Prove that the number $\underbrace{111\ldots 11}_{1997}\underbrace{22\ldots 22}_{1998}5$ (which has 1997 of 1-s and 1998 of 2-s) is a perfect square.

Click for solution It is same as $\frac{10^{1997}-1}{9}\cdot 10^{1999}+20\left(\frac{10^{1998}-1}{9}\right)+5$ Letting $\frac{10^{1997}-1}{9}=a$ , we get $\frac{10^{1998}-1}{9}=10\cdot \frac{10^{1997}-1}{9}+1=10a+1$ Also $10^{1999}=10^{1997}\cdot 100=100(9a+1)$ So the expression become $100a(9a+1)+20(10a+1)+5=25(6a+1)^2$ which is obviously a perfect square since $a$ is integer .

Let $ABCDE$ be a convex pentagon such that $AB=AE=CD=1$, $\angle ABC=\angle DEA=90^\circ$ and $BC+DE=1$. Compute the area of the pentagon. Greece

Click for solution just a hint: take P on the ray DE such that EP = BC. Then you'll have DP=1, and the triangle EAP is equal to BAC, so the area of ACDP is equal to the area of the original pentagon, and DP=CD=1, AP=AC, and you complete the solution...

Find all pairs of positive integers $ (x,y)$ such that \[ x^y = y^{x - y}. \] Albania

Do there exist 16 three digit numbers, using only three different digits in all, so that the all numbers give different residues when divided by 16? Bulgaria

Click for solution The answer is no. Suppose the answer would be affirmative. Obviuously the three digits cannot be all of the same parity, because we would not be able to cover all the residues modulo 2 Thus there exists at least one even and one odd digit. Suppose WLOG that we have one odd digit and two even digits (the other case can be treated in the same way, replacing the word even with the word odd everywhere). Thus the 8 odd residues modulo 16 must be covered by odd numbers, that is the last digit is the odd one. But there are 9 ways to choose the first two digits, and out of those at least 4 must give odd residue when divided by 8, because the number $\overline{xy}\cdot 10$ ( $\overline{xyz}=\overline{xy}\cdot 10 +z\ \Rightarrow\ \overline{xy}\cdot 10$ ) must pass through all the even residues modulo 16. The last 4 numbers which give odd residue when divided by 8 must be formed having the last digit y, and odd number, but there are only 3 such numbers, which leads to a contradiction. Thus the problem is solved. I am not sure, but I belive this was my solution in the contest.