Find the sum of all the integers with such that divides .
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Put a-1=x,b-1=y,c-1=z. Now to find 1<= x<y<z for which 1/x +1/y +1/z +1/xy + 1/yz+1/zx=p is also a integer. we observed that x>=1,y>=2 and z>=3 P<=1+1/2+1/3+1/2+1/3+1/6=17/6<3 Thus p=1 or 2. Taking two cases each of p=1 and p=2 we get triples (a,b,c)={3,5,15},{2,4,8} So total sum of all values =3+5+15+2+4+8=37. Hope u enjoyed this problem of IMO1992.