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.
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1$\begingroup$ Welcome to Puzzling.SE. Why don't you take the tour to earn your first badge. $\endgroup$– u-ndefinedJul 12, 2018 at 9:36
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$\begingroup$ When we are a thousand, you are the same as when we were a hundred, you mean? $\endgroup$– Mr PieJul 12, 2018 at 10:02
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1$\begingroup$ @user477343 Yes. $\endgroup$– LopoloJul 12, 2018 at 10:15
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$\begingroup$ @Lopolo thanks for confirming. I was going to link the word "before" with "four", with "before". $\endgroup$– Mr PieJul 12, 2018 at 10:16
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4 Answers
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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 $6$.
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1$\begingroup$ I read your answer as rot13(fvk snpgbevny). Perhaps you could make it less ambiguous? $\endgroup$– user44966Jul 12, 2018 at 18:35
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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,
$$2^{100} \implies 2^4=16$$
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$\begingroup$ How does this account for the
when I you are a thousand, I am same as i was before? $\endgroup$ Jul 12, 2018 at 11:23 -
$\begingroup$ @anonymous2 this answer came before the edit. I don't know if the user has noticed... $\endgroup$– Mr PieJul 12, 2018 at 11:34
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$\begingroup$ Oh, gotcha. Bother these "2 hours ago" thingames. $\endgroup$ Jul 12, 2018 at 12:21
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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
whoin reference to a certain personfication of the number $8$; i.e., $7$ ate $9$.
Answer after included edit
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
whobecause the "average" person works a nine-to-five job; andwhois 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
Nsince 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 $10$.
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I guess, the answer is
16. The context involves two persons, my guess the number system is binary.
Pattern => 2 to the power of me is him.
2 to the power of 1 is 2.
2 to the power of (2 => 10 in binary) is 4.
2 to the power of (4 -> 100 in binary) is 16.
