|A1. Find a13/(1 - 3a1 + 3a12) + a23/(1 - 3a2 + 3a22) + ... + a1013/(1 - 3a101 + 3a1012), where an = n/101.|
|A2. Find all permutations a1, a2, ... , a9 of 1, 2, ... , 9 such that a1 + a2 + a3 + a4 = a4 + a5 + a6 + a7 = a7 + a8 + a9 + a1 and a12 + a22 + a32 + a42 = a42 + a52 + a62 + a72 = a72 + a82 + a92 + a12.|
|A3. ABC is a triangle. The angle bisector at A meets the side BC at X. The perpendicular to AX at X meets AB at Y. The perpendicular to AB at Y meets the ray AX at R. XY meets the median from A at S. Prove that RS is perpendicular to BC.|
|A4. If m < n are positive integers prove that nn/(mm (n-m)n-m) > n!/( m! (n-m)! ) > nn/( mm(n+1) (n-m)n-m).|
|A5. Given a permutation s0, s2, ... , sn of 0, 1, 2, .... , n, we may transform it if we can find i, j such that si = 0 and sj = si-1 + 1. The new permutation is obtained by transposing si and sj. For which n can we obtain (1, 2, ... , n, 0) by repeated transformations starting with (1, n, n-1, .. , 3, 2, 0)?|
To avoid possible copyright problems, I have changed the wording, but not the substance, of the problems.
© John Scholes
14 April 2002