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01 = 9 1 . . 1 · · · 10 . . 01 10 . . 01 . . 10 . . 01 . 81 9 8 9 8 8 8 8 8 The second and third factors are composed of 9 units, so the sum of their digits is divisible by 9, that is, each is a multiple of 9. Hence 1081 − 1 is divisible by 93 = 729, as is 1081n − 1 for any n. 6. In the isosceles triangle ABC (AC = BC) point O is the circumcenter, I the incenter, and D lies on BC so that lines OD and BI are perpendicular. Prove that ID and AC are parallel. e. O = I) the statement is obvious.

Let i = 1 if xi occurs in A but not in B, −1 if xi occurs in B but not in A, and 0 otherwise; then i xi is divisible by n3 . 47 3. Let x, y be real numbers. Show that if the set {cos(nπx) + cos(nπy)|n ∈ N} is finite, then x, y ∈ Q. Solution: Let an = cos nπx and bn = sin nπx. Then (an + bn )2 + (an − bn )2 = 2(a2n + b2n ) = 2 + (a2n + b2n ). If {an + bn } is finite, it follows that {an − bn } is also a finite set, and hence that {an } is finite, since an = 1 [(an + bn ) + (an − bn )], 2 and similarly {bn } is finite.

E. N ≤ 77. In particular, if N = 16000, this cannot be the case. 5. Show that in the arithmetic progression with first term 1 and ratio 729, there are infinitely many powers of 10. Solution: We will show that for all natural numbers n, 1081n − 1 is divisible by 729. 9 = 9 . . 9 · · · 10 . . 01 10 . . 01 . . 10 . . 01 = 9 1 . . 1 · · · 10 . . 01 10 . . 01 . . 10 . . 01 . 81 9 8 9 8 8 8 8 8 The second and third factors are composed of 9 units, so the sum of their digits is divisible by 9, that is, each is a multiple of 9.

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Mathematical Olympiads, Problems and Solutions from Around the World, 1996-1997


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