Lego gear ratios

Python - 204

Ok, I'll go first:

def p(n,a=[1]*9):
 n=int(n)
 for i in(2,3,5):
    while n%i<1:n/=i;a=[i]+a
 return a,n
(x,i),(y,j)=map(p,raw_input().split(':'))
print[' '.join(`a*8`+':'+`b*8`for a,b in zip(x,y)if a!=b),'IMPOSSIBLE'][i!=j]

edit:

To 'optimize' the output, this can be added before the print statement,

for e in x:
 if e in y:x.remove(e);y.remove(e)

bringing the total up to 266 characters, I believe.

Perl - 310 306 294 288 272

I'm a little rusty with perl and never did a code-golf... but no excuses. Char-count is without line-breaks. Using perl v5.14.2 .

($v,$n)=<>=~/(.+):(.+)/;
($x,$y)=($v,$n);($x,$y)=($y,$x%$y)while$y;
sub f{$p=shift;$p/=$x;for(5,3,2){
while(!($p%$_)){$p/=$_;push@_,$_*8}}
$o="IMPOSSIBLE"if$p!=1;
@_}
@a=f($v);@b=f($n);
if(!$o){for(0..($#b>$#a?$#b:$#a)){
$a[$_]||=8;
$b[$_]||=8;
push@_,"$a[$_]:$b[$_]"}}
print"$o@_\n"

I'm looking forward to critics and hints. It's not so easy to find tips and tricks for code-golf (in perl).

C, 246 216 213 bytes

In a (futile) attempt to beat my own Prolog solution, I completely rewrote the C solution.

b,c,d;f(a,b,p){while(c=a%5?a%3?a%2?1:2:3:5,d=b%5?b%3?b%2?1:2:3:5,c*d>1)c<2|b%c?d<2|a%d?p&&printf("%d:%d ",8*c,8*d):(c=d):(d=c),a/=c,b/=d;c=a-b;}main(a){scanf("%d:%d",&a,&b);f(a,b,0);c?puts("IMPOSSIBLE"):f(a,b,1);}

My original C solution (246 bytes):

#define f(c,d) for(;a%d<1;a/=d)c++;for(;b%d<1;b/=d)c--;
b,x,y,z;main(a){scanf("%d:%d",&a,&b);f(x,2)f(y,3)f(z,5)if(a-b)puts("IMPOSSIBLE");else
while((a=x>0?--x,2:y>0?--y,3:z>0?--z,5:1)-(b=x<0?++x,2:y<0?++y,3:z<0?++z,5:1))printf("%d:%d ",a*8,b*8);}

It was a nice exercise to prove it can be done without building lists.