# Math problem o;



## frogmanson (Jun 29, 2010)

Is there anyway to do this without a computer?

possible "blank numbers" are : 1, 2, 3, 4, 5, 6, 7, 8, 9

here: \( 

\frac{a}{b\times c} + \frac{d}{e\times f} + \frac{g}{h\times i} = 1
\)

Thank you Rinfiyks xD


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## vcuber13 (Jun 29, 2010)

Do we use all numbers once?


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## Jebediah54 (Jun 29, 2010)

Is it (blank over blank) times blank, etc.

Or blank over (blank times blank), etc.?


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## LewisJ (Jun 29, 2010)

That was my question as well - although the fractions adding to one suggests the latter option, I'm not sure if the former is possible at all. And I also want to know if we must use each number - I would think we do, otherwise it is not challenging at all.


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## CubesOfTheWorld (Jun 29, 2010)

(1/8 X 2) + (1/8 X 2) + (1/4 X 2) = 1

I have a bad feeling that I wasn't supposed to do it this way.


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## PeterNewton (Jun 29, 2010)

wtf use variables. this is so ambiguous.


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## Kurbitur (Jun 29, 2010)

x * x + x / x *x + x / x *x = 1


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## Rinfiyks (Jun 29, 2010)

\( \frac{a}{b\times c} + \frac{d}{e\times f} + \frac{g}{h\times i} = 1 \)

I assume you mean this?


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## CubesOfTheWorld (Jun 29, 2010)

PeterNewton said:


> wtf use variables. this is so ambiguous.



I'm only going into 7th grade. I have never done a problem such like this. We would usually have something like this for extra credit.


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## frogmanson (Jun 29, 2010)

Rinfiyks said:


> \( \frac{a}{b\times c} + \frac{d}{e\times f} + \frac{g}{h\times i} = 1 \)
> 
> I assume you mean this?



Yes thank you i meant that o;


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## hawkmp4 (Jun 29, 2010)

If three fractions add up to 1, and you're not using negative numbers, what does that say about each fraction?


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## Rinfiyks (Jun 29, 2010)

frogmanson said:


> Rinfiyks said:
> 
> 
> > \( \frac{a}{b\times c} + \frac{d}{e\times f} + \frac{g}{h\times i} = 1 \)
> ...



Okay.

Here's a solution according to my very inefficient program. There is only one it would seem. (Ignoring different permutations of the same solution - else there are 48.)


Spoiler



\( \frac{1}{3\times 6} + \frac{5}{8\times 9} + \frac{7}{2\times 4} = 1 \)


I'm not too sure how you'd manage to get that without a computer, but there's probably a really elegant way which someone will post soon


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## frogmanson (Jun 29, 2010)

OMG thank you


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## vcuber13 (Jun 29, 2010)

The closest I got was \( 
\frac{1051}{1008}
\)


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## Rinfiyks (Jun 29, 2010)

Finding a solution with a computer is not the challenge!
I don't consider this to be truly solved until it is done by hand and not brute force 
I'll have a go later, should be a fun time waster


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## vcuber13 (Jun 29, 2010)

I find it harder when you already know the answer though.


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## Erdos (Jun 29, 2010)

I tried it. This is one of the convoluted algebraic problems where you can only do this by trial-and-error. However, this isn't to say that you try every single permutation. There are of course some shortcuts via logic and induction. I was going to explain my logic, but it took more than a page so I just gave up. Here's a link if you're still interested:
http://mathforum.org/library/drmath/view/56829.html
It's probably more than 3 pages I'd assume.


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## frogmanson (Jun 29, 2010)

if you ever solve it by hand, could you post it here please and thank you xD


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## Erdos (Jun 29, 2010)

^ Read above.


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## Rinfiyks (Jun 29, 2010)

Erdos said:


> I tried it. This is one of the convoluted algebraic problems where you can only do this by trial-and-error. However, this isn't to say that you try every single permutation. There are of course some shortcuts via logic and induction. I was going to explain my logic, but it took more than a page so I just gave up. Here's a link if you're still interested:
> http://mathforum.org/library/drmath/view/56829.html
> It's probably more than 3 pages I'd assume.



This is a slightly different problem. This thread's asking for two single digit numbers multiplied together on the denominator.
That link's fractions have two numbers on the denominator that are the digits of a two digit number.


