The Search for an 'A' Whose Powers All Start with 9

[Albert1]

Prove it is possile to find a number $A$
Where $A\in N ,\,\,A>1$ and
the first digit of $A,A^2,A^3,...,A^{2015} $ is 9
and find one such number $A$

 

[kaliprasad]

Prove it is possile to find a number $A$
Where $A\in N ,\,\,A>1$ and
the first digit of $A,A^2,A^3,...,A^{2015} $ is 9
and find one such number $A$

Spoiler

if we choose $A = x * 10^n$ form some n then we have for any k ( x < 1)

$A^{k} = x^k * 10^{nk}$
for $A^k$ to start with 9 we choose

$1 > x^k >= .9$
if for k = 1 to 2015 above condition to be met then we must have

$x^2015 >= .9$ or $x >= .9^{\frac{1}{2015}}$
we can find the x from above and multiply by 10 repeatedly till we find all 9's before the decimal and one more digit and add 1.

for example say upto A and $A^2$
$x= .9^.5 = .948$
so we can take A = 95
$95 = 95, 95^2 = 9025$ both leading 9

one such number shall be 99995 because $.9^{\frac{1}{2015}} = .99994...$