2nd Junior Balkan Mathematical Olympiad Problems 1998

1.  The number 11...122...25 has 1997 1s and 1998 2s. Show that it is a perfect square.

2.  The convex pentagon ABCDE has AB = AE = CD = 1 and angle ABC = angle DEA = 90^o and BC + DE = 1. Find its area.

3.  Find all positive integers m, n such that mⁿ = n^m-n.

4.  Do there exist 16 three digit numbers, using only three different digits in all, so that the numbers are all different mod 16? 

Solutions

Problem 1

The number 11...122...25 has 1997 1s and 1998 2s. Show that it is a perfect square.

Solution

Same as Sweden 1981/1. 

Problem 3

Find all positive integers m, n such that mⁿ = n^m-n.

Answer

(m,n) = (9,3) or (8,2).

Solution

Thanks to José Nelson Ramírez

Obviously m ≥ n. Hence m-n ≥ n. So m ≥ 2n. Put d = gcd(m,n), and m = dM, n = dN. Then M ≥ 2N. We have (dM)^dN = dN^dM-dN, so M^dN = N^dM-dNd^dM-2dN. So if N > 1, then its prime factors divide M. Contradiction. So N = 1. Hence M^d = d^dM-2d, so M = d^M-2.

If M = 3, then d = 3, giving the solution m = 9, n = 3. If M = 4, then d = 2 giving the solution m = 8, n = 2. If M > 4, then d>1. But dM-2 ≥ 2M-2 > M. Contradiction. So there are no solutions for M > 4.