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Solve $frac1x+frac1y= frac12007$
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up vote
3
down vote
favorite
The number of positive integral pairs $(x<y)$ such that $frac1x+frac1y= frac12007$
The answer is 7 where as i am getting 6.
The ordered pair are (2676,8028),(2230,20070),(2016,449568),(2010,1344690),(2008,4030056)&(2008,4028049).
I cannot find my mistake.
systems-of-equations diophantine-equations
edited Jul 30 at 12:50
Harry Peter
5,45311438
asked Jul 24 at 9:36
Samar Imam Zaidi
1,063316
add a comment |Â
up vote
3
down vote
favorite
The number of positive integral pairs $(x<y)$ such that $frac1x+frac1y= frac12007$
The answer is 7 where as i am getting 6.
The ordered pair are (2676,8028),(2230,20070),(2016,449568),(2010,1344690),(2008,4030056)&(2008,4028049).
I cannot find my mistake.
systems-of-equations diophantine-equations
edited Jul 30 at 12:50
Harry Peter
5,45311438
asked Jul 24 at 9:36
Samar Imam Zaidi
1,063316
2
How can $1/2008+1/4030056$ and $1/2008+1/4028049$ both be $1/2007$?
– Henning Makholm
Jul 24 at 9:43
There are solutions with $x=2034, 2088$ (and $4014$, but you may not accept that)
– Henry
Jul 24 at 9:57
2
How can we find your mistake if you don't tell us how you calculated them?
– miracle173
Jul 24 at 9:59
I checked in excel
– Samar Imam Zaidi
Jul 24 at 10:28
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
The number of positive integral pairs $(x<y)$ such that $frac1x+frac1y= frac12007$
The answer is 7 where as i am getting 6.
The ordered pair are (2676,8028),(2230,20070),(2016,449568),(2010,1344690),(2008,4030056)&(2008,4028049).
I cannot find my mistake.
systems-of-equations diophantine-equations
edited Jul 30 at 12:50
Harry Peter
5,45311438
asked Jul 24 at 9:36
Samar Imam Zaidi
1,063316
The number of positive integral pairs $(x<y)$ such that $frac1x+frac1y= frac12007$
The answer is 7 where as i am getting 6.
The ordered pair are (2676,8028),(2230,20070),(2016,449568),(2010,1344690),(2008,4030056)&(2008,4028049).
I cannot find my mistake.
systems-of-equations diophantine-equations
edited Jul 30 at 12:50
Harry Peter
5,45311438
asked Jul 24 at 9:36
Samar Imam Zaidi
1,063316
edited Jul 30 at 12:50
Harry Peter
5,45311438
edited Jul 30 at 12:50
Harry Peter
5,45311438
edited Jul 30 at 12:50
Harry Peter
5,45311438
5,45311438
asked Jul 24 at 9:36
Samar Imam Zaidi
1,063316
asked Jul 24 at 9:36
Samar Imam Zaidi
1,063316
asked Jul 24 at 9:36
Samar Imam Zaidi
1,063316
1,063316
2
How can $1/2008+1/4030056$ and $1/2008+1/4028049$ both be $1/2007$?
– Henning Makholm
Jul 24 at 9:43
There are solutions with $x=2034, 2088$ (and $4014$, but you may not accept that)
– Henry
Jul 24 at 9:57
2
How can we find your mistake if you don't tell us how you calculated them?
– miracle173
Jul 24 at 9:59
I checked in excel
– Samar Imam Zaidi
Jul 24 at 10:28
add a comment |Â
2
How can $1/2008+1/4030056$ and $1/2008+1/4028049$ both be $1/2007$?
– Henning Makholm
Jul 24 at 9:43
There are solutions with $x=2034, 2088$ (and $4014$, but you may not accept that)
– Henry
Jul 24 at 9:57
2
How can we find your mistake if you don't tell us how you calculated them?
– miracle173
Jul 24 at 9:59
I checked in excel
– Samar Imam Zaidi
Jul 24 at 10:28
2
2
How can $1/2008+1/4030056$ and $1/2008+1/4028049$ both be $1/2007$?
