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 "2021 Fall AMC 12B Problems/Problem 16"

(Created page with "== Problem == Suppose <math>a</math>, <math>b</math>, <math>c</math> are positive integers such that <cmath>a+b+c=23</cmath> and <cmath>\gcd(a,b)+\gcd(b,c)+\gcd(c,a)=9.</cmat...")
 
(→‎Solution)
Line 6: Line 6:
  
 
==Solution==
 
==Solution==
−
Let <math>\gcd(a,b)=x</math>, <math>\gcd(b,c)=y</math>, <math>\gcd(c,a)=z</math>. WLOG, let <math>x\ley\lez</math>. We can split this off into cases:  
+
Let <math>\gcd(a,b)=x</math>, <math>\gcd(b,c)=y</math>, <math>\gcd(c,a)=z</math>. WLOG, let <math>x \le y \le z</math>. We can split this off into cases:  
  
 
x=1, y=1, z=7: let a = 7A, c=7C, we can try all possibilities of A and C to find that a=7, b=9, c=7 is a solution.  
 
x=1, y=1, z=7: let a = 7A, c=7C, we can try all possibilities of A and C to find that a=7, b=9, c=7 is a solution.  

Revision as of 15:55, 24 November 2021

Problem

Suppose $a$, $b$, $c$ are positive integers such that \[a+b+c=23\] and \[\gcd(a,b)+\gcd(b,c)+\gcd(c,a)=9.\] What is the sum of all possible distinct values of $a^2+b^2+c^2$?

$(\textbf{A})\: 259\qquad(\textbf{B}) \: 438\qquad(\textbf{C}) \: 516\qquad(\textbf{D}) \: 625\qquad(\textbf{E}) \: 687$

Solution

Let $\gcd(a,b)=x$, $\gcd(b,c)=y$, $\gcd(c,a)=z$. WLOG, let $x \le y \le z$. We can split this off into cases:

x=1, y=1, z=7: let a = 7A, c=7C, we can try all possibilities of A and C to find that a=7, b=9, c=7 is a solution.

x=1, y=2, z=6: No solutions. By y and z, we know that a, b, and c have to all be even. Therefore, x cannot be equal to 1.

x=1, y=3, z=5: C has to be both a multiple of 3 and 5. Therefore, c has to be a multiple of 15. The only solution for this is a=5, b=3, c=15.

x=1, y=4, z=4: No solutions. By y and z, we know that a, b, and c have to all be even. Therefore, x cannot be equal to 1.

x=2, y=2, z=5: No solutions. By x and y, we know that a, b, and c have to all be even. Therefore, z cannot be equal to 1.

x=2, y=3, z=4: No solutions. By x and z, we know that a, b, and c have to all be even. Therefore, y cannot be equal to 1.

x=3, y=3, z=3: No solutions. As a, b, and c have to all be divisible by 3, a+b+c has to be divisible by 3. This contradicts the sum $a+b+c=23$.

Putting these solutions together, we have $(7^2+9^2+7^2)+(5^2+3^2+15^2)=179+259=\boxed{\textbf{(B) }438}$

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12B_Problems/Problem_16&oldid=166179"