IMO 2002 SL N1

Origin: UZB | Category: Number Theory

Problem

What is the smallest positive integer t such that there exist integers x1, x2, . . . , xt with x3 1 + x3 2 + \cdot \cdot \cdot + x3 t = 20022002?

Solution

Consider the given equation modulo 9. Since each cube is congruent to either −1, 0 or 1, whereas 20022002 \equiv42002 = 4 \cdot 64667 \equiv4 (mod 9), we conclude that t \geq4. On the other hand, 20022002 = 2002 \cdot (2002667)3 = (103 + 103 + 13 + 13)(2002667)3 is a solution with t = 4. Hence the answer is 4.