Is there a power of $2$ which begins, in base 10, with $999...$?

Question

Is there a power of $2$ which begins, in base 10, with $999...$?

I don't know how to do it, maybe I must apply Dirichlet principle but in which way?

Answer 1

The first power of $2$ that begins with $999$ is $2^{2621}$. The number itself is $789$ digits long. Done with the following python script:

x,y=1,0
while 1:
    if str(x)[:3]=="999":
        print(x,y)
        break
    x,y=x*2,y+1

Answer 2

To find many answers, you can start looking for $n,m$ such that $2^n \approx 10^m$, which is equivalent to $n\log2\approx m\log10$.

In other words, we look for rational approximations of $\frac{\log10}{\log2}$, in which case $n$ would be the numerator.

Using continued fractions, for

$$\frac{\log10}{\log2} \approx 3 + \frac1{3 + \frac1{9 + \frac1{2 + \frac1{2 + \frac1{4 + \frac16}}}}} = \frac{13301}{4004}$$

we find $n = 13301$, for which $2^n$ starts with 9999.

Note that since the sequence of continued fractions converges to the value, there is no limit on the number of leading 9's a power of 2 can have.

Answer 3

Let $a = 2 ^ n$ and $b = 10 ^ m$ and $a <b$.

Note: $1024> 1000$ that comes from $128> 125$.

The sequence of powers of 2 is: $$1,2,4,8,16,32,64,128,256, ...$$

Whose first digits are: $$1,2,4,8,1,3,6,1,2,5,1,2,4,8,1,3,6,1,2,5,1, ...$$

At the moment you do not see nines ...

And, in addition, it seems that the sequence is repeated every 10 positions with: $$1,2,4,8,1,3,6,1,2,5$$

But the reality is that that first one and the following digits grow slowly towards the next digit ...

And the 8 becomes 9 with $2 ^ {53} = 2 ^ 3 \cdot 2 ^ {50} = 9,007,199,254,740,992$

Everything mentioned above is related to the logarithm in base 10 of 2, which is: $0.3010299956639811952...$

And as you can see it is a real number (transcendent number) that is very close to $3/10$. Hence comes the "almost" repetition of ten digits mentioned above.

With these ingredients we can now propose the equation to solve to find a power number of two that starts with 999 when writing it in decimal:

$$2 ^ n = 10 ^ a \cdot b$$

$$b \in [0.999, 1); a, n \in N$$

Taking base 10 logarithms:

$$n * 0.3010299956639811952... = a + c$$

$$c \in [-0.00043451..., 0)$$

Which is easy to solve taking n as 100,000:

$$2 ^ {100,000} = 0.999002093 ... \cdot 10 ^ {30103} = 999.002093 ... \cdot 10 ^ {30100}$$