100 of 100: One Hundred and One

$101^n = 1.01^n \times 100^n$

So the smallest integer n for which $101^n$ starts with a 2 can be found from:

$1.01^n >= 2$

$n \times \ln(1.01) >= \ln(2)$

$n >= \frac {\ln(2)}{\ln(1.01)} = 69.66$

The smallest power of 101 that doesn't start with a 1 is 70.

Examine the first few powers of 101. $101^1=100^1+1^1$ $101^2=(100+1)^2=100^2+2 \times 100 +1^2=10201$ $101^3=(100+1)^3=100^3+3 \times 100^2 +3 \times 100 +1^3=1030301$ As you can see, the leading digit comes from $100^1$ , $100^2$ , $100^3$ and so on. Now, we'll look at the binomial expansion of $100^n$ . $101^n=(100+1)^n$ $101^n=100^n+\color{#D61F06}{\binom{n}{1} 100^{n-1} +\binom{n}{2} 100^{n-2} +\binom{n}{3} 100^{n-3} + \ldots +\binom{n}{n} 100^{n-n}}$ $101^n=100^n+\color{#D61F06}{\xi}$ So, the leading digit of the first few powers of $101$ is ${\color{#3D99F6}1}$ because $100^n$ is dominant in that region. Hence, we deduce the leading digit will be changed when ${\color{#D61F06}\xi}$ becomes equal to or greater than $100^n$ . (For example, we need to add 10000 with 10000 to get a number whose leading digit is not 1.) Therefore, the necessary condition is ${\color{#D61F06}\xi \geq 100^n}$ Now, $101^n=100^n+{\color{#D61F06}\xi} \geq 100^n+{\color{#D61F06}100^n}$ $101^n \geq 2 \times 100^n$ $\bigg(\frac{101}{100}\bigg)^n \geq 2$ Taking natural logarithm, (As it is a monotonically increasing function, inequality will still hold true.) $n (\ln 101 -\ln100) \geq \ln 2$ $n \geq \frac{\ln 2}{\ln 101 -\ln 100} \approx 69.66$ If we force $n$ to be an integer, then $n \geq 70$ So, the smallest power of $101$ that doesn't start with a $\color{#3D99F6}1$ is $70$ .

Not pure maths. But imagine a bank paying 1% interest on a £1 deposit (yes I'm English!) forever. After n years you'd have £1*1.01^n. When you get to £2 in the account, that's when 101^n starts with 2!

The function $a^n$ is monotonically increasing in $n$ for all $a>1$ . If we take $a=1.01$ instead of $101$ , then that does not affect the leading digit (the number is just scaled down by a factor $100^n$ ). The function $1.01^n$ will exceed $2$ for some sufficiently large $n$ . The first time it does so, it will still be less than $3$ (because if $1.01^n \ge 3$ , then $1.01^{n-1} \ge 2$ , which contradicts the assumption that it $1.01^{n-1}<1$ ).

Hence, the leading digit will differ from 1 at some point. In fact, it is easy to see that every digit will be leading at some point, cycling through $1, 2, \ldots, 9, 1, 2, \ldots$ .

I used a way that doesn't need $ln$ or whatever that is quite complicated (to me). But I saw a pattern to it.

To make things simpler, we can first observe the pattern of the first few powers of 11:

$\begin{aligned} 11^0 &=& 1 \\ 11^1 &=& 11 \\ 11^2 &=& 121 \\ 11^3 &=& 1331 \\ 11^4 &=& 14641 \\ 11^5 &=& 165051 \\ \end{aligned}$

If you observe carefully, you can find a pattern of Pascal's Triangle in it.

Note: for $11^5$ , you can find the answer by adding $1 \times 10^5 + 5 \times 10^4 + 10 \times 10^3 + 10 \times 10^2 + 5 \times 10^1 + 1 \times 10^0$

Now, observe the pattern for the first few powers of 101 and you will find the pattern too.

$\begin{aligned} 101^1 &=& {\color{#D61F06}{1}}01 \\ 101^2 &=& {\color{#D61F06}{1}}0201 \\ 101^3 &=& {\color{#D61F06}{1}}030301 \\ 101^4 &=& {\color{#D61F06}{1}}04060401 \\ &\vdots& \end{aligned}$

Now, obviously, sometime soon (we'll get to that later) the second number of the Pascal's Triangle will get to 100.

