(ak+1):(a+k)=(bk+1):(b+k+2)=(ck+2):(c+k+3)=n;a+b+c=12;k=multipkle of 2?

Homework.

I don’t do it.

It’s not homework.Is for a math contest.Please help

math contest? wow

If its a contest and you win can I have the prize?

no I don’t win this was a subject for a contest and i want to know what’s the right answer

What is how much wood a woodchuck could chuck if a woodchuck could chuck wood?

I’d track down the guy/gal who won the contest and ask ‘em how they did it…

oh come on I want the answers.No one got this right

then do it ?

Not a single math teacher at your school? I know school districts are laying off but that’s ridiculous!...

Um…lemme get this straight: you want the answer for a math contest…and supposedly NO ONE got it right…so you think us BRAINIACS in Fluther are going to do this for you?!

Yeah. I gotta step out for a while and pick up a lotto ticket. I think my odds for winning just went up

So the question is, must k be a multiple of 2? I’m assuming a,b,c,k,n are all integers? Positive integers?
Give me a sec, and I’ll see what I can do.

I think I’m going to need you to clarify precisely what the question is. Do you need to find what a,b,c and k are? Is k a multiple of 2, or do you need to figure out if it has to be a multiple of 2 for all choices of a,b,c?

I need to find a,b.c,k,n

are a,b,c all positive?

yes

Are the colons indicative of a ratio?

What are the colons?

Yes, they are ratios.

@Christian95 Is it possible that there are no solutions? Is that one of the options?

Here’s why i ask. In your first ratio, ak+1 is an odd number (since k is even), so this means that a+k is also odd and that n is odd too (rewrite ak+1 = n(a+k) to see why that must be true). So n and a are odd numbers.

By a similar argument, bk+1 is odd, so b+k+2 is odd, and hence b is odd.

Now if a and b are odd, and a+b+c=12, then c must be even. The last ratio can be rewritten ck+2=n(c+k+3)=even=(odd)(odd). That’s a problem. So it looks to me that there can be no integer solutions here.

Maybe I made a mistake somewhere, or maybe you had a typo when you entered the problem. But that looks right to me.

@finkelitis :Why is a+k necessarily odd? I get that k is odd, that is stated. So the numerator in that fist stanza is odd (even)*(anything) + 1 = odd. But the denominator, a+k could be even (if k is even and a is even, the result is even. if k is even and a is odd, then the result is odd).

@Les : k is even, actually. The way to think about it is that (ak+1)/(a+k)=n, or better, (ak+1)=(a+k)n. Now since k is even, ak is even as well, which means ak+1 is odd. This means that a+k and n are both odd as well—if either one of them were even, then ak+1 would have to be even too, since an even number times an even or odd number yields an even number. So that’s why you know that a+k and n are both odd, and since a+k is odd and k is even, then a is odd.

I have my husband the super genius working on it. He’s like a super serious computer nerd.