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## frogmanson (Jun 29, 2010)

thank you  sorry my computer is laggging


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## Erdos (Jun 29, 2010)

Rinfiyks said:


> Erdos said:
> 
> 
> > I tried it. This is one of the convoluted algebraic problems where you can only do this by trial-and-error. However, this isn't to say that you try every single permutation. There are of course some shortcuts via logic and induction. I was going to explain my logic, but it took more than a page so I just gave up. Here's a link if you're still interested:
> ...


The logic is the same when doing this problem because it's the exact same concept.


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## aronpm (Jun 29, 2010)

PAUL?!


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## qqwref (Jun 29, 2010)

Well, if you want the denominators to combine nicely, 5 and 7 have to be on top, since they are primes and you don't have any other numbers with a prime factor of 5 or 7. The remaining denominators are all made up of 1, 2, 3, 4, 6, 8, 9 which means they are all of the form 2^A 3^B. So that saves us some time on our trial and error. We're still going to have to do a lot of it, though, unless someone has a clever idea for the next step. I'd rather find an elegant solution to this.


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## Erdos (Jun 29, 2010)

There is no elegant solution in problems like this. It can only be done by trial-and-error.


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## blah (Jun 29, 2010)

5 and 7 go on top.

Edit: qqwref


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## frogmanson (Jun 30, 2010)

so the only method of doing this by hand is trail and error?


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## cmhardw (Jun 30, 2010)

I solved the problem by hand, and though my method is trial and error, I think it at least uses a good way to find an educated guess of where to start. It took about an hour to find this solution.



Spoiler



I first rewrote:
\( 
\frac{a}{b\times c} + \frac{d}{e\times f} + \frac{g}{h\times i} = 1
\)

as:
\( 
\frac{a}{b\times c} + \frac{d}{e\times f} = 1 - \frac{g}{h\times i}
\)

and then as:
\( 
\frac{a\times e\times f + b\times c\times d}{b\times c\times e\times f} = \frac{h\times i - g}{h\times i}
\)

By a similar reasoning as Michael I figured I would try having 1, 5, and 7 in the numerators of the original equation. These are the only three numbers in the list that are relatively prime to every other number in the list, including each other. This would make it a bit messier to get common denominators later on so I decided to put them into the numerators.

Because no variables are decided yet, I arbitrarily chose a=1, d=5, g=7. Since the addition of the fractions is commutative it does not really matter which order I assign these numbers, but my choice will affect the assignments of the other 6 variables.

With my substitutions the above equation becomes:
\( 
\frac{e\times f + 5\times b\times c}{b\times c\times e\times f} = \frac{h\times i - 7}{h\times i}
\)

The first thing I noticed is that \( h\times i > 7 \) must be true. This is because the sum of 3 positive fractions must give an answer of 1, so taking 1 and subtracting one of the fractions must still be positive.

I wrote out a list for all possibilities of h and i using the remaining numbers and got:
\( \{2,4\} \) ; \( \{2,6\} \) ; \( \{2,8\} \) ; \( \{2,9\} \) ; \( \{3,4\} \) ; \( \{3,6\} \) ; \( \{3,8\} \) ; \( \{3,9\} \) ; \( \{4,6\} \) ; \( \{4,8\} \) ; \( \{4,9\} \) ; \( \{6,8\} \) ; \( \{6,9\} \) ; \( \{8,9\} \)

This gives me 14 choices to analyze, but the problem is that each of these 14 choices has effectively 6 possible ways to arrange the other numbers. Rather than do 84 calculations just to see if my choices for a, d, and g were potentially correct I used geometric mean to estimate it.

I imagined that I would replace each of b, c, e, f with the geometric mean of all 4 numbers. Let's call \( G \) the geometric mean of b, c, e, and f. I am now interested in looking at the equation:

\( 
\frac{G^2 + 5\times G^2}{G^4} \approx \frac{h\times i - 7}{h\times i} 
\)

\( 
\frac{6}{G^2} \approx \frac{h\times i - 7}{h\times i} 
\)

And now, depending on the choice of h and i there is only one possible value of \( G^2 \). This left me with 14 possible choices, and the choice of 2 and 4 for h and i was the one that gave the two sides of the equation the closest approximate values.

Letting h, i be 2 and 4 this reduces to:
\( 
\frac{e\times f + 5\times b\times c}{b\times c\times e\times f} = \frac{1}{8}
\)

And now there are effectively 6 choices to look at for how to arrange the groups of e, and f as well as the group b, and c. The one that works out exactly is b, c are the numbers 3 and 6. e and f are the numbers 8 and 9.