– Henning Makholm
Jul 24 at 9:43
How can $1/2008+1/4030056$ and $1/2008+1/4028049$ both be $1/2007$?
– Henning Makholm
Jul 24 at 9:43
There are solutions with $x=2034, 2088$ (and $4014$, but you may not accept that)
– Henry
Jul 24 at 9:57
There are solutions with $x=2034, 2088$ (and $4014$, but you may not accept that)
– Henry
Jul 24 at 9:57
2
2
How can we find your mistake if you don't tell us how you calculated them?
– miracle173
Jul 24 at 9:59
How can we find your mistake if you don't tell us how you calculated them?
– miracle173
Jul 24 at 9:59
I checked in excel
– Samar Imam Zaidi
Jul 24 at 10:28
I checked in excel
– Samar Imam Zaidi
Jul 24 at 10:28
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
5
down vote
accepted
$$2007x + 2007 y = xy$$
$$0=xy-2007-2007y$$
$$2007^2=xy-2007x-2007y+2007^2$$
$$(2007^2)=(x-2007)(y-2007)$$
$$3^4cdot 223^2=(x-2007)(y-2007)$$
beginalign3^4cdot 223^2 &=(3^0) cdot (3^4cdot 223^2)\
&=(3^1) cdot (3^3cdot 223^2)
\
&=(3^2) cdot (3^2cdot 223^2)
\
&=(3^3) cdot (3^1cdot 223^2)\
&=(3^4) cdot (3^0cdot 223^2)\
&=(3^0cdot 223) cdot (3^4cdot 223)\
&=(3^1cdot 223) cdot (3^3cdot 223)
endalign
I hope you can recover $x$ and $y$ from here.
answered Jul 24 at 9:47
Siong Thye Goh
77.4k134795
There is also $(3^2cdot 223) cdot (3^2cdot 223)$ though this may be excluded by $x lt y$
– Henry
Jul 24 at 9:59
Thanking you for the prompt reply.
– Samar Imam Zaidi
Jul 24 at 10:28
@Henry: “may be†excluded??
– TonyK
Jul 30 at 13:02
@TonyK "can be" or "should be" if you prefer, but it is another factorisation of $3^4cdot 223^2$
– Henry
Jul 30 at 15:25
I would prefer “isâ€Â.
– TonyK
Jul 30 at 15:59
add a comment |Â
up vote
6
down vote
Write the equations as,
$$(x+y)2007=xy$$
$$xy-2007x-2007y+2007^2=2007^2$$
$$(x-2007)(y-2007)=2007^2$$
also, $2007=3^2.223$
Can you continue?
answered Jul 24 at 9:47
prog_SAHIL
773217
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
$$2007x + 2007 y = xy$$
$$0=xy-2007-2007y$$
$$2007^2=xy-2007x-2007y+2007^2$$
$$(2007^2)=(x-2007)(y-2007)$$
$$3^4cdot 223^2=(x-2007)(y-2007)$$
beginalign3^4cdot 223^2 &=(3^0) cdot (3^4cdot 223^2)\
&=(3^1) cdot (3^3cdot 223^2)
\
&=(3^2) cdot (3^2cdot 223^2)
\
&=(3^3) cdot (3^1cdot 223^2)\
&=(3^4) cdot (3^0cdot 223^2)\
&=(3^0cdot 223) cdot (3^4cdot 223)\
&=(3^1cdot 223) cdot (3^3cdot 223)
endalign
I hope you can recover $x$ and $y$ from here.
answered Jul 24 at 9:47
Siong Thye Goh
77.4k134795
There is also $(3^2cdot 223) cdot (3^2cdot 223)$ though this may be excluded by $x lt y$
– Henry
Jul 24 at 9:59
Thanking you for the prompt reply.
– Samar Imam Zaidi
Jul 24 at 10:28
@Henry: “may be†excluded??
