Equations! Find the smallest value...

[TheDrBraniac]

In the following equation,

 

[math]\frac{x(a)}{2}[/math] = [math]\frac{x(b)}{3}[/math]= [math]\frac{x©}{4}[/math],

 

what is the smallest possible value of:

 

[math]\frac{a+b+c+x}{2}[/math]

 

 

In my case, I got 9.5. Are there any lower answers?

[timo]

I don't see how the question makes any sense. You probably missed to include some restrictions, e.g. what type of numbers a, b and c are and what type of function x is. And x seeming to be a function depending on some parameter in the first equation but not in the term you want to minimize is pretty suspicious, too.

[TheDrBraniac]

What I am trying to say is that x, a, b and c are whole, composite, natural and positive numbers. They are algebraic values. You need to find out the smallest possible value of that (the bottom equation) in such a way that it satisfies the above equation.

 

I bracketed it so as to prevent confusion between x(algebraic value) and x(operation). It means multiplied into.

[timo]

I have some doubts about your solution, then. The first composite naturals are 4, 6 and 8.

a=b=c=x=4 is not a solution. Having one of them equal 6 and the others being 4 would give a value of 9. Having 2 sixes and two twos already gives 10. Having three fours and 1 eight also gives a result of 10. All other combinations should give even higher results.

 

If you leave out the "composite" criterion then there is a solution with a lower value.

[Amr Morsi]

timo, have you considered the first equality in your calculations? I think we should use 4,6,8 for a,b,c and 4 for x.

[timo]

There is no need to consider equation 1 for my statement: If there is no solution that satisfies x+a+b+c = 19 then there is no solution that satisfies both, x+a+b+c=19 and equation 1. I did not give the solution because that did not seem to be intended by TheDrBrainac; he would probably have given his solution, then. But as a remark: The question about the smallest value is rather trivial to solve by using brute force; ~10 lines of computer code.

[the tree]

I bracketed it so as to prevent confusion between x(algebraic value) and x(operation). It means multiplied into.

But it doesn't. Seriously, standard notation exists for a reason. [imath]x \times a[/imath] or [imath]x \cdot a[/imath] or even [imath]xa[/imath] aren't ambiguous but [imath]x(a)[/imath] definitely looks like [imath]x[/imath] refers to a function.

 

Anyway, back to the question.

 

The ratio for a:b:c is already set out as 1:1.5:2. There's no doubt that the smallest composite numbers matching that are going to be 4:6:8, and the value chosen for x doesn't affect the restriction so the smallest composite number, 4, may as well be chosen as Amir suggested

 

So the smallest value I found for the expression to be minimised was 11 and I'd be pretty damn suspicious if any other values were given.

 

(if the composite criterion were lifted, then 5, but that makes the question even more trivial)

 

In my case, I got 9.5.

I guess what we're saying, is we don't believe you. Can you give any values for a,b,c,x that would satisfy the criteria you gave?

[the tree]

Fancy telling us how? With the given criteria (and abuse of notation) - I'm sticking by 11 and I'm really not going to believe another answer unless I actually see it.

 

edit Note that you've claimed to have found four composite numbers whose sum is 15, bearing in mind that the smallest composite number is 4 and four 4s makes 16 - either you're interpreting the "problem" differently to everyone else or you've made a very big mistake.

[ewmon]

Minimizing the second equation means minimizing a+b+c+x. All instances of x cancel out of the first equation leaving a/2 = b/3 = c/4. Thus, the lowest values of a, b and c are 4, 6 and 8, respectively. The lowest value for x is simply 4, thus the second equation equals (4+6+8+4)/2 which equals 11. No?

[shyvera]

The question is all about minimizing [math]a+b+c[/math] since [math]x[/math] is independent of [math]a,\ b[/math] and [math]c.[/math] From the first equation, [math]b=\tfrac32a[/math] and [math]c=2a.[/math] Hence

 

[math]a+b+c\ =\ a+\frac32a+2a\ =\ \frac92a[/math]

The minimum value of this is [math]18[/math] corresponding to [math]a=4.[/math] Moreover [math]a=4\ \implies\ b=6[/math] and [math]c=8,[/math] i.e. choosing [math]a=4[/math] also conveniently makes [math]b[/math] and [math]c[/math] composite. Hence the minimum value of

 

[math]\frac{a+b+c+x}2[/math]

is [math]\frac{18+4}2=11.[/math]

 

Hopefully this proves that the tree is right.

[the tree]

It's worth considering that Vijer wont be interested, much as TheDrBraniac wasn't. Not that it's a hugely interesting problem.

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Merged post follows:

Consecutive posts merged

Oh that's interesting, Vijer's post appears to be deleted - was this whole thing just some really slow trolling?