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 "2001 JBMO Problems/Problem 1"

m
(→‎Solution2)
 
(One intermediate revision by the same user not shown)
Line 3: Line 3:
 
Solve the equation <math>a^3 + b^3 + c^3 = 2001</math> in positive integers.
 
Solve the equation <math>a^3 + b^3 + c^3 = 2001</math> in positive integers.
  
−
==Solution==
+
==Solution1==
  
 
Note that for all positive integers <math>n,</math> the value <math>n^3</math> is congruent to <math>-1,0,1</math> [[modulo]] <math>9.</math>  Since <math>2001 \equiv 3 \pmod{9},</math> we find that <math>a^3,b^3,c^3 \equiv 1 \pmod{9}.</math> Thus, <math>a,b,c \equiv 1 \pmod{3},</math> and the only numbers congruent to <math>1</math> modulo <math>3</math> are <math>1,4,7,10.</math>
 
Note that for all positive integers <math>n,</math> the value <math>n^3</math> is congruent to <math>-1,0,1</math> [[modulo]] <math>9.</math>  Since <math>2001 \equiv 3 \pmod{9},</math> we find that <math>a^3,b^3,c^3 \equiv 1 \pmod{9}.</math> Thus, <math>a,b,c \equiv 1 \pmod{3},</math> and the only numbers congruent to <math>1</math> modulo <math>3</math> are <math>1,4,7,10.</math>
Line 15: Line 15:
 
<br>
 
<br>
 
In summary, the only solutions are <math>\boxed{(10,10,1),(10,1,10),(1,10,10)}.</math>
 
In summary, the only solutions are <math>\boxed{(10,10,1),(10,1,10),(1,10,10)}.</math>
 +
 +
==Solution2==
 +
You can also watch remainders modulo 7 which are also -1,0,1. The rest is almost identical as in solution 1.
  
 
==See Also==
 
==See Also==

Latest revision as of 10:01, 28 April 2024

Problem

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

Solution1

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 $\boxed{(10,10,1),(10,1,10),(1,10,10)}.$

Solution2

You can also watch remainders modulo 7 which are also -1,0,1. The rest is almost identical as in solution 1.

See Also

2001 JBMO (Problems • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4
All JBMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2001_JBMO_Problems/Problem_1&oldid=218792"