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A vampire enters a town and bites 2 person every night. The bitten 2 persons grow into two big vampires and combined bite 4 persons the next night. These 4 vampires combined bite 6 people the next night and the 6 bite 8 and so on.......how many vampires will be there on the 5th night?

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closed as off-topic by Deusovi, Fabich, Gareth McCaughan, Engineer Toast, Mike Earnest Jun 17 '16 at 18:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Deusovi, Fabich, Gareth McCaughan, Engineer Toast, Mike Earnest
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 6
    $\begingroup$ So, so far we have answers saying 291, 161, 115, 41, 5843, and 243. I think this is a sign that the question is not clearly enough stated. Could you make it clearer and more explicit exactly what is happening each night? $\endgroup$ – Gareth McCaughan Jun 17 '16 at 11:30
  • $\begingroup$ The math doesn't seem to follow a pattern. It should be 2 biting 2, then 4 biting 4, then 8 biting 8, right? Where does the six come from? $\endgroup$ – Engineer Toast Jun 17 '16 at 16:00
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Notations

Lvl0 = Original vampire
Lvl(N) = vampires bitten by the Lvl(n-1) vampires.

In brackets you will find the number of vampires from previous day.
After that the number of vampires they produce.

Night 0:

1 * Lvl0

Night 1:

(1 Lvl0) + 2 Lvl1.

Night 2:

(1 Lvl0 + 2 Lvl1.) + 2 Lvl1 + 2*4 Lvl 2

Night 3:

(1 Lvl0 + 2 Lvl1. + 2 Lvl1 + 2*4 Lvl 2) + 2 Lvl1 + 2*4 Lvl 2 + + 2*4*6 Lvl3

Night 4:

(1 Lvl0 + 2 Lvl1. + 2 Lvl1 + 2*4 Lvl 2 + 2 Lvl1 + 2*4 Lvl 2 + 2*4*6 Lvl 3) + 2 Lvl1. + 2*4 Lvl 2 + 2*4 Lvl 2 + 2*4*6 Lvl 3 + 2*4 Lvl 2 + 2*4*6 Lvl 3 + 2*4*6*4 Lvl 4

Night 5:

(1 Lvl0 + 2 Lvl1. + 2 Lvl1 + 2*4 Lvl2 + 2 Lvl1 + 2*4 Lvl2 + 2*4*6 Lvl3 + 2 Lvl1. + 2*4 Lvl2 + 2*4 Lvl2 + 2*4*6 Lvl3 + 2*4 Lvl2 + 2*4*6 Lvl3 + 2*4*6*8 Lvl4 ) + 2 Lvl1. + 2 *4 Lvl2 + 2 *4 Lvl2 + 2*4*6 Lvl3 + 2 *4 Lvl2 + 2*4*6 Lvl3 + 2*4*6*8 Lvl4 + 2*4 Lvl2 + 2*4*6 Lvl3 + 2*4*6 Lvl3 + 2*4*6*8 Lvl4 + 2*4*6 Lvl3 + 2*4*6*8 Lvl4 + 2*4*6*8*10 Lvl5

So the total after 5 nights:

5843

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  • $\begingroup$ In my defense, the word "combined" wasn't there when I answered the question $\endgroup$ – Marius Jun 17 '16 at 11:56
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My calculation gave the following table :

enter image description here So we have a total of 291 vampire by the end of night 5.

To further defend what I did here is the way I computed each cell :

enter image description here This only compute the number of group of a given level (lvl) on a given night (n). It is then quite easy to get the real number of vampire as you just have to multiply it by their level time two (the lvl 0 vampire being some kind of an exception (no offense, he must be feeling pretty lonely)).

But it seems like the question should be reworded. does the 2 "LVL2" vampire bites 4 persons each ? or do they bite 4 person in total ? It is not quite clear.

