... Starting with2002 2002 2002 = 103+ 103+ 13+ 13and using once again, we find that2002 = 667 × 3 + 1 2002 2002= 2002 ×( 2002 667)3=(10 × 2002 667)3+(10 × 2002 667)3+( 2002 667)3+( 2002 667)3.Comments1. ... x32+ … +x3t= 2002 2002 ?Solution. The answer is .t = 4We first show that is not a sum of three cubes by considering numbers modulo 9.Thus, from , and we find that 2002 2002 2002 ≡ 4 (mod ... modulo 9.Thus, from , and we find that 2002 2002 2002 ≡ 4 (mod 9) 43≡ 1 (mod 9) 2002 = 667 × 3 + 1 2002 2002≡ 4 2002 ≡ 4 (mod 9),whereas, from , for any integer , we see that .x3≡ 0 ±1 (mod...