x ; y ; z → simple numbers , y<x<z , y+x+z=68 , y ∙ x + x ∙ z + z ∙ y = 1121 , y ∙ x = ?


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Question Number 125743 by MathSh last updated on 13/Dec/20

$x; y; z \to s i m p l e n u m b e r s,$ $y < x < z,$ $y + x + z = 68,$ $y \cdot x + x \cdot z + z \cdot y = 1121,$ $y \cdot x = ?$
Commented byMJS_new last updated on 13/Dec/20

$define ♮ simple number ε$
Commented byMathSh last updated on 13/Dec/20

$S o l u t i o n p l e a s e s i r$
Commented byMJS_new last updated on 13/Dec/20

$what is a ♮ simple number ε ? we cannot solve$ $without knowing this$
Commented byMathSh last updated on 13/Dec/20

$D i v i s i b l e b y i t s e l f a n d 1 S i r$
Commented byHer_Majesty last updated on 13/Dec/20

$y = 2 b e c a u s e t h e s u m o f 2 p r i m e s \neq 2 c a n o n l y$ $b e e v e n$ $x + z = 66 \Rightarrow z = 66 - x$ $2 x + x z + 2 z = 1121 \Rightarrow - x^{2} + 66 x + 132 = 1121$ $\Rightarrow x = 23 \land z = 43$
Commented byMJS_new last updated on 13/Dec/20

$ok I was not sure if you mean p r i m e s or maybe$ $w h o l e n u m b e r s .$ $if x, y, z are prime numbers then I guess they$ $are > 0 . then if their sum is even \Rightarrow one of$ $them must equal 2 . y < x < z \Rightarrow y = 2$ $now we have$ $(1) 2 + x + z = 68 \Leftrightarrow x + z = 66$ $(2) 2 x + x z + 2 z = 1121$ $2 (x + z) + x z = 1121$ $x z = 989 = 23 \times 43$ $\Rightarrow x = 23 y = 2 z = 43$ $\Rightarrow x y = 46$
Commented byMathSh last updated on 13/Dec/20

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