Yet Another Maths Problem

[OlderButNotWiser]

Well as it is the day for maths questions, how about this one

ABBB/BBBC = ABB/BBC = AB/BC = A/C

What is ABC, (All different and adding up to 13)

[factor-fiction]

Well we are clearly not using algebraic conventions here in that, for example, AB must mean 10A+B rather than AxB

[Prudie]

Well that has infinite solutions if they are not whole numbers. And if it's whole numbers and you mean ABBB is AxBxBXB then again it could be all the combinations of 4 numbers that sum to 13 - or is Factor on the right lines?

[factor-fiction]

I can't find any solutions so my assumption that AB = 10A +B etc must be wrong.
I can't even find a solution that allows AB/BC to equal A/C given the constraint that A, B and C are all different and add up to 13

[factor-fiction]

Please could you give the exact wording of the problem as given in this quiz

[Prudie]

I'm confused. let's try A=2, B=4 and C=7
Then AB is 2x4=8
Then BC is 4x7=28
So AB/BC is 8/28 which is 4/7 which is obviously A/C
What am I missing, as I said try a different selection, it will always work ???
There must be something missing

[Prudie]

Doh, sorry a typo, 8/28 is 2/7 which is A/B

[Cloverjo]

Prudie, 8/28 is not equal to 4/7

[Cloverjo]

Typed my last post before i saw your last.

[Prudie]

At least you read the original then :-)

[factor-fiction]

I agree, prudie. That's why I thought that, for example , if A =2 and B= 4 then AB must be 24 rather than the algebraic result of 2x4=8.
If AB did mean AxB then each given fraction could be simplified to A/C. That would give around 60 possible solutions given the constraint that A,b and C are different and add up to 13

[Gizmonster]

I've assumed the same as FF, insofar as AB = 10A + B and not A x B.
My working out may be completely wrong but I've managed to get 2 equations with 2 unknowns, but I can't solve them.
The 2 equations are:

110B^2 - 99C^2 + 22BC + 1287C - 1430B = 0

and:

10B^2 - 9C^2 +2BC - 130B + 117C = 0

.....stumped now :(

[Gizmonster]

In the above equations, BC = B x C

[factor-fiction]

I suppose you can also add in A= 13-(B+C), Gizmonster, unless you've already done that

[factor-fiction]

Sorry, you've clearly done that already to eliminate A

[Gizmonster]

Yea that's what I did - substituted A = 13 - B - C, to try and get 2 equations with 2 unknowns. Usually quite easy to solve .... hmmm :(

[factor-fiction]

Putting to one side the "ABBB/BBBC = ABB/BBC = AB/BC " aspect
I can't find a solution to AB/BC = A/C

Using algebraic terminology we need a solution to :
C(10A+B)= A(10B+C) where A= 13-(B+C).
I'm not sure there is one

[factor-fiction]

I also tried using a spreadsheet.
I could find only these possible values of A, B and C from the constraints that A, B and C were different, A+b+c =13 and C can't be 0 (since that would make A/C of infinite value)
a b c
0 4 9
0 5 8
0 6 7
0 7 7
0 8 5
0 9 4
1 3 9
1 4 8
1 5 7
1 7 5
1 8 4
1 9 3
2 3 8
2 4 7
2 5 8
2 7 4
2 8 3
3 1 9
3 2 8
3 4 6
3 5 7
3 6 4
3 8 2
3 9 1
4 0 9
4 1 8
4 2 7
4 3 6
4 6 3
4 7 2
4 8 1
5 0 8
5 1 7
5 2 6
5 3 4
5 4 3
5 6 2
5 7 1
6 0 7
6 2 5
6 3 4
6 4 3
6 5 2
7 0 6
7 1 5
7 2 4
7 4 2
7 5 1
8 0 5
8 1 4
8 2 3
8 3 2
8 4 1
9 0 4
9 1 3
9 3 1

[Gizmonster]

Going nowhere with this.
I've just noticed that my top equation is simply the bottom equation multiplied by 11, so now I've only got 1 equation with 2 unknowns :(

[factor-fiction]

I think we need to see the exact wording of the problem- something may have been lost in translation

[Prudie]

You two deserve a medal. OBNW it is mildly niggling that you've posted a question and not responded in coming up to 3 hours - is this the exact wording? I'm worried factor and Giz won't get any sleep.