# Math Help - Equation problem help

1. ## Equation problem help

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1$

Note that $a,b,c$ and $d$ are natural numbers and you have to find all the possibilities. I only found that $a=b=c=d=4$ or $a=b=c=6$ and $d=2$.

2. Originally Posted by brainmixer
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1$

Note that $a,b,c$ and $d$ are natural numbers and you have to find all the possibilities. I only found that $a=b=c=d=4$ or $a=b=c=6$ and $d=2$.

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=1$

first, a symetry exists among a, b, c, d
so, suppose a=<b=<c=<d
in this way, 4/a>=1
a>=1
well, you can suppose a=1 or a=2 or a=3 or a=4

* if a=1
there is no solutions
* if a=2
then, $\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{1}{2}$
or b=<c=<d
so 3/b>=1/2
b=<6
in this way, b=1 or b=2 or b=3 or b=4 or b=5 or b=6
**if b=1 : no solutions
**if b=2 : no solutions
**if b=3 : $\frac{1}{c}+\frac{1}{d}=\frac{1}{6}$
or c=<d
so 2/c=>1/6
c<=12
in this way, c=1 or c=2 or c=3 or c=4 or c=5 or c=6 or c=7 or c=8 or c=9 or c=10 or c=11 or c=12
***if c=1 or c=2 or c=3 or c=4 or c=5 or c=6 : no solutions
***if c=7: $\frac{1}{7}+\frac{1}{d}=\frac{1}{6}$
$\frac{1}{d}=\frac{1}{42}$
so, d=42
solution 1 : a=2, b=3, c=7, d=42
***if c=8 : $\frac{1}{8}+\frac{1}{d}=\frac{1}{6}$
$\frac{1}{d}=\frac{1}{24}$
so, d=24
solution 2 : a=2, b=3, c=8, d=24
***if c=9 : $\frac{1}{9}+\frac{1}{d}=\frac{1}{6}$
$\frac{1}{d}=\frac{1}{18}$
so, d=18
solution 3 : a=2, b=3, c=9, d=18
***if c=10 : $\frac{1}{10}+\frac{1}{d}=\frac{1}{6}$
$\frac{1}{d}=\frac{1}{15}$
so, d=15
solution 4 : a=2, b=3, c=10, d=15
***if c=11 : no solution
***if c=12 : $\frac{1}{12}+\frac{1}{d}=\frac{1}{6}$
$\frac{1}{d}=\frac{1}{12}$
so, d=12
solution 5 : a=2, b=3, c=12, d=12
**if b=4:
.....

it seems too long....but for sure it's an exhaustive way to solve this problem.