I've just joined the forum for help on this math homework question my 12 year old son was given. As far as I'm aware he hasn't been given any lessons in how to do it.
I probably spent between 2 and 3 hours trying various techniques to create a fraction that equaled 1/3 but didn't manage to crack it. For instance I'd try 4000/12,000 or 5000/15000 or 6000/18000 etc etc, and see if I could arrange the remaining numerals as a fraction that equaled 1/3 then I could just replace the zeros with that (hope that made sense), but unless I missed an option I could not find a match.
So basically the numerals from 1 to 9 can only be used once and be arranged as a fraction, e.g. 1234/56789
I gave the question to my supervisor at work also. He's done an engineering degree but he wasn't sure how to do this either.
Any ideas ?
I've seen this problem before, given to students at this level, probably intended to be solved by orderly trial and error, with a little intelligence added in choosing what to try. (This gives lots of practice multiplying, and a little practice thinking in an orderly way.) Unfortunately, even for the best of us it probably ends up being more "trial" (and more "error") than we want, and becomes frustrating. To my mind, difficult puzzles should never be assigned for everyone to do, but those who like such challenges can thrive on them.
Note that engineers, who expect there to be a mathematical method to any problem, tend to be more frustrated than those who expect trial and error to be good enough. On the other hand, often trial and error makes sense as a first method even when there is something better, just as a way to become familiar with how a problem works.
But here's how I'd approach it, at that level.
We want abcd/ghijk = 1/3; clearly there will be 4 digits on top and 5 on the bottom. This implies 3*abcd = ghijk, which is easier to work with.
It might be of use to note that ghijk is a multiple of 3, so the sum of its digits is a multiple of 3; but that will only limit what the last digit you pick can be, so it may not help a lot.
We do know there has to be a "carry" into the fifth place, so "a" can't be less than 3 (3*3 plus a 1 carried in would work); that cuts down the possible numbers to try a little. At the other end, if "a" is 9, "g" will be 2, which is not a problem. But this shows that "g" will be either 1 or 2. So if we ever have both a 1 and a 2 (in either number), we can quit immediately.
With this pre-thinking done, I'd just start with small numbers starting with 3 and avoiding 1 (which would be in the product), and see what happens. I'd hope to discover more tricks to be aware of as I go along.
3*3245 = 97
35, no good - product too small; but also, can't have 5 at the end!
3*3456 = 1
0368, no good - need to be even bigger to avoid the zero; skip to 11000/3
3*3678 = 1
1034, no good - oops, better skip to 12345/3
3*4235 = 1
27
05, no good - watch that 2, avoid 5 at the end.
and so on. I'm taking it slowly here (with a spreadsheet I'd be speeding through it, but not thinking nearly as much) in order to learn as I go. After a few more tries, I might come up with a better idea and reformulate my method.