Find all possible solutions to the system of equations:

(i) n(p) + n(q) = p;
(ii) p+q+ n(r) = r;
(iii) n(p) + n(q) + n(r) = q - 4;

where, each of p, q and r are positive integers and n(p), n(q) and n(r) respectively denote the number of digits of the integers p, q and r.

From i we get p>=2 and with i and iii we get n(r)=q-p-4. Hence q>=p+5>=7. With i and ii we get r=2q-4 and n(r)<=n(2q)<=n(q)+1 and n(p)<=n(q). With iii we get 3*n(q)+1>=q-4 from which q<=3n(q)+5.

Since 10^(n(q)-1) <= q it's obvious that this equation can only hold true if n(q)=1 or n(q)=2. If n(q)=1 then q would be 7 or 8. If q was 7 then p=2 and r=10 which doesn't satisfy ii. If q=8 we get a triplet which satisfies all equations (2,8,12). If n(q)=2 then q=10 or 11. If q=10 then r=16 and p<=5. Not satisfying iii we try q=11 from which r=18 and p<=6. This doesn't satisfy iii either.

So the only triplet satisfying the given equations is (2,8,12)

Posted by atheron on 2007-02-10 17:17:02