So my final answer is that:
\( a=1 \)
\( b,c = \{3,6\} \)
\( d=5 \)
\( e,f = \{8,9\} \)
\( g=7 \)
\( h,i = \{2,4\} \)

And all together it's:
\( 
\frac{1}{3\times 6} + \frac{5}{8\times 9} + \frac{7}{2\times 4} = 1
\)

which is true.


Chris


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## frogmanson (Jun 30, 2010)

WOW Chris thanks for your time! this will help me


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## Lucas Garron (Jun 30, 2010)

Rinfiyks said:


> Spoiler
> 
> 
> 
> \( \frac{1}{3\times 6} + \frac{5}{8\times 9} + \frac{7}{2\times 4} = 1 \)



I agree, and I'm also too lazy to do it by hand. 


```
Select[Permutations[Range[9]], ((Inner[Power, #, {1, -1, -1}, Times] & /@Partition[#, 3])//Total) == 1 &]
```



Spoiler



{{1, 3, 6, 5, 8, 9, 7, 2, 4}, {1, 3, 6, 5, 8, 9, 7, 4, 2},
{1, 3, 6, 5, 9, 8, 7, 2, 4}, {1, 3, 6, 5, 9, 8, 7, 4, 2},
{1, 3, 6, 7, 2, 4, 5, 8, 9}, {1, 3, 6, 7, 2, 4, 5, 9, 8},
{1, 3, 6, 7, 4, 2, 5, 8, 9}, {1, 3, 6, 7, 4, 2, 5, 9, 8},
{1, 6, 3, 5, 8, 9, 7, 2, 4}, {1, 6, 3, 5, 8, 9, 7, 4, 2},
{1, 6, 3, 5, 9, 8, 7, 2, 4}, {1, 6, 3, 5, 9, 8, 7, 4, 2},
{1, 6, 3, 7, 2, 4, 5, 8, 9}, {1, 6, 3, 7, 2, 4, 5, 9, 8},
{1, 6, 3, 7, 4, 2, 5, 8, 9}, {1, 6, 3, 7, 4, 2, 5, 9, 8},
{5, 8, 9, 1, 3, 6, 7, 2, 4}, {5, 8, 9, 1, 3, 6, 7, 4, 2},
{5, 8, 9, 1, 6, 3, 7, 2, 4}, {5, 8, 9, 1, 6, 3, 7, 4, 2},
{5, 8, 9, 7, 2, 4, 1, 3, 6}, {5, 8, 9, 7, 2, 4, 1, 6, 3},
{5, 8, 9, 7, 4, 2, 1, 3, 6}, {5, 8, 9, 7, 4, 2, 1, 6, 3},
{5, 9, 8, 1, 3, 6, 7, 2, 4}, {5, 9, 8, 1, 3, 6, 7, 4, 2},
{5, 9, 8, 1, 6, 3, 7, 2, 4}, {5, 9, 8, 1, 6, 3, 7, 4, 2},
{5, 9, 8, 7, 2, 4, 1, 3, 6}, {5, 9, 8, 7, 2, 4, 1, 6, 3},
{5, 9, 8, 7, 4, 2, 1, 3, 6}, {5, 9, 8, 7, 4, 2, 1, 6, 3},
{7, 2, 4, 1, 3, 6, 5, 8, 9}, {7, 2, 4, 1, 3, 6, 5, 9, 8},
{7, 2, 4, 1, 6, 3, 5, 8, 9}, {7, 2, 4, 1, 6, 3, 5, 9, 8},
{7, 2, 4, 5, 8, 9, 1, 3, 6}, {7, 2, 4, 5, 8, 9, 1, 6, 3},
{7, 2, 4, 5, 9, 8, 1, 3, 6}, {7, 2, 4, 5, 9, 8, 1, 6, 3},
{7, 4, 2, 1, 3, 6, 5, 8, 9}, {7, 4, 2, 1, 3, 6, 5, 9, 8},
{7, 4, 2, 1, 6, 3, 5, 8, 9}, {7, 4, 2, 1, 6, 3, 5, 9, 8},
{7, 4, 2, 5, 8, 9, 1, 3, 6}, {7, 4, 2, 5, 8, 9, 1, 6, 3},
{7, 4, 2, 5, 9, 8, 1, 3, 6}, {7, 4, 2, 5, 9, 8, 1, 6, 3}}


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## Erdos (Jul 1, 2010)

A lot of this logical "guess work" in the trial-and-error of that solution seems to be based off already knowing the answer. For example, why not try 2 and 3 as numerators, then go over their possible results, and find that they don't work in conjunction with other possible numerator values?