So the answer is $\textbf{NO}$

So, when? (Ah, we're here)

I can't think of a way to find it without using $ln$ but we can find an approx. value for it.

Let's first observe the first 5 digits (Why 5 and not 3? Well it won't be so accurate. You'll get to an answer of 100) and ignore the other digits. You will find this pattern:

$First\ five\ digits\ of\ 100^{n} = 10000 + 100n + \frac{n(n+1)}{2}$

We must now solve $\begin{aligned} 10000 + 100n + \frac{n(n+1)}{2} \geqslant 20000 \\ -10000 + 100n + \frac{n(n+1)}{2} \geqslant 0 \\ n^2 +201n - 20000 \geqslant 0 \\ \end{aligned}$ We then find the value of $n$ using the Quadratic Formula $x = \frac { - b \pm \sqrt{ b^2 - 4ac } } { 2a}$ and we find

$\begin{aligned} n \geqslant \frac{1}{2} ( \sqrt{120421} -201) \\ n \leqslant \frac{1}{2} ( - \sqrt{120421} -201) \\ \end{aligned}$

Since we want the positive value, we choose $n \geqslant \frac{1}{2} ( \sqrt{120421} -201)$ which is approx. $n \geq 72.9942$ . The correct solution is actually 70. We are 3 apart from the answer

I saw that the 3rd number grows by 1 for each power, therefore i thought that by 101^100 the first number had to become a 2, but then i realized i goes quicker than that.

My student and I discovered a beautiful method for this last month while on a pattern finding expedition! - Build a triangle with 101 at the top in the middle - With each new row add a 1 on either end - To find the numbers in the middle, take the sum of the two numbers above and diagonal. Like so:

• As digits get carried it still works, it just gets sticky. Take a look at the 5th row for how this works:

• And then go on as long as you want!

• Since the rounding will keep spilling forward, the first digit will eventually be rounded up to 2 and so on and so forth. Also, the last 2 digits will always be 01. Great problem to end on!

The way I see it is the following: $(1+1/100)^{100}$ is approximately $e$ . So $101^{100}$ is about $e \times 100^{100}$ who has a $2$ as a first digit

One way of seeing this is by considering logarithms. A number $n$ starts in a "1" if $\{0.00 \leq \log_{10} n\} < 0.301.$ Here, $\{x\}$ stands for the fractional part of $x$ .

Now $\log_{10} 101^n = n\log_{10} 101 = 2.00 + \epsilon$ where $\epsilon < 1$ . Thus we may write $\log_{10} 101^n = 2n + n\epsilon,$ and as long as $n < 1/\epsilon$ , $\{\log_{10} 101^n\} = n\epsilon.$ Since $\epsilon$ is small, we can find a value of $n$ such that $0.301 < n\epsilon < 1.00$ , which means that $101^n$ starts in a digit not equal to 1.

Bonus: The minimum value of $n$ where this happens is the first integer greater than $\frac{\log 2}{\epsilon} = \frac{\log 2}{\log 1.01} = 69.7...$ Thus, $101^{70}$ does not start in 1; its first few digits are $2\,006\,763\,368\,395\,385\,\dots$

101^n = 100^n (1 + 0.01)^n If n is big then (1 + 0.01)^n = 1 + 0.01 x n And obviously, we can find value n to make 1 + 0.01 x n >= 2

199*101 = 20099. So, any number of the form 199.......... which is a power of 101 is likely to be a solution.

By trial and error, we find (101)^70= 2........141 digits. and (101)^69=198.......

So 70 seems to be the smallest power of 101 for which the leading digit is not a 1

We can also apply Knorecker's Approximation Theorem!!!

Look at the sequence in the problem. The third digit of each line is: 1, 2, 3, 4. This appears to increase by 1 on each iteration. Is this the case?

Think of it this way: Multiplying a number x by 101 is equivalent to adding x to 100x. 100x is just x shifted left two places. The third digit of 100x will line up with the first digit of x. As long as the first digit is 1, this will increase the third digit by 1 on each iteration. On the tenth iteration, it'll spill over into the second digit, and by the 100th iterations, it'll hit the first digit, increasing it to 2.

