2019 H3_Solution

Su ggested Solution Remarks 1i      2 222 2 22236 23 6 4 9 2367 xyz xyz xyz        1ii 22 2 222 2 2367 1 236Since 2 3 6 7 , by observation, , , . 777 xyz xyz xyz        1iii 2 1 Suppose 1. n i i x   22 2 11 1 1 1 nn n n ii ii ii i i x xn x n x n                 Since if we let 1 ix n  for all 1 in , we yield 2 2 1 1 1 n i i xn n   and 1 1n i i x nn n   , the maximum possible value of 1 n i i x   is n . 1iv Suppose there are n squares of lengths xi contained in the unit square. Their total area is 2 1 1 n i i x   , and their total perimeter is 1 18 4 n i i x   . By part (iii), 1 18 20.254 n i i xn n     . Hence, there must be more than 20 such squares. 2ia 841Number of ways 165 41   [Bijection with a string of eight 0’s (objects ) and three 1’s (partitions). For example, the string 00100010100 would correspond to the combination of 2 A’s, 3 C’s, 1 G and 2 T’s.] 2ib 44 1Number of ways 35 41   [Place one 0 in each box and the remaining four 0’s into the boxes.] 2iia 8Number of sequences 4 65536 2iib 7Number of sequences 4 3 8748  2iic 88 8 8 444Number of sequences 4 3 2 1 40824321           Principle of inclusion / exclusion. 3ia 1 11 1 Given 1 and 0. 11 2 1 1... ...111 1 3 2 1 1 for all ii i i i xa ia i i i i ixx x x xi i ii ii i xi i                                   3ib From (3i(a)), 22 2 11 1 11 1 22 nn n i in in in x in     .

3ic  12 12 22 2 2 21 12 1 21 2 1 2 11 1... 22 lim lim is unbounded.2 nn n nn n iii i ii ii ii nnii nxxx x x x n nxx                     3iia Since   111 1 ii i i i iaix i x i x i x a x i        , summing from im to in yields the following.        11 2 1 11 11 2 1 ... 1 1 nn ii i m m m m im im nn n m a x ixi x mx m x m x mx nxn x nxm x                        Alternative Solution Let Pn be the statement  1 1 11 n in i ax n x     . 12 1 1 1 1RHS of P 2 1 2 1 LHS of P2 P is true. axa a x       Assume Pk is true for some k   .    1 k+1 1 11 11 1 22 1 1 1 LHS of P 11 11 121 2 RHS of P P is true P is true. kk ii k ii kk k kk k kk axax a x kx a x ka x kakx x x k                            Since P1 is true and  1P is true P is truekk  , Pn is true for all n   .     1 1 1 1 11 1 11 i f 1 Hence, 11 1 i f 1 1. n inn i i nm im ii n m ii nm xn x m x xx n x m x m nxm x                       Alt. Sol. We first ask ourselves what is the partial sum from 1 to n, and from there we would be able to find the sum from m to n. By observing the equation