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Difference between revisions of "2002 IMO Shortlist Problems/N1"
 
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==Solution==
 
==Solution==
  
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{{solution}}
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Observe that <math>2002^{2002}\equiv 4^{2002}\equiv 64^{667}\cdot 4\equiv 4\pmod{9}</math>. On the other hand, each cube is congruent to 0, 1, or -1 modulo 9. So a sum of at most three cubes modulo 9 must among <math>0,\pm 1,\pm 2,\pm 3</math> none of which are congruent to 4. Therefore <math>t\geq 4</math>.
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To show that '''4''' is the minimum value of <math>t</math>, note that
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<math>(10\cdot 2002^{667})^3+(10\cdot 2002^{667})^3+(2002^{667})^3+(2002^{667})^3=2002^{2002}</math>
  
 
*[[2002 IMO Shortlist Problems]]
 
*[[2002 IMO Shortlist Problems]]
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[[Category:Olympiad Number Theory Problems]]

Latest revision as of 14:53, 24 March 2007

Problem

What is the smallest positive integer $t$ such that there exist integers $x_1,x_2,\ldots,x_t$ with

$x^3_1+x^3_2+\,\ldots\,+x^3_t=2002^{2002}$?

Solution

Observe that $2002^{2002}\equiv 4^{2002}\equiv 64^{667}\cdot 4\equiv 4\pmod{9}$. On the other hand, each cube is congruent to 0, 1, or -1 modulo 9. So a sum of at most three cubes modulo 9 must among $0,\pm 1,\pm 2,\pm 3$ none of which are congruent to 4. Therefore $t\geq 4$.


To show that 4 is the minimum value of $t$, note that $(10\cdot 2002^{667})^3+(10\cdot 2002^{667})^3+(2002^{667})^3+(2002^{667})^3=2002^{2002}$

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