Need Help Solving 1/a + 1/ab + 1/abc = 25/84

[get_physical]

Homework Statement

Not sure if I'm posting in the right area. Stuck on this question.

1/a + 1/ab + 1/abc = 25/84
Find the value of a+b+c

Homework Equations

The Attempt at a Solution

I tried making abc = 84 then bc+c +1 = 25

but still couldn't solve it.

 

[rcgldr]

I think you're supposed to assume that a, b, c are integers, which limits the number of possibilities to integer factors of 84, since as you mentioned, abc = 84.

 

[get_physical]

I think you're supposed to assume that a, b, c are integers, which limits the number of possibilities to integer factors of 84, since as you mentioned, abc = 84.

yes I am going to assume a, b, c are integers... but I cannot make the assumption that abc=84 because abc = 84x where x is an integer.

 

[phyzguy]

I think there are multiple solutions. I found 10 solutions if negative integers are allowed, 3 if you restrict to positive integers.

 

[FeDeX_LaTeX]

Homework Statement

Not sure if I'm posting in the right area. Stuck on this question.

1/a + 1/ab + 1/abc = 25/84
Find the value of a+b+c

Homework Equations

The Attempt at a Solution

I tried making abc = 84 then bc+c +1 = 25

but still couldn't solve it.

Assuming $a,b,c \in \mathbb{Z}^{+}$, that gets you some of the solutions, since $bc + c + 1 = 25 \implies c(b + 1) = 24$ giving you some case-checking to do.

You may also wish to consider making the substitution $bc = 24 - c$, or see how changing $a$ affects $bc$.

 

[get_physical]

yes I got bc+c+1=25⟹c(b+1)=24
but is there some way to do it from here without guess and check?

where should i sub in bc=24-c?

 

[phyzguy]

yes I got bc+c+1=25⟹c(b+1)=24
but is there some way to do it from here without guess and check?

where should i sub in bc=24-c?

I don't think there is any way to find the solutions except to restrict the range of possible integers, then exhaustively search through them for possible solutions. As you say, "guess and check".

As the original problem is posed, even restricted to positive integers, I find three solutions.

 

[rcgldr]

but is there some way to do it from here without guess and check?

not reallly, but as mentioned earlier, assuming that a, b, c are integers and that abc = 84, then a, b, c are each one of the numbers or the product of a pair of the numbers selected from the factors of 84, which are 2,2,3,7.

 

[get_physical]

Yes, a, b, c, are positive integers and there's actually another condition... a>= b>= c

 

[Twich]

c = 84 / (25ab-84b-84) ; a,b,c <>0

if a, b, c are real number there are many solutions and a , b , c must not be zero.

if c is integer then 25ab-84b-84 should be the member of {1,-1, 2, -2, 3, -3, 4, -4, 6, -6, 7, -7, 14, -14, 21, -21, 42,-42,84,-84}

so you have 20 case to test

case 1 : 25ab-84b-84 = -84 , 25ab -84b = 0 then a =0 that prohibit .
...

 

[FeDeX_LaTeX]

Yes, a, b, c, are positive integers and there's actually another condition... a>= b>= c

Are you sure? I don't think there are any solutions with those constraints.

 

[phyzguy]

Are you sure? I don't think there are any solutions with those constraints.

I agree. There are no solutions with the conditions you've given. You'd better check the problem again.

 

[get_physical]

Oh I'm very sorry - definitely need more sleep. it's a<= b<= c

 

[phyzguy]

Oh I'm very sorry - definitely need more sleep. it's a<= b<= c

So with those conditions there is exactly one solution. Have you found it yet?