# Powers of $$2$$'s After Another

 $$n$$ $$\hspace{10mm} 2^n$$ Concatenation of the powered numbers Divisibility checked 0 $$\hspace{5mm} 2^0=1$$ $$\hspace{25mm} 1$$ $$\hspace{8mm} 1\, \big|\, 1$$ 1 $$\hspace{5mm} 2^1=2$$ $$\hspace{25mm} 12$$ $$\hspace{8mm} 2\, \big|\, 12$$ 2 $$\hspace{5mm} 2^2=4$$ $$\hspace{25mm} 124$$ $$\hspace{8mm} 4\, \big|\, 124$$ 3 $$\hspace{5mm} 2^3=8$$ $$\hspace{25mm} 1248$$ $$\hspace{8mm} 8\, \big|\, 1248$$ 4 $$\hspace{5mm} 2^4=16$$ $$\hspace{25mm} 124816$$ $$\hspace{8mm} 16\, \big|\, 124816$$

As we get greater and greater numbers in column 3 of the table by concatenation (i.e. 12481632, 1248163264, ...) for $$n>4,$$ will the divisibility in the last column still hold?

×