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Algebra Difficulty 5.0 AIME Find the answer

Solve the equation a3+b3+c3=2001a^3 + b^3 + c^3 = 2001 in positive integers.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Note that for all positive integers n,n, the value n3n^3 is congruent to −1,0,1-1,0,1 modulo 9.9. Since 2001≡3(mod9),2001 \equiv 3 \pmod{9}, we find that a3,b3,c3≡1(mod9).a^3,b^3,c^3 \equiv 1 \pmod{9}. Thus, a,b,c≡1(mod3),a,b,c \equiv 1 \pmod{3}, and the only numbers congruent to 11 modulo 33 are 1,4,7,10.1,4,7,10.
WLOG , let a≥b≥c.a \ge b \ge c. That means a3≥b3,c3a^3 \ge b^3, c^3 and 3a3≥2001.3a^3 \ge 2001. Thus, a3≥667,a^3 \ge 667, so a=10.a = 10.
Now b3+c3=1001.b^3 + c^3 = 1001. Since b3≥c3,b^3 \ge c^3, we find that 2b3≥1001.2b^3 \ge 1001. That means b=10b = 10 and c=1.c = 1.
In summary, the only solutions are (10,10,1),(10,1,10),(1,10,10).\boxed{(10,10,1),(10,1,10),(1,10,10)}.

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