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]

I'm off now Prudie, and will look back tomorrow, but I'm pretty sure the problem is not stated correctly

[jim360]

I thought that there might be some point in relaxing the condition that the numbers are in base 10, but apparently I get "solutions" that don't seem to work, so...

[jim360]

2,6,5.

[factor-fiction]

Can you illustrate how that works please Jim?

[Graham-W]

In the next to last formula AB/BC=A/C 26/65=2/5 Doesn't work. ??

[BlackadderV]

I can see that AB would be 12/30 = A/C = 2/5, but I cant'be bothered to do the big sums.

[Graham-W]

In the next to last formula AB/BC=A/C 26/65=2/5 Doesn't work. ??

Ah. It's multiplied. Then it works!

[factor-fiction]

It works if AB means A x B and BC means BxC, etc, but that is bound to be the case for other values of A,B and C too

[Graham-W]

In the next to last formula AB/BC=A/C 26/65=2/5 Doesn't work. ??

Ah. It's multiplied. Then it works!

2*6/6*5=2/5

[factor-fiction]

Maybe jim just means I missed 265 off my list

[jim360]

Sorry, I was just checking my method against other solutions in different bases.

Your interpretation is correct, and if we consider a number in a general base x then some fairly simple algebra gives that:

[(A-B)C]/[(C-B)A]=x

this turns out to be the only unique equation that emerges; eliminating B then gives:

[(2A+C-x-3)C]/[(2C+A-x-3)A]=x

which is an equation that has a simple right-hand side making for easy checking. In base 10 the equation holds true for A=2, C=5, so that B=6. This then gives:

2666/6665=266/665=26/65 = 2/5

which can be checked as all giving 0.4.

(I have so far also found the solution A=1, B=5, C=3 in base 6, using the same equation).

[BlackadderV]

No it is algebraic, just done it on a calculator. Using Jim's figures and multiplying, all four answers are 0.4

[Gizmonster]

2666/6665
Multiply top and bottom lines by 1333 and divide by 133 gives:
266/665
Multiply top and bottom lines by 133 and divide by 13 gives:
26/65
Divide top and bottom lines by 13 gives:
2/5 Well done - how did you solve it, was it just brute force ??

[jim360]

26/65 = (13*2)/(13*5) = 2/5, so it would work.

[factor-fiction]

jim is right- for some reason I somehow missed the 265 combination off my list.

[factor-fiction]

A spreadsheet solves it quite easily- it just failed for me because I somehow managed to missed out the one combination that does work!

[jim360]

Yes, it's an interesting little problem. A fully general solution involves a quadratic equation, where you have to take the square root of a quintic(!) polynomial in the base, and includes one free parameter, that you then have to choose such that A, B C are distinct positive integers less than 13 (or x+3 for the "base" x).

I'll have to check that my algebra is fully correct before I publish the solution; Excel seems to suggest that I might have made a slip in that general solution. Oh for Mathematica's "solve" function...

[BlackadderV]

One thing I do remember from my maths O level days, 50 years ago, is that in equations letters standing for numbers are always in lower case, never upper.

[jim360]

Not really sure what you mean by that. You can use large letters to represent numbers, although there are all sorts of weird conventions governing how you should use them I guess, so maybe they were different back in't day.

anyway I fixed the error, a stray minus sign (of course, curse those things!). I suppose now I ought to find a full general solution, ie when A+B+C = px+q ...

[BlackadderV]

Yes, I'm old. One always wrote (a + b) (a -b), never (A + B) (A -B)