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## cmhardw (Jul 1, 2010)

Erdos said:


> A lot of this logical "guess work" in the trial-and-error of that solution seems to be based off already knowing the answer. For example, why not try 2 and 3 as numerators, then go over their possible results, and find that they don't work in conjunction with other possible numerator values?



I did spend quite a lot of time trying out numerators and seeing if I could get the rest of the numbers to make this work. I didn't use the geometric mean idea until I had used 1, 5, and 7 as the numerators. One thing I noticed when using the 5 and 7 as denominator factors is that the answer of adding up all the fractions tended to differ quite a bit from the 1 we needed. That's when I started thinking of how 5 and 7 were relatively prime to every other number in the list (1 is also) and I figured I would try to use those 3 numbers as the numerators.

Also, had 1, 5, and 7 been incorrect for the numerators (i.e. the problem been setup or phrased slightly differently, or the fractions added to a different answer) I probably would not have discovered this for quite some time. Had I been using 2 and 4 for h and i, and found all possibilities of the other numbers to fail then I probably would have tried other values for h and i for some time (and *not* have changed the values of the numerators until convincing myself that they were probably wrong).

So yes, with the method I used I basically just got lucky that the choices I was making meant that things worked out fairly well. I definitely concede that had I made different choices, or had the problem been slightly more difficult (with one of the numbers that is relatively prime to everything in the list being needed as a denominator factor), then my method may have taken quite a lot more time than it did.

That's why I admitted that my method is still classified as a "trial and error" method, and thus subject to the luck of which guesses I make at which time.

Chris


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## blah (Jul 1, 2010)

There's a completely logical way to show that 5 and 7 *must* be two of the numerators. At least that's what I did. I think that's what qqwref did too. (I still haven't seen the solution )


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## KrazyFK (Jul 1, 2010)

Trying to find a non-trial-and-error based solution, here is my work so far:

I can show that 5 and 7 are two of the numerators, and that the remaining numerator must be less than or equal to 3. BTW, during the LaTeX, i'm not putting in multiplication signs, so don't be confused by the lack of them. Also, I just previewed this post, and even with LaTeX it is a bit messy, but not impossible to follow if you are interested in the problem. Please try to understand the method.

First, suppose, for contradiction, that one of the numerators is not 5, so it is on one of the denominators. Then:

\( \frac{a}{5c} + \frac{d}{ef} + \frac{g}{hi} = 1 \)

\( a(efhi) + d(5chi) + g(5cef) = 5cefhi \)

But now the LHS is not divisible by 5 and the RHS is, contradiction. A similar method shows that 7 also must be a numerator. So the problem is reduced to:

\( \frac{5}{ab} + \frac{7}{cd} + \frac{e}{fg} = 1 \)

Multiplying out denominators:

\( 5(cdfg) + 7(abfg) + e(abcd) = abcdfg \)

Applying the AM-GM inequality: (Google it if you don't know what it is) 

\( abcdfg \)

\( = 5(cdfg) + 7(abfg) + e(abcd) \)

\( \ge 3\sqrt[3]{35e(abcdfg)^2} \)

Now, using the fact that \( e(abcdfg) = 10368 \) so: \( e = \frac{10368}{abcdfg} \) on our last line we get:

\( 3\sqrt[3]{35e(abcdfg)^2} \)

\( = 3\sqrt[3]{35(10368) \frac{(abcdfg)^2}{abcdfg}} \)

\( = 3\sqrt[3]{362880abcdfg} \)

So our final inequality is:

\( abcdfg \ge 3\sqrt[3]{362880abcdfg} \)

\( (abcdfg)^3 \ge 9797760abcdfg \)

\( (abcdfg)^2 \ge 9797760 \)

\( \frac{(10368)^2}{e^2} \ge 9797760 \)

\( e^2 \le \frac{107495424}{9797760} = \frac{384}{35} \)

\( e \le \sqrt{\frac{384}{35}} \)

\( e \le 3.31 \cdots \)

Since we are dealing with integers:

\( e \le 3 \)

I will try to continue from here, I haven't had another look at it yet. Perhaps someone else would like to carry on?


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