -> T(n,d) [0 .. 2^d-1]
?           # search first element in array where T(n,d) is true -> f(d)


## Mathematica 38 36

(For[i=1,DivisorSum[++i,1&]!=#,];i)&


Usage:

(For[i=1,DivisorSum[++i,1&]!=#,];i)&@200


Result:

498960


Edit

Some explanation:

DivisorSum[n,form] represents the sum of form[i] for all i that divide n.

As form[i] I am using the function 1 &, that returns always 1, so effectively computing the sum of the divisors in a terse way.