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

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==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>. |

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+ | 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]] | *[[2002 IMO Shortlist Problems]] |

## Latest revision as of 14:53, 24 March 2007

## Problem

What is the smallest positive integer such that there exist integers with

## Solution

Observe that . 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 none of which are congruent to 4. Therefore .

To show that **4** is the minimum value of , note that