Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2002 IMO Shortlist Problems/N1"

 
 
(2 intermediate revisions by 2 users not shown)
Line 3: Line 3:
 
==Solution==
 
==Solution==
  
−
{{solution}}
+
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>.
  
−
*[[2002 IMO Shortlist]]
+
 
 +
To show that '''4''' is the minimum value of <math>t</math>, note that
 +
<math>(10\cdot 2002^{667})^3+(10\cdot 2002^{667})^3+(2002^{667})^3+(2002^{667})^3=2002^{2002}</math>
 +
 
 +
*[[2002 IMO Shortlist Problems]]
 +
 
 +
 
 +
[[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 $(10\cdot 2002^{667})^3+(10\cdot 2002^{667})^3+(2002^{667})^3+(2002^{667})^3=2002^{2002}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_IMO_Shortlist_Problems/N1&oldid=14169"