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Given $12a + 18b + 27c = 227$, how can we prove that $a, b, c$ can never be integers? I don't have many ideas. Can someone give me some ideas?

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3 Answers 3

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$12a+18b+27c=3(4a+6b+9c)$

$\implies \frac{12a+18b+27c}3=4a+6b+9c$ which is an integer if $a,b,c$ are all integers

$\implies 3$ divides $(12a+18b+27c)$ if $a,b,c$ are all integers

but $227\equiv2\pmod 3\not\equiv0\implies 3$ does not divide $227$

So, it leads to clear contradiction.

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  • $\begingroup$ @AndréNicolas, thanks for your help to make things perfect. Ya, $3\not\mid a\implies a\equiv\pm 1\pmod 3$ $\endgroup$ Commented Mar 30, 2013 at 15:40
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$3\;$ divides all the summands but not the rhs $= 227$ . Hence no integer solution.

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    $\begingroup$ +1 As simple, short and elegant as that... $\endgroup$
    – DonAntonio
    Commented Mar 30, 2013 at 15:50
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Generally, if $\rm\:i,j,k,n\in \Bbb Z\:$ then $\rm\: i\,a + j\,b + k\,c = n\:$ has a solution $\rm\:a,b,c\in \Bbb Z\iff gcd(a,b,c)\mid n.\:$ The necessity $\rm\:(\Rightarrow)\:$ is clear, and the sufficiency $(\Leftarrow)$ follows from Bezout, i.e. integers of form $\rm\:i\,a + j\,b + k\,c\:$ are closed under subtraction, so closed under remainder, so closed under gcd.

Your example fails the criterion: $\rm\: (12,18,27) = 3(4,6,9)=3,\:$ but $\rm\:mod\ 3\!:\ 227\equiv 2\!+\!2\!+\!7\equiv 2\not\equiv 0.$

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