Question

Aflati numerele naturale a, b, c daca 2^a+4^b+8^c=16^100

Answer 1

 

$\displaystyle\bf\\2^a+4^b+8^c=16^{100}\\\\2^a+\Big(2^2\Big)^{b}+\Big(2^3\Big)^c=\Big(2^4\Big)^{100}\\\\2^a+2^{2b}+2^{3c}=2^{400}\\\\2^a+2^{2b}+2^{3c}=2^{398+2}\\\\2^a+2^{2b}+2^{3c}=2^{398}\times2^2\\\\2^a+2^{2b}+2^{3c}=4\times2^{398}\\\\2^a+2^{2b}+2^{3c}=2^{398}+2^{398}+\underbrace{\bf2^{398}+2^{398}}_{\^Ii ~adunam}\\\\2^a+2^{2b}+2^{3c}=2^{398}+2^{398}+2\times2^{398}\\\\2^a+2^{2b}+2^{3c}=2^{398}+2^{398}+2^{398+1}$

$\displaystyle\bf\\2^a+2^{2b}+2^{3c}=2^{398}+2^{398}+2^{399}\\\\2^a=2^{398}\implies \boxed{\bf a=398}\\\\2^{2b}=2^{398}\implies 2b=398\implies b=\frac{398}{2}\implies\boxed{\bf b=199}\\\\2^{3c}=2^{399}\implies 3c=399\implies c=\frac{399}{3}\implies\boxed{\bf c=133}$

$\displaystyle\bf\\\\\implies\boxed{\bf2^{398}+4^{199}+8^{133}=16^{100}}$