0
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A party is being held at a local mansion. The host is very rich and his success is because of one thing — his famous recipe for Spaghetti!

The only guests that may attend are people who correctly reply to the guard at the door.

Here's where you come in. You and a friend are trying to steal this recipe. You sneak by and listen to the passwords.

The first guest arrives. The security says "2", and the guest replies "4".

The second guest arrives. The security says "3", the guest replies "4".

The third guest arrives. The security says "5", the guest replies "4".

The fourth guest arrives. The security says "8", the guest replies "8".

Your friend thinks he's got it all figured out so he walks up to the door and the security says "4". He replied "4", and is tarred and feathered and sent away in shame.

Another guest arrives, security says "6" and he says "6" to get in.

You walk up and get a "7" from security. What is your response?

Thanks for the ideas, credits to:

Part 1 was created by warspyking and is found here: The Security to the Party

Part 2 in Mew's version The Security to the Party [Part 2]

Part 3 in skv's version The Security to the Party [Part 3]

Part 4 in JNF's version The Security to the Party [Part 4]

Tip 1

Oops! I'm sorry. The tip WAS wrong!
Real tip:
It's related to electronic display screen. Draw the number like this:

 -
| |
 -
| |
 -

And COMPARE WITH the answer.

Tip 2

2 more example:
security: 1 guest: 0
security: 4 guest: 3

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  • $\begingroup$ @Mathias711 It's a mistake. $\endgroup$ – user3804649 Nov 11 '14 at 7:40
  • $\begingroup$ Yeah, everyone has got the same mistake. No problemo, they are all correct now :P $\endgroup$ – Mathias711 Nov 11 '14 at 7:40
  • $\begingroup$ This is the fith party. The guards know your friend by now. $\endgroup$ – SQB Nov 11 '14 at 7:41
  • $\begingroup$ @SQB This is the fifth spaghetti cook. $\endgroup$ – user3804649 Nov 11 '14 at 7:43
  • 1
    $\begingroup$ Since this is apparently quite a busy party, fake an emergency phone call and walk away quickly without giving an answer, change your appearance, come back, and make sure not to get involved with the guard again until you know all the questions and correct answers. :) $\endgroup$ – hvd Nov 11 '14 at 11:01
3
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I say

1.

Because:

I think the answer is the number of corners in the digital display representation of it, with a T shape counting twice, i.e. towards both sides.

Example for explanation:

For "3" we count: 1: the upper right corner connects the top bar and the top right bar. 2: the top right and middle bar. 3: middle and lower right bar. (2+3 make said T shape). 4: lower right and bottom bar.

Edit: A more mathematical way to express this is:

answer = number of unique pairs (h,v) where h is a horizontal bar, v an adjacent vertical bar and both are lit when displaying the number given by the guard.

Edit: forgot an important detail in the previous (hidden) paragraph.

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  • $\begingroup$ Why is this NOT marked as accepted? $\endgroup$ – warspyking Nov 13 '14 at 18:25
  • $\begingroup$ Wait. I'm searching for the earliest answer. EDIT: Yes, this's it. $\endgroup$ – user3804649 Nov 14 '14 at 5:26
  • $\begingroup$ Hmmm... it appears I over thought it... $\endgroup$ – Warlord 099 Nov 14 '14 at 14:50
6
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Do you want a stupid answer? I got one that fits perfectly and is almost definitely not what you are looking for.

The Series

What do we have here? There is a pattern to be extrapolated; probably several hundred thousand different expressions that would generate the series.

 The series
/----------\
 2   =>   4 
 3   =>   4 
 4   !!   4 
 5   =>   4 
 6   =>   6 
 7   =>   ? 
 8   =>   8 
\----------/

It looks arbitrary when first glanced at, yet there's something to it. The significance of the 4 is obvious, as it's the only clue we have as to what doesn't work. The amount of fours repeating is also pretty nauseating, so I've decided to try something interesting.

Pattern?

