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Question-209062




Question Number 209062 by hardmath last updated on 01/Jul/24
Answered by som(math1967) last updated on 01/Jul/24
 x^y^z  =1   x,y,z ∈N ⇒1^y^z  =1 ⇒x=1   y^z^x  =125⇒y^z =5^3 ⇒y=5⇒z=3   z^y^x  =3^5^1  =243   9x+10y−18z=9+10×5−18×3  =5
xyz=1x,y,zN1yz=1x=1yzx=125yz=53y=5z=3zyx=351=2439x+10y18z=9+10×518×3=5

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