# Show that there exists values of $x$ whose first digit is not $1$

Let $x$ be a positive integer. Show that if $x, x^2, x^3, \dots, x^n$ all start with the same digit, and $n$ is a positive integer, there exist values of $x$ whose first digit is not $1$.

I presume we are working in base $10$ for this question. I am wondering how to get the first digit out of the powers of $x$ since we will need to know the greatest power of ten dividing each power of $x$.

• What does "there exists values of $x$ whose first digit is not $1$" mean? $2$ doesn't begin with $1$....does that prove your claim? – lulu May 26 '16 at 15:48
• @lulu No, it's not false. It says "there exists." – Puzzled417 May 26 '16 at 15:49
• Then I don't understand the question. Of course there exist values of $x$ whose first digit is not $1$. – lulu May 26 '16 at 15:50
• @lulu They mean "show that there exists values of $x$ that satisfy the given conditions". – Noble Mushtak May 26 '16 at 15:54
• @lulu - The question asks for a way to prove that there are solutions which do NOT start with 1 beginning from the two statements above – K. Hoffmann May 26 '16 at 15:55

Hint:

Look at powers of $99$. $99^n$ for any positive integer $1 \leq n \leq 10$ has a first digit of $9$.

Look at powers of $999$. $999^n$ for any positive integer $1 \leq n \leq 105$ has a first digit of $9$.

Now, if you're given a value of $n$, what kind of values of $x$ would you look for to satisfy the given conditions?

Here's another way to think of it: $$.99^k > .9 \ \text{for all} \ 1 \leq k \leq 10$$ $$.999^k > .9 \ \text{for all} \ 1 \leq k \leq 105$$

You want to find the following for some $n$: $$y^k > .9 \ \text{for all} \ 1 \leq k \leq n$$

If you substitute $n$ into that inequality, you get $y^n > .9$. How can you solve that inequality for $y$ and once you do, how can you convert the decimal $y$ back to a suitable integer $x$?

• The other hint I would give is to apply the binomial theorem. – user7530 May 26 '16 at 16:06
• @NobleMushtak How do you determine the range of $n$? – Puzzled417 May 26 '16 at 16:36
• @Puzzled417 For any $n$, you can always find a positive integer $x$ such that $x, x^2, x^3, ..., x^n$ all begin with the digit $9$. I've added another hint above that might help you. – Noble Mushtak May 26 '16 at 20:03
• @NobleMushtak We get $y > 10^{\frac{\log(0.9)}{n}}$ right? – Puzzled417 May 26 '16 at 21:07
• @Puzzled417 Yes, that's right. Now, if you choose $10^{\frac{\log 0.9}{n}} < y < 1$, you need to convert that to an integer $x$ by multiplying $y$ by a power of $10$ and then rounding to the nearest integer so that the first digit is $9$ (i.e. make sure it does not round up to $10$). We know that such an integer exists because $0.9 < y < 1$, so the first digit in $y$ is $9$, but if we multiply by $10$ enough, there should be some non-$9$ digit so that we can round to the nearest integer without rounding to the next power of $10$. – Noble Mushtak May 26 '16 at 21:14