Deciding to preformulate each of those left-hand input numbers into an expression of the form xn where n is a prime and x the coefficient gives us this series:

 The series again
/----------------\
  1 * 2   =>   4 
  1 * 3   =>   4 
  2 * 2   !!   4 
  1 * 5   =>   4 
  2 * 3   =>   6 
  1 * 7   =>   ? 
  4 * 2   =>   8 
\----------------/

This has the drawback of giving a variable representation to some numbers; 6 can be both 2 * 3 and 3 * 2, for example, so I've decided to put the smaller number to the left. Since we have the series, we can see that we can disregard the prime and still have rising coefficients in some sort of logical order:

 Third series
/------------\
  1   =>   4 
  1   =>   4 
  2   !!   4 
  1   =>   4 
  2   =>   6 
  1   =>   ? 
  4   =>   8 
\------------/

   [Note]: This series has
   little to do with the first
   and as such they aren't
   equivalent.

This kind of series lends itself better to the eyes, and so we can cut out some duplicate entries, giving series #10923853Sqrt[-1]:

/----------\
 1   =>   4
 2   !!   4
 1   =>   ?
 2   =>   6
 4   =>   8
\----------/

From this we can expect x=8 => 10, probably following the pattern of x=n^a => 4+2*a. As such, we can reasonably expect that seven will imply four as the answer, as (n^0)*7 => 4+2*0 = 4. A few further bits of this series would be:

/----------\
 4   =>   6
 9   =>   8
 10  =>   6
 11  =>   4
 12  =>   8
 13  =>   4
\----------/

Pfft, so much effort for something so wrong and pointless ^^

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  • 2
    $\begingroup$ It's definitely not going to be the answer, but it's definitely getting +1 for just being bonkers $\endgroup$ – Joe Nov 11 '14 at 13:21
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    $\begingroup$ And I had to explain why there are so many fours: it's to explain why your friend gave the wrong number. $\endgroup$ – user3804649 Nov 12 '14 at 13:18
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My response is 4.

The formula I used is 2+2f where f is the number of factors in the prime factorization of x.
For 2: The prime factors in 2 are: 2. 2+(2*1) = 4
For 3: The prime factors in 3 are: 3. 2+(2*1) = 4
For 4: The prime factors in 4 are: 2*2. 2+(2*2) = 6.
For 5: The prime factors in 5 are: 5. 2+(2*1) = 4
For 6: The prime factors in 6 are: 2*3. 2+(2*2) = 6
For 7: The prime factors in 7 are: 7. 2+(2*1) = 4.
For 8: The prime factors in 8 are: 2*2*2. 2+(2*3) = 8.

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  • $\begingroup$ You'll have to use an HTML line break: <br> $\endgroup$ – Martin Ender Nov 11 '14 at 17:02
  • $\begingroup$ Holy potato, I got the same answer using insane troll logic $\endgroup$ – Darkgamma Nov 11 '14 at 17:05
  • $\begingroup$ @Darkgamma Yeah... I didn't follow your post at all lol $\endgroup$ – Warlord 099 Nov 11 '14 at 17:07
  • $\begingroup$ @Warlord099, I can't even follow it myself, don't worry :D Yours makes much more sense $\endgroup$ – Darkgamma Nov 11 '14 at 17:08
  • $\begingroup$ It probably isn't the expected reasoning (or even the correct answer), but I was just happy to find something that worked. $\endgroup$ – Warlord 099 Nov 11 '14 at 17:14
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In light of new information, I say the answer is 2.

Looking at the digital display I could find the following information:
x - | a ......... x is input; - is number of crossbars; | is number of vertical bars; a is answer
1 0 2 0
2 3 2 4
3 3 2 4
4 1 3 3
5 3 2 4
6 3 3 6
7 1 2 ?
8 3 4 8

From this it appears:

That the number of crossbars is a multiplier for the number of vertical bars.
0 bars = x0
1 bar = x1
3 bars = x2
Why 3 bars is a x2 multiplier is unknown to me. Could it be because of the number of "on" bits to make up the number in binary?

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0
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I think the answer is 1.

Count the amount of corners.

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  • $\begingroup$ Answer correct. Time - too late. $\endgroup$ – user3804649 Nov 14 '14 at 5:27

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