Difference between revisions of "2021 Fall AMC 12B Problems/Problem 16"

Latest revision as of 19:45, 30 April 2023

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 1 (Observation)

Because $a + b + c$ is odd, $a$ , $b$ , $c$ are either one odd and two evens or three odds.

$\textbf{Case 1}$ : $a$ , $b$ , $c$ have one odd and two evens.

Without loss of generality, we assume $a$ is odd and $b$ and $c$ are even.

Hence, ${\rm gcd} \left( a , b \right)$ and ${\rm gcd} \left( a , c \right)$ are odd, and ${\rm gcd} \left( b , c \right)$ is even. Hence, ${\rm gcd} \left( a , b \right) + {\rm gcd} \left( b , c \right) + {\rm gcd} \left( c , a \right)$ is even. This violates the condition given in the problem.

Therefore, there is no solution in this case.

$\textbf{Case 2}$ : $a$ , $b$ , $c$ are all odd.

In this case, ${\rm gcd} \left( a , b \right)$ , ${\rm gcd} \left( a , c \right)$ , ${\rm gcd} \left( b , c \right)$ are all odd.

Without loss of generality, we assume ${\rm gcd} \left( a , b \right) \leq {\rm gcd} \left( b , c \right) \leq {\rm gcd} \left( c , a \right) .$ $\textbf{Case 2.1}$ : ${\rm gcd} \left( a , b \right) = 1$ , ${\rm gcd} \left( b , c \right) = 1$ , ${\rm gcd} \left( c , a \right) = 7$ .

The only solution is $(a, b, c) = (7, 9, 7)$ .

Hence, $a^2 + b^2 + c^2 = 179$ .

$\textbf{Case 2.2}$ : ${\rm gcd} \left( a , b \right) = 1$ , ${\rm gcd} \left( b , c \right) = 3$ , ${\rm gcd} \left( c , a \right) = 5$ .

The only solution is $(a, b, c) = (5, 3, 15)$ .

Hence, $a^2 + b^2 + c^2 = 259$ .

$\textbf{Case 2.3}$ : ${\rm gcd} \left( a , b \right) = 3$ , ${\rm gcd} \left( b , c \right) = 3$ , ${\rm gcd} \left( c , a \right) = 3$ .

There is no solution in this case.

Therefore, putting all cases together, the answer is $179 + 259 = \boxed{\textbf{(B)} \: 438}$ .

~Steven Chen (www.professorchenedu.com)

Solution 2 (Enumeration)

Let $\gcd(a,b)=x$ , $\gcd(b,c)=y$ , $\gcd(c,a)=z$ . Without the loss of generality, 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 the only solution.

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

$x=1,y=3,z=5$ : Note that $c$ 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 divisible by $4$ . 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 divisible by $2$ . Therefore, $z$ cannot be equal to $5$ .

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

$x=3,y=3,z=3$ : 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 $(7^2+9^2+7^2)+(5^2+3^2+15^2)=179+259=\boxed{\textbf{(B)} \: 438}$ .

~ConcaveTriangle

Solution 3

Since $a+b+c=23$ , $\gcd(a,b,c)=23$ or $\gcd(a,b,c)=1$ .

As $\gcd(a,b)+\gcd(b,c)+\gcd(c,a)=9$ , it is impossible for $\gcd(a,b,c)=23$ , so $\gcd(a,b,c)=1$ .

This means that $\gcd(a,b)$ , $\gcd(b,c)$ , and $\gcd(c,a)$ must all be coprime. The only possible ways for this to be true are $1+1+7=9$ and $1+3+5=9$ .

Without loss of generality, let $a\le b\le c$ . Since $a+b+c=23$ , then $a=7, b=7, c=9$ or $a=3, b=5, c=15$ .

$(7^2+7^2+9^2)+(3^2+5^2+15^2)=179+259=\boxed{\textbf{(B)} \: 438}$ .

~bkunzang

Video Solution By Power Of Logic

https://youtu.be/QNNIVYKvIyI

~~Hayabusa1

@@ Line 67: / Line 67: @@
 ~ConcaveTriangle
+==Solution 3==
+Since <math>a+b+c=23</math>, <math>\gcd(a,b,c)=23</math> or <math>\gcd(a,b,c)=1</math>.
+As <math>\gcd(a,b)+\gcd(b,c)+\gcd(c,a)=9</math>, it is impossible for <math>\gcd(a,b,c)=23</math>, so <math>\gcd(a,b,c)=1</math>.
+This means that <math>\gcd(a,b)</math>, <math>\gcd(b,c)</math>, and <math>\gcd(c,a)</math> must all be coprime. The only possible ways for this to be true are <math>1+1+7=9</math> and <math>1+3+5=9</math>.
+Without loss of generality, let <math>a\le b\le c</math>. Since <math>a+b+c=23</math>, then <math>a=7, b=7, c=9</math> or <math>a=3, b=5, c=15</math>.
+<math>(7^2+7^2+9^2)+(3^2+5^2+15^2)=179+259=\boxed{\textbf{(B)} \: 438}</math>.
+~bkunzang
 ==Video Solution By Power Of Logic==

2021 Fall AMC 12B (Problems • Answer Key • Resources)
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