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

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It is possible to find a natural number A greater than 1 such that the first digit of A and its powers A^2, A^3, ..., A^2015 is 9. The discussion explores mathematical properties and examples of numbers that meet this criterion. A specific example of such a number is identified, demonstrating that the condition can be satisfied. The conversation also delves into the logarithmic properties that underpin the leading digit of powers of numbers. Ultimately, the existence of such a number A is confirmed through mathematical reasoning.
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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 $A$
 
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Albert said:
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 $A$

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 $.9^{\frac{1}{2015}} = .99994...$
 
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