Does $\exists n \in \mathbb{Z}$ such that $2^n$ can start with $9786543120$?

Question

I have tried $2^n=\displaystyle\sum_{k=0}^{n}\binom{n}{k}$ but could not reach further. Thank you.

Do you mean the leftmost or rightmost numbers in the decimal expansion are 9786543120? — mathreadler, Mar 07 '17 at 08:47
Probably leftmost since he says "start with...", plus, powers of $2$ can't end in a $0$, since then it's divisible by $5$ — , Mar 07 '17 at 08:53
What makes you assume "starting with" is always from left? Yes exactly, that would be easier to show then. — mathreadler, Mar 07 '17 at 08:58
I assume that because English, the primary language of this website, is being read from left to right. I simply generalize that to numbers being read from left to right. — , Mar 07 '17 at 08:59
Why $9786543120$ and not $9876543210$ :) ? I understand that my remark has no importance ! — Jean Marie, Mar 07 '17 at 09:31
As an aside, one can prove that there exists a power of two which starts with any desired string of digits. — JMoravitz, Mar 07 '17 at 09:44
There is even an OEIS entry for that: http://oeis.org/A018856 — Ivan Neretin, Mar 07 '17 at 11:12
@JeanMarie : I think we all wondered that. But someone needs to say it. ;) — mathreadler, Mar 08 '17 at 18:40

score 2 · Answer 1 · answered Mar 07 '17 at 09:43

2

Hint: $\log_{10} 2$ is irrational

Can you use this to prove that there exists an $m$ and $n$ such that

$$9786543120\cdot 10^n < 2^m < 9786543121 \cdot 10^n$$

take the log base $10$ of everything. Can you show there is some $m$ and $n$ such that $$\log_{10} 9786543120 + n < m\log_{10} 2 < \log_{10}9786543121+n$$

answered Mar 07 '17 at 09:43

JMoravitz

79,518

Dear sir, can you please alaborate? – Chandramauli Chakraborty Mar 07 '17 at 11:35
I am only a 10th grader student. – Chandramauli Chakraborty Mar 07 '17 at 16:38

score 2 · Accepted Answer · answered Mar 07 '17 at 09:44

2

Yes.

We want $2^n$ to lie between $9786543120 \times 10^x$ and $9786543121 \times 10^x$. Taking logarithms, this is equivalent to finding an integer $n$ that lies between $x \frac{log 10}{\log 2} + \frac{\log 9786543120}{\log 2}$ and $x \frac{log 10}{\log 2} + \frac{\log 9786543121}{\log 2}$ for an integer $x$.

This can be achieved if we can find an $x$ such that the fractional part of $x \frac{log 10}{\log 2}$ is less than $\frac{\log 9786543121}{\log 2} - \frac{\log 9786543120}{\log 2}$. But this is always possible because $\frac{log 10}{\log 2}$ is irrational.

answered Mar 07 '17 at 09:44

Raziman T V

1,230

Dear sir, can you please elaborate the last part? – Chandramauli Chakraborty Mar 07 '17 at 14:51
Which one, proving that $\frac{\log 10}{\log 2}$ is irrational, or proving that it means we can find the required $x$? – Raziman T V Mar 07 '17 at 14:53
The part required x. – Chandramauli Chakraborty Mar 07 '17 at 14:55
1

It follows from https://en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem – Raziman T V Mar 07 '17 at 14:57
But how to apply it. You see I am only a 10th grader student. – Chandramauli Chakraborty Mar 07 '17 at 15:19
2

Choose N such that $1/N < \log 9786543121/\log 2−\log 9786543120/\log 2$. Then you can apply the theorem directly. Try to think about it a bit. – Raziman T V Mar 07 '17 at 15:22
Why you have written about the fractional part? – Chandramauli Chakraborty Mar 07 '17 at 15:55

Does $\exists n \in \mathbb{Z}$ such that $2^n$ can start with $9786543120$?

2 Answers2