43rd IMO 2002 shortlist Problem N1 Express 2002 2002 as the smallest possible number of (positive or negative) cubes.
All cubes are 0, 1, or -1 mod 9. But 2002 = 4 mod 9, so 2002 3 = 1 mod 9, hence 2002 2001 = 1 mod 9 and 2002 2002 = 4 mod 9. So we need at least 4 cubes. Now note that 2002 = 10 3 + 10 3 + 1 3 + 1 3 . Multiplying through by N 3 , where N = 2002 667 , gives 2002 2002 as a sum of 4 cubes. 43rd IMO shortlist 2002 (C) John Scholes jscholes@kalva.demon.co.uk 8 Aug 2003 Last corrected/updated 8 Aug 03