Problema (nível 4): Na barra abaixo, a soma dos números contidos em três casas con­se­cu­ti­vas é sempre igual a $36$. Qual é o valor de $x-y$? (image1)


Conteúdo referenciado:
[image1]
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Resolução: Solução 1: Como a soma dos números contidos em três casas consecutivas é igual a $36$, concluímos que de três em três casas os números repetem-se. Com efeito, se os números $a$, $b$, $c$ e $d$ estiverem, por esta ordem, em quatro casas consecutivas, então $a+b+c=36$ e $b+c+d=36$ o que implica $a=d$. Assim, (image2) $$\downarrow$$ (image3) e, consequentemente, $y=19$ e $19+x+11=36$, isto é, $x=6$. Portanto $x-y=-13$. $\,$$\,$ Sociedade Portuguesa de Matemática $\;$ XVII ONM - Categoria B (10$^{\makebox{\underline{\bf{\footnotesize o}}}}$, 11$^{\makebox{\underline{\bf{\footnotesize o}}}}$ e 12$^{\makebox{\underline{\bf{\footnotesize o}}}}$ anos) 1$^{\makebox{\underline{\bf{\footnotesize a}}}}$ Eliminatória $\;\;$ 11/11/98 (table1) $\,$$\,$ Solução 2: Denotemos por $a,b,c,d,e,f,g$ e $h$, os números a seguir indicados (image4) Como $$ \left\{ (image5) \right. $$ então $f=19$, $g=6$ e $h=11$. Assim, deduz-se sucessivamente $e=6$, $y = 19$, $d=11$, $c=6$, $b=19$, $a=11$ e $x=6$. Finalmente, $x-y=-13$.


Conteúdo referenciado:
[image2]
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[image3]
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[table1]
\begin{tabular}{|p{7,7cm}|} \hline SUGEST\~{O}ES para a resolu\c{c}\~{a}o dos problemas\\ \hline \end{tabular}
[image4]
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[image5]
\begin{array}{c} 11+f+g=36 \\ f+g+h=36 \\ g+h+19 = 36 \end{array}