# Not a very simple riddle

When you are one, I am two. When you are ten, I am four. When you are hundred, who am I ?

Edit: Also, when you are thousand, I am same as i was before.

Edit_2: I can't move. I can't see the end. No matter how big you are, i am always small.

• Welcome to Puzzling.SE. Why don't you take the tour to earn your first badge. Jul 12, 2018 at 9:36
• When we are a thousand, you are the same as when we were a hundred, you mean? Jul 12, 2018 at 10:02
• @user477343 Yes. Jul 12, 2018 at 10:15
• @Lopolo thanks for confirming. I was going to link the word "before" with "four", with "before". Jul 12, 2018 at 10:16

I believe that given all the clues,

You are $6$.
Your rule is mapping $n$ to the last digit of $2^n$:
$2^1=2$, $2^{10}=1024$.
As it is a digit, it is always small!
Finally, the last digit of powers of 2 is a cycle of length 4: 2,4,8,6,2,... Since 100 and 1000 both are multiple of 4, $2^{100}$ and $2^{1000}$ have same last digit $6$.

– user44966
Jul 12, 2018 at 18:35

Could you be

Sixteen $16$?

It appears that the pattern is,

$$2^x$$ where $x$ is in binary.

Thus we have

\begin{align}2^1&=1 \\ 2^{10}\implies 2^2&=4\end{align}

Therefore,

$$2^{100} \implies 2^4=16$$

• Perhaps the who am I? part uses a "who" because Rot13$[$gur ahzore 'fvkgrra' vapyhqrf "grra" juvpu vf fubeg sbe "grrantre".$]$ Jul 12, 2018 at 9:49
• How does this account for the when I you are a thousand, I am same as i was before? Jul 12, 2018 at 11:23
• @anonymous2 this answer came before the edit. I don't know if the user has noticed... Jul 12, 2018 at 11:34
• Oh, gotcha. Bother these "2 hours ago" thingames. Jul 12, 2018 at 12:21

Could you be

Eight $8$?

It appears that the pattern is described as the ratio,

$$10^n:2^{n+1}\tag{n\geqslant 0}$$

Thus we have

\begin{align}10^0=1&\implies 2^{0+1}=2^1=2 \\ 10^1=10&\implies 2^{1+1}=2^2=4.\end{align}

Therefore,

$$10^2=100\implies 2^{2+1}=2^3=8$$

Who am I?

$8$? You say who in reference to a certain personfication of the number $8$; i.e., $7$ ate $9$.

Could you be

Nine $9$?

It appears that the pattern is as follows:

If we are a number represented by $a$, then you are $${\small\text{The first digit of this:}} \ (a-3)^2-2.$$

Supporting examples:

\begin{align}1\implies (1-3)^2-2 &= (-2)^2-2 \\ &=4-2 \\ &=2\tag{{\small\text{the first digit is}} \ 2} \\ \\ 10\implies (10-3)^2-2 &= 7^2-2 \\ &=49-2 \\ &=47\tag{{\small\text{the first digit is}} \ 4}\end{align}

Therefore,

\begin{align} 100\implies (100-3)^2-2 &= 97^2-2 \\ &= 9409-2 \\ &=9407\tag{{\small\text{the first digit is}} \ 9} \\ \\ 1000\implies (1000-3)^2-2 &= 997^2-2 \\ &=994009-2 \\ &=994007\tag*{\bigg(\begin{align}&{\small\text{the first digit is}} \ 9\\ &{\small\text{too, like before.}}\end{align}\bigg)}\end{align}

Who am I?

$9$? You say who because the "average" person works a nine-to-five job; and who is a reference to how "nin" is included in the word nine, meaning "an affectionate name for a grandmother".

Title:

Not a very simple riddle

The title has a capital letter N since that is the first letter of nine.

Edit 2:

I can't move. I can't see the end. No matter how big you are, I am always small.

Unsure about the first two lines, but for the last two lines, you are always the first digit no matter how big a number we are; i.e., you are always less than $10$.