Powers of \(2\)'s After Another

\(n\)\(\hspace{10mm} 2^n\)Concatenation of the powered numbersDivisibility 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?


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