Difference between revisions of "2001 JBMO Problems/Problem 1"

Revision as of 20:48, 11 August 2018

Problem

Solve the equation $a^3 + b^3 + c^3 = 2001$ in positive integers.

Solution

Note that for all positive integers $n,$ the value $n^3$ is congruent to $-1,0,1$ modulo $9.$ Since $2001 \equiv 3 \pmod{9},$ we find that $a^3,b^3,c^3 \equiv 1 \pmod{9}.$ Thus, $a,b,c \equiv 1 \pmod{3},$ and the only numbers congruent to $1$ modulo $3$ are $1,4,7,10.$

WLOG, let $a \ge b \ge c.$ That means $a^3 \ge b^3, c^3$ and $3a^3 \ge 2001.$ Thus, $a^3 \ge 667,$ so $a = 10.$

Now $b^3 + c^3 = 1001.$ Since $b^3 \ge c^3,$ we find that $2b^3 \ge 1001.$ That means $b = 10$ and $c = 1.$

In summary, the only solutions are $\boxed{(10,10,1),(10,1,10),(1,10,10)}.$

@@ Line 11: / Line 11: @@
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-Now <math>b^3 + c^3 = 1001.</math>  Since <math>b^3 \ge c^3,</math> we find that <math>2b^3 \ge 1001.</math>  Thus, <math>b = 10</math> and <math>c = 1.</math>
+Now <math>b^3 + c^3 = 1001.</math>  Since <math>b^3 \ge c^3,</math> we find that <math>2b^3 \ge 1001.</math>  That means <math>b = 10</math> and <math>c = 1.</math>
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Difference between revisions of "2001 JBMO Problems/Problem 1"

Revision as of 20:48, 11 August 2018

Problem

Solution

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