[tex]\displaystyle\bf\\IV)\\a)\\\\2^a+3^b=\overline{ab}\\\\\text{Nu avem formula pentru suma de puteri cu baze si exponenti diferiti}\\\text{pentru a ajunge la o ecuatie.}\\\text{Vom rezolva prin incercari.}\\\\a~si~b~sunt~cifre.\\2^a~~si~~3^b~~trebuie~sa~fie~numere~de~maxim~2~cifre.\\\implies a\leq6~~si~~b\leq4\\\\ Prin~cautari~am~ajuns~la:~~a=4~~si~~b=3\\\\\boxed{\bf2^4+3^3=16+27=43}[/tex]
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IV)
b)
[tex]\displaystyle\bf\\ Trebuie~sa~aratam~ca:\\\\\Big(\overline{abc}\Big)^{1340}<\Big(\overline{xyz}\Big)^{2010}~~~pt.~orice~numere~\overline{abc}~si~\overline{xyz}.\\\\Cel~mai~mic~numar~~\overline{xyz}~este~100.\\Cel~mai~mare~numar~~\overline{abc}~este~999.\\\\100^{2010}=\Big(10^2\Big)^{2010}=10^{2\times2010}=\boxed{\bf10^{4020}}\\\\999^{1340}~~nu~poate~fi~calculat,~dar~~999^{1340}<1000^{1340}\\\\1000^{1340}=\Big(10^3\Big)^{1340}=10^{3\times1340}=\boxed{\bf10^{4020}}[/tex]
[tex]\displaystyle\bf\\Observam~ca~1000^{1340}=100^{2010}=10^{4020}\\\\Dar~999^{1340}<1000^{1340}=100^{2010}\\\\\implies~\boxed{\bf999^{1340}<100^{2010}}\\\\\implies~\Big(\overline{abc}\Big)^{1340}<\Big(\overline{xyz}\Big)^{2010}~orcare~ar~fi~cifrele~~a,b,c~~si~~x,y,z[/tex]
