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Question Number 151911 by mathdanisur last updated on 24/Aug/21

Determine the digit a and prime numbers x;y;z such that x<y, z<1000 and x + y^(2a) = z

$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{digit}\:\boldsymbol{\mathrm{a}}\:\mathrm{and}\:\mathrm{prime} \\ $$ $$\mathrm{numbers}\:\boldsymbol{\mathrm{x}};\boldsymbol{\mathrm{y}};\boldsymbol{\mathrm{z}}\:\mathrm{such}\:\mathrm{that}\:\boldsymbol{\mathrm{x}}<\boldsymbol{\mathrm{y}},\:\boldsymbol{\mathrm{z}}<\mathrm{1000} \\ $$ $$\mathrm{and}\:\:\mathrm{x}\:+\:\mathrm{y}^{\mathrm{2}\boldsymbol{\mathrm{a}}} \:=\:\mathrm{z} \\ $$

Commented byRasheed.Sindhi last updated on 24/Aug/21

2a means 20,21,22,...,29? Or 2a=2×a? ′digit a′ means decimal digit?

$$\mathrm{2}\boldsymbol{\mathrm{a}}\:{means}\:\mathrm{20},\mathrm{21},\mathrm{22},...,\mathrm{29}?\:{Or}\:\mathrm{2}\boldsymbol{\mathrm{a}}=\mathrm{2}×\boldsymbol{\mathrm{a}}? \\ $$ $$'\mathrm{digit}\:\boldsymbol{\mathrm{a}}'\:{means}\:{decimal}\:{digit}? \\ $$

Commented bymathdanisur last updated on 24/Aug/21

Ser, 2a=2xa a=0,1,2,3,4,5,6,7,8,9 2a=2∙2 ...

$$\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{2a}=\mathrm{2xa}\: \\ $$ $$\mathrm{a}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9} \\ $$ $$\mathrm{2a}=\mathrm{2}\centerdot\mathrm{2}\:... \\ $$

Commented bymathdanisur last updated on 24/Aug/21

Solution dear Ser please

$$\mathrm{Solution}\:\mathrm{dear}\:\mathrm{Ser}\:\mathrm{please} \\ $$

Commented byRasheed.Sindhi last updated on 24/Aug/21

x + y^(2a) = z; x<y, z<1000;x,y,z∈P ^• a=0: x+1=z⇒x=2,z=3 ∀y(<1000)∈P (Two primes having difference 1 are only 2 & 3) ^• a=1: 2+3^2 =11

$$\mathrm{x}\:+\:\mathrm{y}^{\mathrm{2}\boldsymbol{\mathrm{a}}} \:=\:\mathrm{z};\:\boldsymbol{\mathrm{x}}<\boldsymbol{\mathrm{y}},\:\boldsymbol{\mathrm{z}}<\mathrm{1000};\mathrm{x},\mathrm{y},\mathrm{z}\in\mathbb{P} \\ $$ $$\:^{\bullet} \mathrm{a}=\mathrm{0}:\:\mathrm{x}+\mathrm{1}=\mathrm{z}\Rightarrow\mathrm{x}=\mathrm{2},\mathrm{z}=\mathrm{3}\:\:\forall\mathrm{y}\left(<\mathrm{1000}\right)\in\mathbb{P}\: \\ $$ $$\left(\mathrm{Two}\:\mathrm{primes}\:\mathrm{having}\:\mathrm{difference}\:\mathrm{1}\:\mathrm{are}\right. \\ $$ $$\left.\mathrm{only}\:\mathrm{2}\:\&\:\mathrm{3}\right) \\ $$ $$\:^{\bullet} \mathrm{a}=\mathrm{1}:\:\mathrm{2}+\mathrm{3}^{\mathrm{2}} =\mathrm{11} \\ $$

Commented bymathdanisur last updated on 24/Aug/21

(x,y,z,a)∈{(0,2,p,3)(2,3,11,1)(2,3,83,2) p=prime

$$\left({x},{y},{z},{a}\right)\in\left\{\left(\mathrm{0},\mathrm{2},{p},\mathrm{3}\right)\left(\mathrm{2},\mathrm{3},\mathrm{11},\mathrm{1}\right)\left(\mathrm{2},\mathrm{3},\mathrm{83},\mathrm{2}\right)\right. \\ $$ $${p}={prime} \\ $$

Commented byMJS_new last updated on 24/Aug/21

0 is not a prime number

$$\mathrm{0}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number} \\ $$

Commented bymathdanisur last updated on 24/Aug/21

in the statement a is digit i.e is in the set {0,2,...,9}

$$\mathrm{in}\:\mathrm{the}\:\mathrm{statement}\:\mathrm{a}\:\mathrm{is}\:\mathrm{digit} \\ $$ $$\mathrm{i}.\mathrm{e}\:\mathrm{is}\:\mathrm{in}\:\mathrm{the}\:\mathrm{set}\:\left\{\mathrm{0},\mathrm{2},...,\mathrm{9}\right\} \\ $$

Commented byMJS_new last updated on 24/Aug/21

(x,y,z,a)∈{(0,2,p,3)(2,3,11,1)(2,3,83,2)

$$\left({x},{y},{z},{a}\right)\in\left\{\left(\mathrm{0},\mathrm{2},{p},\mathrm{3}\right)\left(\mathrm{2},\mathrm{3},\mathrm{11},\mathrm{1}\right)\left(\mathrm{2},\mathrm{3},\mathrm{83},\mathrm{2}\right)\right. \\ $$ $$ \\ $$

Commented byMJS_new last updated on 24/Aug/21

you don′t seem to understand anything.

$$\mathrm{you}\:\mathrm{don}'\mathrm{t}\:\mathrm{seem}\:\mathrm{to}\:\mathrm{understand}\:\mathrm{anything}. \\ $$

Commented byRasheed.Sindhi last updated on 24/Aug/21

(x,y,z,a)∈{(2,p,3,0)(2,3,11,1)(2,3,83,2)}

$$\left({x},{y},{z},{a}\right)\in\left\{\left(\mathrm{2},{p},\mathrm{3},\mathrm{0}\right)\left(\mathrm{2},\mathrm{3},\mathrm{11},\mathrm{1}\right)\left(\mathrm{2},\mathrm{3},\mathrm{83},\mathrm{2}\right)\right\} \\ $$

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