– TonyK
Jul 30 at 13:02
@TonyK "can be" or "should be" if you prefer, but it is another factorisation of $3^4cdot 223^2$
– Henry
Jul 30 at 15:25
I would prefer “isâ€Â.
– TonyK
Jul 30 at 15:59
add a comment |Â
up vote
5
down vote
accepted
$$2007x + 2007 y = xy$$
$$0=xy-2007-2007y$$
$$2007^2=xy-2007x-2007y+2007^2$$
$$(2007^2)=(x-2007)(y-2007)$$
$$3^4cdot 223^2=(x-2007)(y-2007)$$
beginalign3^4cdot 223^2 &=(3^0) cdot (3^4cdot 223^2)\
&=(3^1) cdot (3^3cdot 223^2)
\
&=(3^2) cdot (3^2cdot 223^2)
\
&=(3^3) cdot (3^1cdot 223^2)\
&=(3^4) cdot (3^0cdot 223^2)\
&=(3^0cdot 223) cdot (3^4cdot 223)\
&=(3^1cdot 223) cdot (3^3cdot 223)
endalign
I hope you can recover $x$ and $y$ from here.
answered Jul 24 at 9:47
Siong Thye Goh
77.4k134795
There is also $(3^2cdot 223) cdot (3^2cdot 223)$ though this may be excluded by $x lt y$
– Henry
Jul 24 at 9:59
Thanking you for the prompt reply.
– Samar Imam Zaidi
Jul 24 at 10:28
@Henry: “may be†excluded??
– TonyK
Jul 30 at 13:02
@TonyK "can be" or "should be" if you prefer, but it is another factorisation of $3^4cdot 223^2$
– Henry
Jul 30 at 15:25
I would prefer “isâ€Â.
– TonyK
Jul 30 at 15:59
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
$$2007x + 2007 y = xy$$
$$0=xy-2007-2007y$$
$$2007^2=xy-2007x-2007y+2007^2$$
$$(2007^2)=(x-2007)(y-2007)$$
$$3^4cdot 223^2=(x-2007)(y-2007)$$
beginalign3^4cdot 223^2 &=(3^0) cdot (3^4cdot 223^2)\
&=(3^1) cdot (3^3cdot 223^2)
\
&=(3^2) cdot (3^2cdot 223^2)
\
&=(3^3) cdot (3^1cdot 223^2)\
&=(3^4) cdot (3^0cdot 223^2)\
&=(3^0cdot 223) cdot (3^4cdot 223)\
&=(3^1cdot 223) cdot (3^3cdot 223)
endalign
I hope you can recover $x$ and $y$ from here.
answered Jul 24 at 9:47
Siong Thye Goh
77.4k134795
$$2007x + 2007 y = xy$$
$$0=xy-2007-2007y$$
$$2007^2=xy-2007x-2007y+2007^2$$
$$(2007^2)=(x-2007)(y-2007)$$
$$3^4cdot 223^2=(x-2007)(y-2007)$$
beginalign3^4cdot 223^2 &=(3^0) cdot (3^4cdot 223^2)\
&=(3^1) cdot (3^3cdot 223^2)
\
&=(3^2) cdot (3^2cdot 223^2)
\
&=(3^3) cdot (3^1cdot 223^2)\
&=(3^4) cdot (3^0cdot 223^2)\
&=(3^0cdot 223) cdot (3^4cdot 223)\
&=(3^1cdot 223) cdot (3^3cdot 223)
endalign
I hope you can recover $x$ and $y$ from here.
answered Jul 24 at 9:47
Siong Thye Goh
77.4k134795
answered Jul 24 at 9:47
Siong Thye Goh
77.4k134795
answered Jul 24 at 9:47
Siong Thye Goh
77.4k134795
answered Jul 24 at 9:47
Siong Thye Goh
77.4k134795
77.4k134795
There is also $(3^2cdot 223) cdot (3^2cdot 223)$ though this may be excluded by $x lt y$
– Henry
Jul 24 at 9:59
Thanking you for the prompt reply.
– Samar Imam Zaidi
Jul 24 at 10:28
@Henry: “may be†excluded??
