MHB Find positive integers x,y,z in 1/x+1/y=7/8(x,y∈N)

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$(1)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{7}{8}(x,y\in N)$
$find$: $x,y$
$(2)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{6}{12}(x,y\in N)$
$find$: $x,y$
$(3)$ $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}(x,y,z\in N)$
$find$: $x,y,z$
 
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Albert said:
$(1)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{7}{8}(x,y\in N)$
$find$: $x,y$
$(2)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{6}{12}(x,y\in N)$
$find$: $x,y$
$(3)$ $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}(x,y,z\in N)$
$find$: $x,y,z$

without loss of generality we can take x <=y <=z as any permutation can give ans

for (1)

$\frac{7}{8} > \frac{1}{x} >= \frac{7}{16}$ or so x = 2 and it does not give y integer so no solution

for (2)

$\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{2}$

or $\dfrac{1}{2} > \dfrac{1}{x} >= \dfrac{1}{4}$ x = 4 or 3 , x=4 gives y =4 and x= 3 gives y = 6 so solution set(4,4), (3,6), (6,3)
 
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kaliprasad said:
without loss of generality we can take x <=y <=z as any permutation can give ans

for (1)

$\frac{7}{8} > \frac{1}{x} >= \frac{7}{16}$ or so x = 2 and it does not give y integer so no solution

for (2)

$\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{2}$

or $\dfrac{1}{2} > \dfrac{1}{x} >= \dfrac{1}{4}$ x = 2 or 3 , x=2 gives y =2 and x= 3 gives y = 6 so solution set(2,2), (3,6), (6,3)
set (x,y)=(2,2) is not a solution ,it should be (x,y)=(4,4)
 
Albert said:
set (x,y)=(2,2) is not a solution ,it should be (x,y)=(4,4)

you are right I have done correction inline . Thanks
 
Albert said:
$(1)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{7}{8}(x,y\in N)$
$find$: $x,y$
$(2)$ $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{6}{12}(x,y\in N)$
$find$: $x,y$
$(3)$ $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}(x,y,z\in N)$
$find$: $x,y,z$
hint of (3)$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}$
let :$\dfrac{6}{24}=\dfrac {a+b+c}{24}$
we see $a+b+c=6$
and $24$ is a multiple of $a,b,$ and $c\, (a,b,c\in N)$
with this little trick ,this kind of question becomes so easy
 
Albert said:
hint of (3)$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}$
let :$\dfrac{6}{24}=\dfrac {a+b+c}{24}$
we see $a+b+c=6$
and $24$ is a multiple of $a,b,$ and $c\, (a,b,c\in N)$
with this little trick ,this kind of question becomes so easy

The above hint is not correct because

x= 6, y = 16, z= 48 is solution not provided by above
 
Albert said:
hint of (3)$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac {1}{z}=\dfrac{6}{24}$
let :$\dfrac{6}{24}=\dfrac {a+b+c}{24}$
we see $a+b+c=6$
and $24$ is a multiple of $a,b,$ and $c\, (a,b,c\in N)$
with this little trick ,this kind of question becomes so easy
[sp]But that method does not find all the solutions, for example $\dfrac6{24} = \dfrac15 + \dfrac1{21} + \dfrac1{420}$.[/sp]
 
yes you are right,that is why I express (3) as:
$\dfrac{6}{24} = \dfrac1{x} + \dfrac1{y} + \dfrac1{z}$
if I express (3) as :
$\dfrac{12}{48} = \dfrac1{x} + \dfrac1{y} + \dfrac1{z}$
and using the same method we will get more solutions,up to now I did not see
any theorem mentioning to find the amounts of solutions of this kind ,may be someone can tell me
(if both numerator and denominator are fixed,then the solutions will be fixed,if both
are multiplied by some factor k,then may be we will get more answers)
 
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Albert said:
any theorem mentioning to find the amounts of solutions of this kind ,may be someone can tell me
These are known as Egyptian fractions.
 
  • #10
Opalg said:
These are known as Egyptian fractions.
Can we get all the amounts of solutions of this kind ?
Is there any formula to express it ?
 

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