Difference between revisions of "2021 Fall AMC 12B Problems/Problem 16"
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<math>x=1,y=1,z=7</math>: let <math>a=7A, c=7C,</math> we can try all possibilities of <math>A</math> and <math>C</math> to find that <math>a=7, b=9, c=7</math> is the only solution. | <math>x=1,y=1,z=7</math>: let <math>a=7A, c=7C,</math> we can try all possibilities of <math>A</math> and <math>C</math> to find that <math>a=7, b=9, c=7</math> is the only solution. | ||
− | <math>x=1,y=2,z=6</math>: No solutions. By <math>y</math> and <math>z</math>, we know that <math>a</math>, <math>b</math>, and <math>c</math> have to all be | + | <math>x=1,y=2,z=6</math>: No solutions. By <math>y</math> and <math>z</math>, we know that <math>a</math>, <math>b</math>, and <math>c</math> have to all be divisible by <math>2</math>. Therefore, <math>x</math> cannot be equal to <math>1</math>. |
<math>x=1,y=3,z=5</math>: C has to be both a multiple of <math>3</math> and <math>5</math>. Therefore, <math>c</math> has to be a multiple of <math>15</math>. The only solution for this is <math>a=5, b=3, c=15</math>. | <math>x=1,y=3,z=5</math>: C has to be both a multiple of <math>3</math> and <math>5</math>. Therefore, <math>c</math> has to be a multiple of <math>15</math>. The only solution for this is <math>a=5, b=3, c=15</math>. | ||
− | <math>x=1,y=4,z=4</math>: No solutions. By <math>y</math> and <math>z</math>, we know that <math>a</math>, <math>b</math>, and <math>c</math> have to all be | + | <math>x=1,y=4,z=4</math>: No solutions. By <math>y</math> and <math>z</math>, we know that <math>a</math>, <math>b</math>, and <math>c</math> have to all be divisible by <math>2</math>. Therefore, <math>x</math> cannot be equal to <math>1</math>. |
− | <math>x=2,y=2,z=5</math>: No solutions. By <math>x</math> and <math>y</math>, we know that <math>a</math>, <math>b</math>, and <math>c</math> have to all be | + | <math>x=2,y=2,z=5</math>: No solutions. By <math>x</math> and <math>y</math>, we know that <math>a</math>, <math>b</math>, and <math>c</math> have to all be divisible by <math>2</math>. Therefore, <math>z</math> cannot be equal to <math>1</math>. |
− | <math>x=2,y=3,z=4</math>: No solutions. By <math>x</math> and <math>z</math>, we know that <math>a</math>, <math>b</math>, and <math>c</math> have to all be | + | <math>x=2,y=3,z=4</math>: No solutions. By <math>x</math> and <math>z</math>, we know that <math>a</math>, <math>b</math>, and <math>c</math> have to all be divisible by <math>2</math>. Therefore, <math>y</math> cannot be equal to <math>1</math>. |
<math>x=3,y=3,z=3</math>: No solutions. As <math>a</math>, <math>b</math>, and <math>c</math> have to all be divisible by <math>3</math>, <math>a+b+c</math> has to be divisible by <math>3</math>. This contradicts the sum <math>a+b+c=23</math>. | <math>x=3,y=3,z=3</math>: No solutions. As <math>a</math>, <math>b</math>, and <math>c</math> have to all be divisible by <math>3</math>, <math>a+b+c</math> has to be divisible by <math>3</math>. This contradicts the sum <math>a+b+c=23</math>. |
Revision as of 17:02, 24 November 2021
Problem
Suppose ,
,
are positive integers such that
and
What is the sum of all possible distinct values of
?
Solution
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