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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 byMJS_new last updated on 13/Dec/20

define ♮simple numberε

$$\mathrm{define}\:\natural\mathrm{simple}\:\mathrm{number}\varepsilon \\ $$

Commented byMathSh last updated on 13/Dec/20

Solution please sir

$${Solution}\:{please}\:{sir} \\ $$

Commented byMJS_new last updated on 13/Dec/20

what is a ♮simple numberε? we cannot solve without knowing this

$$\mathrm{what}\:\mathrm{is}\:\mathrm{a}\:\natural\mathrm{simple}\:\mathrm{number}\varepsilon?\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{solve} \\ $$ $$\mathrm{without}\:\mathrm{knowing}\:\mathrm{this} \\ $$

Commented byMathSh last updated on 13/Dec/20

Divisible by itself and 1 Sir

$${Divisible}\:{by}\:{itself}\:{and}\:\mathrm{1}\:{Sir} \\ $$

Commented byHer_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 byMJS_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 byMathSh last updated on 13/Dec/20

Thank you very much Sir

$${Thank}\:{you}\:{very}\:{much}\:{Sir} \\ $$

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