Difference between revisions of "2008 Indonesia MO Problems/Problem 3"
Victorzwkao (talk | contribs) (Created page with "==Solution 1== Summing up the equation <math>\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}</math> yields the result <math>\frac{a^2b+ab^2+b^2c+bc^2+c^2a+ca^2}{abc}=z\in \mathbb...") |
Victorzwkao (talk | contribs) (→Solution 1) |
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Let <math>b>1</math>. If <math>b=c</math>, then <math>\gcd(b,c)=b>1</math>, and they won't be co-prime. As a result, <math>b</math> is strictly less than <math>c</math>. Since <math>b|c+a</math>, and <math>a=1</math>, we have <math>b|c+1</math>. Similarly, <math>c|b+1</math>. But <math>b<c</math>. Thus, <math>c=b+1</math>. Using the fact that <math>b|c+1</math>, we have <math>b|b+2\implies b|2</math>. Hence, <math>b=2</math>, and <math>c=3</math>. | Let <math>b>1</math>. If <math>b=c</math>, then <math>\gcd(b,c)=b>1</math>, and they won't be co-prime. As a result, <math>b</math> is strictly less than <math>c</math>. Since <math>b|c+a</math>, and <math>a=1</math>, we have <math>b|c+1</math>. Similarly, <math>c|b+1</math>. But <math>b<c</math>. Thus, <math>c=b+1</math>. Using the fact that <math>b|c+1</math>, we have <math>b|b+2\implies b|2</math>. Hence, <math>b=2</math>, and <math>c=3</math>. | ||
− | Plugging in the answers <math>(1,1,1)</math>, <math>(1,2,3)</math>, and <math>(1,1,2)</math> all yield valid results. | + | Plugging in the answers <math>(1,1,1)</math>, <math>(1,2,3)</math>, and <math>(1,1,2)</math> all yield valid results of <math>k=6</math>, <math>k=8</math>, and <math>k=7</math>, respectively. |
Latest revision as of 12:10, 17 September 2024
Solution 1
Summing up the equation yields the result
. Thus,
Since ,
,
, are pairwise relatively prime, this implies that
,
, and
.
, but because
and
,
. Similarly,
and
.
WLOG, .
Suppose .
Since they are pairwise relatively prime, and all ,
,
,
is not equal to
nor
, and
is not equal to
. A strict order of
can be made. Since
and
, we have
. We also know that
and
, which implies that
. Thus, the only way
, while
, is if
.
Plugging in , we get
, and
. Because
, we know that
. Thus, in order for
and
, the only possible way is for
. However, we have previously established that
, and combined with the fact that
,
and
are not co-prime, which is a contradiction.
Hence, .
Case 1:
Let . since
, we have
. The only options will be
and
. This gives us the answers
and
Case 2:
Let . If
, then
, and they won't be co-prime. As a result,
is strictly less than
. Since
, and
, we have
. Similarly,
. But
. Thus,
. Using the fact that
, we have
. Hence,
, and
.
Plugging in the answers ,
, and
all yield valid results of
,
, and
, respectively.