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  • $\begingroup$ 2 vampires bite 4 persons in total $\endgroup$ – TVV Jun 17 '16 at 14:57
  • $\begingroup$ this is taken in account right after calculating the number of different vampire "group". When people choose to say "2 vampires bites 4 person", I prefer saying "two vampire creates a single group of 4 vampires". $\endgroup$ – Joulin Nicolas Jun 17 '16 at 15:06
  • $\begingroup$ Your table great.I was just wondering if there at all was a way to representing the data effectively in 2X2 table. Your approach is just great. Hats off. $\endgroup$ – TVV Jun 19 '16 at 14:34
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new vampires for each night is shown
night0: a lvl0 vampire entered
night1:2
night2:2 4
night3:2 4 4 6
night4:2 4 4 4 6 6 6 8
night5:2 4 4 4 4 6 6 6 6 6 6 8 8 8 8 10

Total: 161 vamps
2's are because of Lvl0 vampire and are of Lvl1
these 2's of Lvl1 combined create 4 Lvl2 vamps and so on..
so.. no. of lvl(n+1) vamps are 2*n*(total Lvl(n) from present until last night)

A neat pattern is also visible viewed this way:
a group of 2 lvl1 vamps is a unit as they always come in pairs of 2(The Sith??? o.O)
a group of 4 lvl2 vamps is a unit as they always come in groups of 4
a group of 6 lvl3 vamps is a unit as they always come in groups of 6
a group of 8 lvl4 vamps is a unit as they always come in groups of 8
a group of 10 lvl5 vamps is a unit as they always come in groups of 10

no of grps of each lvl present at end of each night: (no. of grps are arranged in increasing order of lvl(0-5:left-right))
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

giving 1*1 + 5*2 + 10*4 + 10*6 + 5*8 + 1*10 = 161

I got these ideas as a credit of and

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One the first night $1$ vampire bites $2$ people, making a total of $3$ vampires. On the second night, $3$ bite a total of $6$, making a total of $9$. And so on. On the fifth night there will be $3^5=243$ vampires.

A proof that

on the $n$th night there will be $3^n$ vampires: this is true of the first night, as already stated. If on the $n-1$st night there are $3^{n-1}$ vampires, then, on the $n$th night, each will bite $2$ people. That is, $2*3^{n-1}$ will be bitten and become vampires. Together with the $3^{n-1}$ who were already vampires, this makes $(2+1)*3^{n-1}=3*3^{n-1}=3^n$ vampires. The result follows by induction.

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  • $\begingroup$ 4 vampires bite 6 people not 8 people $\endgroup$ – TVV Jun 17 '16 at 8:55
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    $\begingroup$ Rosie, the problem is simpler than you're making it. The number of new vampires doesn't depend on the number of current vampires. It just increases by 2 every night. :/ $\endgroup$ – Deusovi Jun 17 '16 at 8:57
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I think this is not the answer that you are searching but I think the question must be more specific :P

This is my approach (sorry for my english):

If the first vampire bites EVERY night and the other only NEXT night... and if the people grow into vampires after that night...

Legend

First vampire: v1
Next 2 vampires: v2
Next 4 vampires: v4
Next 6 vampires: v6
Next 8 vampires: v8
Persons bitten: ps

Night 1

v1 arrives the town and bites 2 persons: 1 vampire

Night 2

2ps grow into v2 and they bite 4 ps
v1 bites 2 ps
v1 + v2 = 3 vampires (6 persons bitten)

Night 3

v2 don't bite this night
4ps grow into v4 and they bite 6 ps
2ps grow into v2 and they bite 4 ps
v1 bites 2 ps
9 vampires

Night 4

v2*2 don't bite this night
v4 don't bite this night
6ps grow into v6 and they bite 8 ps
4ps grow into v4 and they bite 6 ps
2ps grow into v2 and they bite 4 ps
v1 bites 2 ps
21 vampires

Night 5

v2*3 don't bite this night
v4*2 don't bite this night
v6 don't bite this night
8ps grow into v8 and they bite 10 ps
6ps grow into v6 and they bite 8 ps
4ps grow into v4 and they bite 6 ps
2ps grow into v2 and they bite 4 ps
v1 bites 2 ps
41 vampires

Cheers

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Assuming that the original vampire bites 2 people each night, and $n$ vampires bite $n - 2$ people each night, I got the following table:

vampire table

Where $Lvl(0)$ is the number of original vampires, and $Lvl(n)$ is the number of vampires converted by the $Lvl(n - 1)$ vampires. So the answer is a total of

115 vampires

In general, the number of vampires on night $n$ is:

$f(n) = 2^{n+2} - 2n - 3$

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