Count how many numbers are divisible by perfect numbers in a given range

Mathematica - 117

Naive approach, linear for the range size

With[{p=#(#+1)/2&/@Select[2^Range@@Floor@Log2@Sqrt@#-1,PrimeQ]},Length@Select[Range@@#,Or@@Divisible[#,p]&]]&@Input[]

The correct way would be constructing numbers from the perfect numbers that are in the given range of course.

C, 125120

a,b,c,p[4]={6,28,496,8128};main(i){for(scanf("%d%d",&a,&b);a<=b;a++)for(i=0;i<4;)if(a%p[i++]==0)c++,i=4;printf("%d",c);}

A little more readable:

a,b,c=0,i,p[4]={6,28,496,8128};
main()
{
    for(scanf("%d%d",&a,&b);a<=b;a++)
        for(i=0;i<4;)
            if(a%p[i++]==0)
            {
                c++;
                i=4;
            }
    printf("%d",c);
}

This works with signed 32bit integers, up to 2^31-1=2147483647.

C, 212209

#include<stdint.h>
uint64_t a,b,c,p[8]={6,28,496,8128,33550336,8589869056,137438691328,0x1fffffffc0000000};main(i){for(scanf("%lld%lld",&a,&b);a<=b;a++)for(i=0;i<8;)if(a%p[i++]==0)c++,i=8;printf("%lld",c);}

Should work up to 2^64-1=18446744073709551615.

Haskell - 157

c x=x==sum[i|i<-[1..x-1],mod x i==0]
d x=any((0==).mod x)$takeWhile(<=x)(filter c[2..])
main=do m<-getLine;n<-getLine;print$length(filter d[read m..read n])

Can work with arbitrarily large numbers given respectively large time.
Input is given on 2 lines.