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)

[OlderButNotWiser]

Sorry about delay in posting this, but didn't check computer when got home last night.
Have checked this with my grandson whose problem this is to solve, but thought that his nan may be able to help, some hopes there!!!

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

A+B+C = 13 and A, B and C are different. What is ABC.

Thanks again and next time he asks i will say go and join Answerbank!!

[Prudie]

How old is your grandson? is it written exactly like this? I only ask because it took some work from graduate and PhD level mathematicians to get the answer. Seems odd.

[OlderButNotWiser]

Thanks for all the answers given here and the trouble it seems to have caused, will definately not be asking any more maths questions. Thanks again.

[Gizmonster]

No no no - please don't stop asking anymore maths questions.

There is genuinely no sarcasm when I say, I enjoyed tearing my hair out last night .... not much fun when they're too easy :P

[Gizmonster]

By the way, I'd also like to know how old your grandson is, as I've an A level in maths and a degree in Materials Science .... but I found this puzzle quite a handful ......

[jim360]

A=2, B=6, C= 5 is the correct solution.

I'd say that my approach was probably overkill, but it was inspired at the time by ff apparently not finding any solutions. Turns out he just overlooked it.

Do please keep posting them. I enjoyed the experience of solving this problem!

[factor-fiction]

Hi blackadder- it is normally the case that lower case letters are used in algebra but sometimes symbols other than letters, usually Greek letters can be used, and sometimes upper case= particularly in formulae such as V= IR.
Besides, this isn't a conventional algebra problem because , A B And C represent digits- so ABB means 100A +10B +B rather than AxBxB.

It's just like one of the puzzles we used to do as children where we were told TRAIN + TRACK = DRIVER and we had to find the value of each letter.

Well done to Jim for solving it. My spreadsheet method seemed an easy way to solve it by calculating and comparing the ABBB, BBBBC, ABB, BBC,AB and BC for each of the sixty or so possible combinations of A,B and C, but for some reason (Sod's law?) on my list of all the combinations the 265 combination was the only one I omitted to include.

Please ask if you have any more, OlderButNotWiser

[Prudie]

I also have this morning a whole page of algebra lying on the floor,done late in the dim light last night. It was working on Giz's method of substituting one variable leaving an equation of 2 which then required trial and error.Had I had my old BBC B I could have knocked up a program in Basic fairly quickly to solve it (if there were a solution) but am not up to faffing about doing it in Excel so I gave up and went to bed. I still think it's an odd question for apparent school homework.

[weecalf]

looks like 2 3 8 to me

[jim360]

I'd like to say that I'm still finding it an interesting problem. I was aware of the 26/65 =2/5 result from several years ago (it also works for 16/64, by the way) but your asking it prompted me to look into the problem seriously.

For example, it doesn't stop there, and indeed it seems that if AB/BC = A/C then you can add an arbitrarily large number of B's and you'll still get A/C at the end (in any base). I've not formally proved this yet (seems like a somewhat tough exercise), but heck, that's an interesting discovery.

In terms of the original problem, I think a general solution is possible: in base ten one finds that, if AB/BC = A/C,

10 = [(A-B)*C]/[(C-B)*A]

which implies that B must be smaller than both A and C, or larger than both. Incidentally, this also forces A, B and C to be different: if A=B then you get 0 on the right-hand side; if C=B you get infinity, and if A=C then you get 1, none of which equals 10. Hence A, B and C must be different.

I think there are ways to impose various other constraints on the values of A, B, C, eg I think you might find that A must be less than C.