– TonyK
Jul 30 at 13:02
@TonyK "can be" or "should be" if you prefer, but it is another factorisation of $3^4cdot 223^2$
– Henry
Jul 30 at 15:25
I would prefer “isâ€Â.
– TonyK
Jul 30 at 15:59
add a comment |Â
There is also $(3^2cdot 223) cdot (3^2cdot 223)$ though this may be excluded by $x lt y$
– Henry
Jul 24 at 9:59
Thanking you for the prompt reply.
– Samar Imam Zaidi
Jul 24 at 10:28
@Henry: “may be†excluded??
– TonyK
Jul 30 at 13:02
@TonyK "can be" or "should be" if you prefer, but it is another factorisation of $3^4cdot 223^2$
– Henry
Jul 30 at 15:25
I would prefer “isâ€Â.
– TonyK
Jul 30 at 15:59
There is also $(3^2cdot 223) cdot (3^2cdot 223)$ though this may be excluded by $x lt y$
– Henry
Jul 24 at 9:59
There is also $(3^2cdot 223) cdot (3^2cdot 223)$ though this may be excluded by $x lt y$
– Henry
Jul 24 at 9:59
Thanking you for the prompt reply.
– Samar Imam Zaidi
Jul 24 at 10:28
Thanking you for the prompt reply.
– Samar Imam Zaidi
Jul 24 at 10:28
@Henry: “may be†excluded??
– TonyK
Jul 30 at 13:02
@Henry: “may be†excluded??
– TonyK
Jul 30 at 13:02
@TonyK "can be" or "should be" if you prefer, but it is another factorisation of $3^4cdot 223^2$
– Henry
Jul 30 at 15:25
@TonyK "can be" or "should be" if you prefer, but it is another factorisation of $3^4cdot 223^2$
– Henry
Jul 30 at 15:25
I would prefer “isâ€Â.
– TonyK
Jul 30 at 15:59
I would prefer “isâ€Â.
– TonyK
Jul 30 at 15:59
add a comment |Â
up vote
6
down vote
Write the equations as,
$$(x+y)2007=xy$$
$$xy-2007x-2007y+2007^2=2007^2$$
$$(x-2007)(y-2007)=2007^2$$
also, $2007=3^2.223$
Can you continue?
answered Jul 24 at 9:47
prog_SAHIL
773217
add a comment |Â
up vote
6
down vote
Write the equations as,
$$(x+y)2007=xy$$
$$xy-2007x-2007y+2007^2=2007^2$$
$$(x-2007)(y-2007)=2007^2$$
also, $2007=3^2.223$
Can you continue?
answered Jul 24 at 9:47
prog_SAHIL
773217
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Write the equations as,
$$(x+y)2007=xy$$
$$xy-2007x-2007y+2007^2=2007^2$$
$$(x-2007)(y-2007)=2007^2$$
also, $2007=3^2.223$
Can you continue?
answered Jul 24 at 9:47
prog_SAHIL
773217
Write the equations as,
$$(x+y)2007=xy$$
$$xy-2007x-2007y+2007^2=2007^2$$
$$(x-2007)(y-2007)=2007^2$$
also, $2007=3^2.223$
Can you continue?
answered Jul 24 at 9:47
prog_SAHIL
773217
answered Jul 24 at 9:47
prog_SAHIL
773217
answered Jul 24 at 9:47
prog_SAHIL
773217
answered Jul 24 at 9:47
prog_SAHIL
773217
773217
add a comment |Â
add a comment |Â
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2
How can $1/2008+1/4030056$ and $1/2008+1/4028049$ both be $1/2007$?
– Henning Makholm
Jul 24 at 9:43
There are solutions with $x=2034, 2088$ (and $4014$, but you may not accept that)
– Henry
Jul 24 at 9:57
2
How can we find your mistake if you don't tell us how you calculated them?
– miracle173
Jul 24 at 9:59
I checked in excel
– Samar Imam Zaidi
Jul 24 at 10:28