Powers of 101

Consider

$\begin{aligned} 101^{100} & = (10^2+1)^{100} \\ & = 10^{200} + 100 \times 10^{198} + 4950 \times 10^{196} + ... + 1 \\ & = 10^{200} + 10^{200} + 0.4950 \times 10^{200} + ... + 1 \\ & = 2.4950 \times 10^{200} + ... + 1 \end{aligned}$ .

Therefore, $101^{100}$ does not begin with 1. No , the powers of 101 do not always begins with 1.

From a program, the beginning digit for respective $n$ is as follows:

$\begin{array} {cc} \text{Range of }n & \text{Beginning digit} \\ 1 \le n \le 69 & 1 \\ 70 \le n \le 110 & 2 \\ 111 \le n \le 139 & 3 \\ 140 \le n \le 161 & 4 \\ 162 \le n \le 180 & 5 \\ 181 \le n \le 195 & 6 \\ 196 \le n \le 208 & 7 \\ 209 \le n \le 220 & 8 \\ 221 \le n \le 231 & 9 \\ 232 \le n \le 301 & 1 \end{array}$

Observe the pattern:

$101^1 = 01|| 01$

$101^2 = 01|| 02|| 01$

$101^3 = 01|| 03|| 03|| 01$

$101^4 = 01|| 04 ||06|| 04|| 01$

$101^5 = 01|| 05 ||10 ||10|| 05|| 01$

$...$

In general, for all positive integers n, $101^n$ is written as

$C^n_0 C^n_1 C^n_2 ... C^n_{n-1} C^n_n$

each every two digits, from left to right.

When n = 100, $C^n_1$ is a three-digit number and carries a digit to 1, making the first digit 2.

$∴ 101^{100}$ starts with 2 instead of 1.

Without computing many values, we can figure this out with logarithms. Because we use the base-10 representation of numbers we can take the base-10 log to know the first digit of a number. Obeserve that the log of a pure power of 10 is always an integer number and if the number is close to a power of 10 it is also almost a whole number. For example $log_{10}(11)=1.041392$ But as soon as the first digit changes to something more than 1 the decimal part of the logarithm becomes greater than $log_{10}(2)=0.3010$ . We can see why that is by "splitting" a number: $log_{10}(200)=log_{10}(2*100)=2+log_{10}(2)$ Because every number less than 10 is less than 1 after applying the base-10 log we can conclude that this relates to the decimal part of the number. $log_{10}(101^{n})=n*log_{10}(101)$ We can already see that the first digit will be greater than 1 eventually, because $log_{10}(101)$ is not an integer, thus multiplying it with a sufficiently large number will make its decimal part greater than $log_{10}(2)=0.3010$

Another way of phrasing this problem is: Does $101^{n}=(1.01 \times 100)^{n}=100^{n} \times 1.01^{n}$ have it's first digit not start with a 1? In other words, we can ignore the $100^{n}$ term and reduce the question to can $1.01^{n}$ start with a digit other than 1 for some integer $n$ . Clearly the limit of $1.01^n$ goes to infinity, and so will not always have a first digit of 1. For instance for $n$ =100 $1.01^{n}$ evaluates to approximately 2.7.

Consider the powers of 1.01.

There is no upper bound on 1.01^n. We can therefore find a value for n, which gives 1.01^n = 3

This is equivalent to a power of 101 beginning with 3 (or any other digit).

In this case, we could use Binomial theorem . We should represent 101 as a sum $100 + 1$ . Therefore:

$(100+1)^n =100^n+C_1^n 100^{n-1} 1^1 + C_2^n 100^{n-2} 1^2+C_3^n 100^{n-3} 1^3+...$

First term will dominate in the beginning. It will determine what is the first digit of the number. Second term could influence the first digit, but it could not reach such heights because it lacks one power of 100. However, as $n$ becomes greater, so $C_n^1$ increase and eventually it will reach magic number 100 or higher. We could attempt to calculate after which power first digit will change from 1 to 2:

$C_n^1 =\frac{n!}{n-1!1!}=\frac{n-1!n}{n-1!1!}=\frac{n}{1!}=n=100$

However, it is underestimation, because we do not include consequent terms. Thus, greater $n$ , greater influence of a consequente terms on a first digit.

The power can be written as $\sum_{k = 0}^n \left(\begin{array}{c} n \\ k \end{array}\right) 10^{2k}.$ The first digit would be the coefficient of $10^{2n}$ , which is 1, unless there is carry from terms farther down the line.

The last example where there is no carry over in the sum at all is $101^8 = 10828567056280801.$ After that comes $101^9 = 1093685272684360901,$ where the coefficients of $10^4$ and $10^5$ are equal to 126.

We will certainly have carry over to the first digit if the second coefficient is 100 or greater. Thus

$101^{100} = \cdots + 100\cdot 10^{198} + 1\cdot 10^{200} = \cdots + 2\cdot 10^{200}.$

Assuming the pattern continues, as it should (and does), the has to come a point where digits must be carried over. This will come sooner in some base systems, and later in others, but it must come

If you look at the numbers the question gives us,

= 101 = 10201 = 1030301 = 104060401

You can see that the numbers in the 3rd column are increasing consistently by one, (1,2,3,4) so eventually they will go up to 9, and then 10, and then the 2nd column which was all zeros before, will be ones, and then eventually the second column too will add up and up (1,2,3,4,5,6,7,8,9,10) and then the first column will have a 2 in it..... and then eventually a 3 and a 4 and a.... anyways no not all powers of 101 will begin with one.

101 10201 1030301 10460401 There is a pattern which is the third digit from the left. The number goes up in 1 therefore at some point the first digit will be two.