Find $a+b+c$ for Integer $a,b,c$ Given $a+b=2004$

[anemone]

For the integers $a, b, c$, we have $\dfrac{2a-b}{c}=\dfrac{2b+c}{a}=\dfrac{-2a-c}{b}$.

If $a+b=2004$, find $a+b+c$.

 

[kaliprasad]

Re: Find a+b+c

For the integers $a, b, c$, we have $\dfrac{2a-b}{c}=\dfrac{2b+c}{a}=\dfrac{-2a-c}{b}$.

If $a+b=2004$, find $a+b+c$.

Spoiler

We have
(2a-b)/c = (-2a-c)/b
As (a/b= c/d = (a+c)/(b+d))
Have both = (2a-b – 2a –c)/(b+c) = -1
Furher from (2b+c)/a = (-2a –c) we get both =2 (b-a)/(b+a) = - 1
So
2(b-a) = - (b+a) = - 2004
Or b – a = - 1002
So a = 1503, b = 501
Now 2b+c = -a or c = - (2b-a) = - 2505
So a + b+c = - 501
(ratio a:b:c = 3:1:-5 and sum = -b)

 

[anemone]

Re: Find a+b+c

Hi kaliprasad, thank for participating and I particularly like the following "trick" that you used in your method to solve the problem!

If $ad=bc$, then

i$\dfrac{a+c}{b+d}= \dfrac{\frac{bc}{d}+c}{b+d}=\dfrac{\frac{c}{d}(b+d)}{b+d}=\dfrac{c}{d}$

$\;\;\;\;\;\;\;\;\;\;\overset{or}= \dfrac{a+\frac{ad}{b}}{b+d}=\dfrac{\frac{a}{b}(b+d)}{b+d}=\dfrac{a}{b}$.

My solution:

Spoiler

From the given equation $\dfrac{2a-b}{c}=\dfrac{2b+c}{a}=\dfrac{-2a-c}{b}$, we have:

$\dfrac{2a-b}{c}=\dfrac{2b+c}{a}$	$\dfrac{2a-b}{c}=\dfrac{-2a-c}{b}$	$\dfrac{2b+c}{a}=\dfrac{-2a-c}{b}$
$2a^2-ab=2bc+c^2$	$(b+c)(2a-b+c)=0$
	But notice that $b+c\ne 0$ because if $b+c=0$, $2a^2-ab=2bc+c^2$ becomes $2a^2-ac=-2c^2+c^2$ $2a^2-ac+c^2=0$ $2a^2-ac+c^2=0$ $2(a-\frac{c}{4})+\frac{7c^2}{8}\ne 0$ for all integers $a, c$. Hence, $2a-b+c=0$	$\dfrac{2b+c}{a}=\dfrac{-(2a+c)}{b}$ $\dfrac{2b+c}{a}=\dfrac{-b}{b}$ $2b+c=-a$ $a+2b+c=0$

Now, solve the equations $2a-b+c=0$ and $a+2b+c=0$ by eliminating the variable $c$ for a and b, we get $a=3b$ and from $a+b=2004$, we obtain $a=1503, b=501, c=-2505$ and therefore $a+b+c=-501$.

 

[Albert1]

Re: Find a+b+c

For the integers $a, b, c$, we have $\dfrac{2a-b}{c}=\dfrac{2b+c}{a}=\dfrac{-2a-c}{b}$.

If $a+b=2004$, find $a+b+c$.

let :
$\dfrac{2a-b}{c}=\dfrac{2b+c}{a}=\dfrac{-2a-c}{b}=\dfrac {1}{k} , k\neq 0$.
then :
a=(2b+c)k----(1)
b=(-2a-c)k---(2)
c=(2a-b)k----(3)
(2)+(3) :b+c=(-b-c)k , so k=-1
(1)+(2)+(3): a+b+c=bk=-b
(1)+(2) :a+b=2a-2b , so a=3b
freom a+b=4b=2004 , b=501
we get a+b+c=-b=-501

 

[CellCoree]

To find $a+b+c$, we first need to solve for $a$ and $b$ in terms of $c$ from the given equations. From the first equation, we can rearrange to get $b=2004-a$. Substituting this into the second equation, we get $\dfrac{2a-(2004-a)}{c}=\dfrac{2(2004-a)+c}{a}$. Simplifying this, we get $a^2+2004a-2004c=0$. Using the quadratic formula, we can solve for $a$ in terms of $c$: $a=\dfrac{-2004 \pm \sqrt{2004^2+4(2004)c}}{2}$. This gives us two possible values for $a$, which are $a=\dfrac{-2004+\sqrt{2004^2+4(2004)c}}{2}$ and $a=\dfrac{-2004-\sqrt{2004^2+4(2004)c}}{2}$.

Similarly, we can solve for $b$ in terms of $c$ by substituting $b=2004-a$ into the third equation. This gives us $\dfrac{2(2004-a)+c}{2004-a}=\dfrac{-2(2004-a)-c}{2004-a}$. Simplifying this, we get $c^2+2004c-2004(2004-a)=0$. Again using the quadratic formula, we can solve for $c$ in terms of $a$: $c=\dfrac{-2004 \pm \sqrt{2004^2+4(2004)(2004-a)}}{2}$. This gives us two possible values for $c$, which are $c=\dfrac{-2004+\sqrt{2004^2+4(2004)(2004-a)}}{2}$ and $c=\dfrac{-2004-\sqrt{2004^2+4(2004)(2004-a)}}{2}$.

Now, to find $a+b+c$, we can substitute the values we found for $a$ and $c$ into the equation $a+b+c$. This gives us four possible values for $a+b+c$:

1. $\dfrac{-2004+\sqrt{2004^2+4(2004)c}}{2