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x-y-z-simple-numbers-y-lt-x-lt-z-y-x-z-68-y-x-x-z-z-y-1121-y-x-




Question Number 125743 by MathSh last updated on 13/Dec/20
x ; y ; z → simple numbers ,  y<x<z ,  y+x+z=68 ,  y ∙ x + x ∙ z + z ∙ y = 1121 ,  y ∙ x = ?
$$\boldsymbol{{x}}\:;\:\boldsymbol{{y}}\:;\:\boldsymbol{{z}}\:\rightarrow\:\boldsymbol{{simple}}\:\boldsymbol{{numbers}}\:, \\ $$$$\boldsymbol{{y}}<\boldsymbol{{x}}<\boldsymbol{{z}}\:, \\ $$$$\boldsymbol{{y}}+\boldsymbol{{x}}+\boldsymbol{{z}}=\mathrm{68}\:, \\ $$$$\boldsymbol{{y}}\:\centerdot\:\boldsymbol{{x}}\:+\:\boldsymbol{{x}}\:\centerdot\:\boldsymbol{{z}}\:+\:\boldsymbol{{z}}\:\centerdot\:\boldsymbol{{y}}\:=\:\mathrm{1121}\:, \\ $$$$\boldsymbol{{y}}\:\centerdot\:\boldsymbol{{x}}\:=\:? \\ $$
Commented by MJS_new last updated on 13/Dec/20
define “simple number”
$$\mathrm{define}\:“\mathrm{simple}\:\mathrm{number}'' \\ $$
Commented by MathSh last updated on 13/Dec/20
Solution please sir
$${Solution}\:{please}\:{sir} \\ $$
Commented by MJS_new last updated on 13/Dec/20
what is a “simple number”? we cannot solve  without knowing this
$$\mathrm{what}\:\mathrm{is}\:\mathrm{a}\:“\mathrm{simple}\:\mathrm{number}''?\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{solve} \\ $$$$\mathrm{without}\:\mathrm{knowing}\:\mathrm{this} \\ $$
Commented by MathSh last updated on 13/Dec/20
Divisible by itself and 1 Sir
$${Divisible}\:{by}\:{itself}\:{and}\:\mathrm{1}\:{Sir} \\ $$
Commented by Her_Majesty last updated on 13/Dec/20
y=2 because the sum of 2 primes ≠2 can only  be even  x+z=66 ⇒ z=66−x  2x+xz+2z=1121 ⇒ −x^2 +66x+132=1121  ⇒ x=23∧z=43
$${y}=\mathrm{2}\:{because}\:{the}\:{sum}\:{of}\:\mathrm{2}\:{primes}\:\neq\mathrm{2}\:{can}\:{only} \\ $$$${be}\:{even} \\ $$$${x}+{z}=\mathrm{66}\:\Rightarrow\:{z}=\mathrm{66}−{x} \\ $$$$\mathrm{2}{x}+{xz}+\mathrm{2}{z}=\mathrm{1121}\:\Rightarrow\:−{x}^{\mathrm{2}} +\mathrm{66}{x}+\mathrm{132}=\mathrm{1121} \\ $$$$\Rightarrow\:{x}=\mathrm{23}\wedge{z}=\mathrm{43} \\ $$
Commented by MJS_new last updated on 13/Dec/20
ok I was not sure if you mean primes or maybe  whole numbers.  if x, y, z are prime numbers then I guess they  are >0. then if their sum is even ⇒ one of  them must equal 2. y<x<z ⇒ y=2  now we have  (1) 2+x+z=68 ⇔ x+z=66  (2) 2x+xz+2z=1121    2(x+z)+xz=1121  xz=989=23×43  ⇒ x=23 y=2 z=43  ⇒ xy=46
$$\mathrm{ok}\:\mathrm{I}\:\mathrm{was}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:\mathrm{you}\:\mathrm{mean}\:{primes}\:\mathrm{or}\:\mathrm{maybe} \\ $$$${whole}\:{numbers}. \\ $$$$\mathrm{if}\:{x},\:{y},\:{z}\:\mathrm{are}\:\mathrm{prime}\:\mathrm{numbers}\:\mathrm{then}\:\mathrm{I}\:\mathrm{guess}\:\mathrm{they} \\ $$$$\mathrm{are}\:>\mathrm{0}.\:\mathrm{then}\:\mathrm{if}\:\mathrm{their}\:\mathrm{sum}\:\mathrm{is}\:\mathrm{even}\:\Rightarrow\:\mathrm{one}\:\mathrm{of} \\ $$$$\mathrm{them}\:\mathrm{must}\:\mathrm{equal}\:\mathrm{2}.\:{y}<{x}<{z}\:\Rightarrow\:{y}=\mathrm{2} \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{have} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{2}+{x}+{z}=\mathrm{68}\:\Leftrightarrow\:{x}+{z}=\mathrm{66} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{2}{x}+{xz}+\mathrm{2}{z}=\mathrm{1121} \\ $$$$ \\ $$$$\mathrm{2}\left({x}+{z}\right)+{xz}=\mathrm{1121} \\ $$$${xz}=\mathrm{989}=\mathrm{23}×\mathrm{43} \\ $$$$\Rightarrow\:{x}=\mathrm{23}\:{y}=\mathrm{2}\:{z}=\mathrm{43} \\ $$$$\Rightarrow\:{xy}=\mathrm{46} \\ $$
Commented by MathSh last updated on 13/Dec/20
Thank you very much Sir
$${Thank}\:{you}\:{very}\:{much}\:{Sir} \\ $$

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