Difference between revisions of "2021 Fall AMC 12B Problems/Problem 16"
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<math>(\textbf{A})\: 259\qquad(\textbf{B}) \: 438\qquad(\textbf{C}) \: 516\qquad(\textbf{D}) \: 625\qquad(\textbf{E}) \: 687</math> | <math>(\textbf{A})\: 259\qquad(\textbf{B}) \: 438\qquad(\textbf{C}) \: 516\qquad(\textbf{D}) \: 625\qquad(\textbf{E}) \: 687</math> | ||
− | ==Solution== | + | ==Solution 1== |
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: | 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: | ||
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-ConcaveTriangle | -ConcaveTriangle | ||
+ | |||
+ | == Solution 2 == | ||
+ | Because <math>a + b + c</math> is odd, <math>a</math>, <math>b</math>, <math>c</math> are either one odd and two evens or three odds. | ||
+ | |||
+ | <math>\textbf{Case 1}</math>: <math>a</math>, <math>b</math>, <math>c</math> have one odd and two evens. | ||
+ | |||
+ | Without loss of generality, we assume <math>a</math> is odd and <math>b</math> and <math>c</math> are even. | ||
+ | |||
+ | Hence, <math>{\rm gcd} \left( a , b \right)</math> and <math>{\rm gcd} \left( a , c \right)</math> are odd, and <math>{\rm gcd} \left( b , c \right)</math> is even. | ||
+ | Hence, <math>{\rm gcd} \left( a , b \right) + {\rm gcd} \left( b , c \right) + {\rm gcd} \left( c , a \right)</math> is even. This violates the condition given in the problem. | ||
+ | |||
+ | Therefore, there is no solution in this case. | ||
+ | |||
+ | <math>\textbf{Case 2}</math>: <math>a</math>, <math>b</math>, <math>c</math> are all odd. | ||
+ | |||
+ | In this case, <math>{\rm gcd} \left( a , b \right)</math>, <math>{\rm gcd} \left( a , c \right)</math>, <math>{\rm gcd} \left( b , c \right)</math> are all odd. | ||
+ | |||
+ | Without loss of generality, we assume | ||
+ | <cmath> | ||
+ | \[ | ||
+ | {\rm gcd} \left( a , b \right) \leq {\rm gcd} \left( b , c \right) \leq {\rm gcd} \left( c , a \right) . | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | <math>\textbf{Case 2.1}</math>: <math>{\rm gcd} \left( a , b \right) = 1</math>, <math>{\rm gcd} \left( b , c \right) = 1</math>, <math>{\rm gcd} \left( c , a \right) = 7</math>. | ||
+ | |||
+ | The only solution for <math>\left( a, b, c \right)</math> is (7, 9, 7). | ||
+ | |||
+ | Hence, <math>a^2 + b^2 + c^2 = 179</math>. | ||
+ | |||
+ | <math>\textbf{Case 2.2}</math>: <math>{\rm gcd} \left( a , b \right) = 1</math>, <math>{\rm gcd} \left( b , c \right) = 3</math>, <math>{\rm gcd} \left( c , a \right) = 5</math>. | ||
+ | |||
+ | The only solution for <math>\left( a, b, c \right)</math> is (5, 3, 15). | ||
+ | |||
+ | Hence, <math>a^2 + b^2 + c^2 = 259</math>. | ||
+ | |||
+ | <math>\textbf{Case 2.3}</math>: <math>{\rm gcd} \left( a , b \right) = 3</math>, <math>{\rm gcd} \left( b , c \right) = 3</math>, <math>{\rm gcd} \left( c , a \right) = 3</math>. | ||
+ | |||
+ | There is no solution in this case. | ||
+ | |||
+ | Therefore, putting all cases together, the answer is <math>179 + 259 = 438</math>. | ||
+ | |||
+ | Therefore, the answer is <math>\boxed{\textbf{(B) }438}</math>. | ||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) |
Revision as of 00:15, 26 November 2021
Problem
Suppose ,
,
are positive integers such that
and
What is the sum of all possible distinct values of
?
Solution 1
Let ,
,
. WLOG, let
. We can split this off into cases:
: let
we can try all possibilities of
and
to find that
is the only solution.
: No solutions. By
and
, we know that
,
, and
have to all be divisible by
. Therefore,
cannot be equal to
.
: C has to be both a multiple of
and
. Therefore,
has to be a multiple of
. The only solution for this is
.
: No solutions. By
and
, we know that
,
, and
have to all be divisible by
. Therefore,
cannot be equal to
.
: No solutions. By
and
, we know that
,
, and
have to all be divisible by
. Therefore,
cannot be equal to
.
: No solutions. By
and
, we know that
,
, and
have to all be divisible by
. Therefore,
cannot be equal to
.
: No solutions. As
,
, and
have to all be divisible by
,
has to be divisible by
. This contradicts the sum
.
Putting these solutions together, we have
-ConcaveTriangle
Solution 2
Because is odd,
,
,
are either one odd and two evens or three odds.
:
,
,
have one odd and two evens.
Without loss of generality, we assume is odd and
and
are even.
Hence, and
are odd, and
is even.
Hence,
is even. This violates the condition given in the problem.
Therefore, there is no solution in this case.
:
,
,
are all odd.
In this case, ,
,
are all odd.
Without loss of generality, we assume
:
,
,
.
The only solution for is (7, 9, 7).
Hence, .
:
,
,
.
The only solution for is (5, 3, 15).
Hence, .
:
,
,
.
There is no solution in this case.
Therefore, putting all cases together, the answer is .
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)