43rd IMO 2002 shortlist

Express 2002²⁰⁰² as the smallest possible number of (positive or negative) cubes.

Solution
All cubes are 0, 1, or -1 mod 9. But 2002 = 4 mod 9, so 2002³ = 1 mod 9, hence 2002²⁰⁰¹ = 1 mod 9 and 2002²⁰⁰² = 4 mod 9. So we need at least 4 cubes.

Now note that 2002 = 10³ + 10³ + 1³ + 1³. Multiplying through by N³, where N = 2002⁶⁶⁷, gives 2002²⁰⁰² 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