16th Irish 2003 Problem A3 Find ( [1 1/2 ] - [1 1/3 ] ) + ( [2 1/2 ] - [2 1/3 ] ) + ... + ( [2003 1/2 - [2003 1/3 ] ).
40842 Solution We have 1 3 = 1, 2 3 = 8, 3 3 = 27, 4 3 = 64, 5 3 = 125, 6 3 = 216, 7 3 = 343, 8 3 = 512, 9 3 = 729, 10 3 = 1000, 11 3 = 1331, 12 3 = 1728, 13 3 > 2003. So S [n 1/3 ] = 7·1 + 19·2 + 37·3 + 61·4 + 91·5 + 127·6 + 169·7 + 217·8 + 271·9 + 331·10 + 397·11 + 276·12 = 17964. Similarly, S [n 1/2 ] = 3·1 + 5·2 + 7·3 + ... + 97·43 + 68·44 = (2 S n 2 + S n) + 68·44 = 43(43+1) (2·43+1)/3 + 43·44/2 + 68·44 = 58806. Comment. Of course one could find the general formula for the cube root sum, but it is faster to do it longhand (you should know the low cubes off by heart). 16th Irish 2003 © John Scholes jscholes@kalva.demon.co.uk 14 June 2003 Last updated/corrected 14 Jun 03