### 43rd IMO 2002 shortlist

**Problem N1**

Express 2002^{2002} 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 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