In fact, it'll happen sooner than that, because lower digits will wind up affecting it too. But this simple reasoning shows us that the first digit will increase eventually, and gives us an upper bound for when.

The easiest way to see it is by looking at powers of 1.01. Each subsequent one adds more than 0.01 so 1.01^100 is more than 2. In fact, that's an approximation for e.

In my head I estimated the smallest such power to be between 60 and 70. The calculator confirmed it was 70.

Answered all 100 questions, what now?

101^n = 100^n * 1.01^n Multiplying by 100 doesn't change the first digit so first digit of 101^n = first digit of 1.01^n, which is trivially not always 1.

Since we want to see left most digit of $101^n, n \in \mathbb{N}$ , we consider $101^n - 100^n$ . This value should at least $100^n$ . So $101^n - 100^n \ge 100^n$ With a little logarithm magic, we have $n \ge 69.66$ , so $n=70$ is the smallest value of $n$ that make $101^n$ doesn't start with 1.

Consider the function f(n) = 101^n. Clearly, f is monotonically increasing and is unbounded. Therefore, there must be a value of n for which the leading digit must be 2. Trial calculations reveal that: 1.99475766x10(^124) = 101^(69) and 2.014705236x10^(125)= (101)^(70). Ed Gray

The question doesn't say "all positive powers of 101." Then 101 to the -1 power does not begin with a 1.

This should not be a 5-star problem. It should be 3 stars or 4 stars.

The following statement is true until 101^69. But hereby, the power begins with 2.

2,00676336839534e+140

I am a middle school student ,and thought about it this way.If you keep multiplying 101 it will slowly creep up.The one at the end will become a bigger number and take over the ones place.

Only up to the 69th power. 101^69 begins with the digits 1986… But 101^70 begins with the digits 2006… The first digit on 101^n is the same as the first digit of (1.01)^n.

Proof:

101=(1.01)∗(102)101=(1.01)∗(102)

101n=((1.01)∗(102))n=((1.01)n)∗(10(2∗n))101n=((1.01)∗(102))n=((1.01)n)∗(10(2∗n))

Eventually n is large enough that 2 < (1.01)^n < 3, in which case the first digit of 101^n will be 2.

My solution is not nearly as "mathematical" as some others, and probably rather simple-minded. Mine has to do with looking at the patterns in the results of each power.

The third digit from the left of each successive power increases by 1. If this pattern continues, then once we reach 101 to the 100th, the 1 as the first digit will change. Actually, it will probably occur sooner since the 5th digit is increasing by 3 with each power, but the question was just whether or not the first digit would always be 1.

$101^{100}=(1.01^{100})100^{100}$ starts with 2 because $1.01^{100}$ is very close to the natural logarithm base $e=2.71....$ . A calculator app on my tablet says $1.01^{100}=2.70481382942152609326719471080$ :-)

$70$ is the smallest power that doesn't start with a $1$ .

There is actually a very simple shortcut to the solution that doesn't require any advanced math to discover. If you look at the third digit of the first four powers of 101, you might notice that these are integers in ascending order. If we ignore everything past the third digit, the powers of 101 begin to look like integers in ascending order, starting with 101. You can therefore determine that the 100th power of 101 can't start with a 1, and if it does, then some power of 101 before the 100th power can't start with a 1.

For the number not to begin with 1, it is necessary for the power to equal n*ln(2) or greater. This first happens at n=70, for which an answer of 2.00676...e140 is obtained. As a follow-up question, what is the next power of 101 to have an answer beginning with 1 once again?

because of the carry adds up to populate the zeroes ..

and the carry gets added up after 10, 9, 8, 7 , 6, 5, 4, 3, 2, 1 additional powers.. in the second left digit

the second left digit will turn to zero with the carry 1 at the power 55 (if i am not wrong) ..

So after power 55, the first digit will no longer be 1.

-- no.. wrong... not 55. 70 is the right answer... as the additional power is of the form: 9, 9, 8, 7, 7, 7, 6, 6, 5, 5, 5, 5, 4, 4, 5, 4, 3, 4, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 2, 3, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, ... that will keep retaining the same second left digit and after which 1 will be added as carry to the same.

seems to be a log curve indeed.