Any idea how to solve analytically? (1 Viewer)

Fizzy_Cyst · Nov 22, 2013

$\frac{A}{BC} + \frac{D}{EF} + \frac{G}{HI} = 1$

The denominators are 2 digit numbers, not the product of 2 single digit numbers.

Can only use the numbers 1-9 once.

Any idea how to solve mathematically? I know how to solve using a permutation solver, but want to know if there is any real process to solve mathematically without just guessing?

anomalousdecay · Nov 22, 2013

$\frac{A}{BC} + \frac{D}{EF}+\frac{G}{HI} = 1$

$\frac{AEFHI+DBCHI+GBCEF}{BCEFHI} = 1$

$AEFHI+DBCHI+GBCEF-BCEFHI=0$

$EFHI(A-BC)+BC(DHI+GEF) = 0$

We know that A does not equal BC and that DHI does not equal -GEF.

I don't know how to continue. I will see if I can figure something out in the mean time.

obliviousninja · Nov 22, 2013

anomalousdecay said:
$\frac{A}{BC} + \frac{D}{EF}+\frac{G}{HI} = 1$

$\frac{AEFHI+DBCHI+GBCEF}{BCEFHI} = 1$

$AEFHI+DBCHI+GBCEF-BCEFHI=0$

$EFHI(A-BC)+BC(DHI+GEF) = 0$

We know that A does not equal BC and that DHI does not equal -GEF.

I don't know how to continue. I will see if I can figure something out in the mean time.

It doesn't look like you can't really get much further from there.

anomalousdecay · Nov 22, 2013

obliviousninja said:
It doesn't look like you can't really get much further from there.

I was thinking maybe if you factorise different methods, get a few inequalities and solve simultaneously.

Any idea how to solve analytically? (1 Viewer)

Fizzy_Cyst

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anomalousdecay

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anomalousdecay

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