Another amicable number problem

Question

Two numbers are said to be 'amicable' or 'friends' if the sum of the divisors of the first is equal to the second, and viceversa. For example, the divisors of 220 are: 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110 which sums up 284. 284 divisors are 1, 2, 4, 71 and 142, which sums 220, thus 220 and 284 are friends. Write a function which returns either true of false if a number is a friend of other number.

Examples:

friend(220) ==> true

friend(7) ==> false

friend(284) ==> true

Function with least number of chars wins.

NOTE: I just found out that there is a similar question, but I believe this one is simpler and perhaps more general.

Answer 1

Haskell, 47

o m=sum[x|x<-[1..m-1],m`mod`x<1]
f m=(o.o)m==m

Call the f function (for "friend").
Didn't provide a main because nobody else is doing so.

Answer 2

Python, 67

a=lambda x:sum(i for i in range(1,x)if x%i<1)
b=lambda x:x==a(a(x))

Answer 3

eTeX, 177 (yes, that language is too verbose)

\let~\ifnum\let\c\newcount\c\X\c\D\c\R\def\!{\advance\D1 ~\D<\X~\X=\numexpr\D*\numexpr
\X/\D\advance\R\D\fi\!\fi}\def\a#1{\X#1 \!\X\R\D0\R0 \!\message{~#1=\R YES\else NO\fi}\end}

Used as etex filename.tex "\a{220}".

Answer 4

Scala, 67

def d(n:Int)=(1 to n-1).filter(n%_==0).sum
def f(n:Int)=d(d(n))==n

Answer 5

Ruby, 95

l=lambda{|n|(1..n-1).select{|e|n.modulo(e)==0}.reduce(:+)}
m=lambda{|n|(l.call(l.call(n))==n)}

Answer 6

Ruby 1.9, 57 characters

s=->n{eval (1...n).select{|a|n%a<1}*?+}
f=->n{s[s[n]]==n}

Pretty similar to all the other solutions here.

Answer 7

J, 24 characters

The pure answer is this:

=(+/&(((0:=|~)#])i.))^:2

It's a monad that takes a scalar and returns 0 or 1 if it's a friend or not. Unfortunately J precedence rules make it impossible to use with no separator from its argument. I'm keeping the character count as such because of the way the question is worded. Anyway, here's a demonstration of three possible ways to use it:

   NB. using a named verb
   f =: =(+/&(((0:=|~)#])i.))^:2
   f 220
1
   NB. using parentheses
   (=(+/&(((0:=|~)#])i.))^:2) 7
0
   NB. using verb "Same"
   =(+/&(((0:=|~)#])i.))^:2 [ 284
1

Obligatory J unscrambling:

• i. 220 returns the list of naturals below 220 (0, 1, 2 to 219)
• (|~i.) 220 divides them all by 220 and returns the remainder
• ((0:=|~)i.) 220 compares to 0 (returns boolean as 0 or 1)
• (((0:=|~)#])i.) 220 returns the natural where the 1s were. In effect: returns the full divisor list.
• (+/&(((0:=|~)#])i.)) 220 sums them.
• ((+/&(((0:=|~)#])i.))^:2) 220 performs the operation twice.
• (=(+/&(((0:=|~)#])i.))^:2) 220 checks we fall back on the initial argument.

Answer 8

Q, 42 33

{x={0+/a(&)0=x mod a:(!)6h$(1+x%2)}/[2;x]}

shorter by not limiting the check for divisors to enumerations up to n/2 but obviously slower as a result:

{x={0+/a(&)0=x mod a:(!)x}/[2;x]}

timing

q)\t {x={0+/a(&)0=x mod a:(!)x}/[2;x]} 2200000
289
q)\t {x={0+/a(&)0=x mod a:(!)6h$(1+x%2)}/[2;x]} 2200000
124

Answer 9

GolfScript (25 chars)

{.{:a,(\{.a\%!*+}/}2*=}:f

a can, of course, be replaced by the name of any other variable whose contents you don't care about.