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A1. S is a square side 1. The points A, B, C, D lie on the sides of S in that order, and each side of S contains at least one of A, B, C, D. Show that 2 ≤ AB2 + BC2 + CD2 + DA2 ≤ 4.
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A2. A sumsquare is a 3 x 3 array of positive integers such that each row, each column and each of the two main diagonals has sum m. Show that m must be a multiple of 3 and that the largest entry in the array is at most 2m/3 - 1.
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A3. The sequence a1, a2, a3, ... is defined by a1 = 1, a2n = an, a2n+1 = a2n + 1. Find the largest value in a1, a2, ... , a1989 and the number of times it occurs.
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A4. n2 ends with m equal non-zero digits (in base 10). What is the largest possible value of m?
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A5. An n-digit number has the property that if you cyclically permute its digits it is always divisible by 1989. What is the smallest possible value of n? What is the smallest such number? [If we cyclically permute the digits of 3701 we get 7013, 137, 1370, and 3701.]
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B1. L is a fixed line, A a fixed point, k > 0 a fixed real. P is a variable point on L, Q is the point on the ray AP such that AP·AQ = k2. Find the locus of Q.
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B2. Each of n people has a unique piece of information. They wish to share the information. A person may pass another person a message containing all the pieces of information that he has. What is the smallest number of messages that must be passed so that each person ends up with all n pieces of information? For example, if A, B, C start by knowing a, b, c respectively. Then four messages suffice: A passes a to B; B passes a and b to C; C passes a, b and c to A; C passes a, b, c to B.
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B3. Let k be the product of the distances from P to the sides of the triangle ABC. Show that if P is inside ABC, then AB·BC·CA ≥ 8k with equality iff ABC is equilateral.
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B4. Show that (n + √(n2 + 1))1/3 + (n - √(n2 + 1))1/3 is a positive integer iff n = m(m2 + 3)/2 for some positive integer m.
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B5.(a) Show that 2nCn < 22n and that it is divisible by all primes p such that n < p < 2n (where 2nCn = (2n)! /(n! n!) ).
(b) Let π(x) denote the number of primes ≤ x. Show that for n > 2 we have π(2n) < π(n) + 2n/log2n and π(2n) < (1/n) 2n+1 log2(n-1). Deduce that for x ≥ 8, π(x) < (4x/log2x) log2(log2x ).
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