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A = 6 · · · 6 3 or a = 4 · · · 4 2 or a = 2 · · · 2 1, n n n for n ≥ 0. The first two possibilities must be rejected for n ≥ 1, since a2 = d would have 2n + 2 digits, which means that c would have to have at least 2n + 2 digits, but we already know that c must have 2n + 1 digits. Thus the only remaining possibilities are a = 3 or a = 2 or a = 2 · · · 2 1, n for n ≥ 0. It is easily seen that they all satisfy the requirements of the problem. 54 N3. Determine all pairs of positive integers (a, b) such that a2 2ab2 − b3 + 1 is a positive integer.

Together with the pure periodicity, we see that max k ≥ m − 1. m−1 On the other hand, if there are m-consecutive zeroes in (ri ), then the definition formula and the pure periodicity force ri = 0 for any i ≥ 0, a contradiction. Thus max k = m − 1. 52 N2. Each positive integer a undergoes the following procedure in order to obtain the number d = d(a): (i) move the last digit of a to the first position to obtain the number b; (ii) square b to obtain the number c; (iii) move the first digit of c to the end to obtain the number d.

S P Q R L Let S be the point such that the triangle QRS is equilateral, where the points P and S lie in the same half-plane bounded by the line QR. Then the point P lies inside the circumcircle of the triangle QRS, which lies inside the circle with centre L and radius √ 3 QR/2. This completes the proof of the lemma. B M A F P C D N E 45 The main diagonals of a convex hexagon form a triangle though the triangle can be degenerated. Thus we may choose two of these three diagonals that form an angle greater than or equal to 60◦ .