GameBanshee Forums

Aegis, probably should have been more specific.

I meant, what number goes where the question mark is located?

1 1 1 1
1 3 5 7
1 5 13 25
1 7 25 ?

So what's the last number?

Nope. It's a definitive answer. Not 55. Theres a pattern there... look harder.

1 1 1 1
1 3 5 7
1 5 13 25
1 7 25 ?

EDIT I mean 63 - I can't count

[ 07-23-2001: Message edited by: Xandax ]

63 - the sum of the three adjoining numbers left/diagonal left/above.

aaarrrhhh!! Xandax beat me to it! I don't know I go off line to feed my kids....

[ 07-23-2001: Message edited by: Yshania ]

This is logic, right? I don't do logic.

Originally posted by Yshania:
<STRONG><snip>
aaarrrhhh!! Xandax beat me to it! I don't know I go off line to feed my kids....
</STRONG>

I'm sorry just because I don't have kids


@Fable: well that is logical

[ 07-23-2001: Message edited by: Xandax ]

@Fable - you surprise me! No really!

@Xandax - I was scribbling on bits of paper in the kitchen while cooking pizza! lol! I bet I had the answer before you! (not!)

[ 07-23-2001: Message edited by: Yshania ]

Score Check:

Xandax: 2
Yshania: 1

Question 4:

A prison guard ran his prison in accordance with certain security precautions which dictated that:
a) There must be twice as many prisoners on the top floor as on the ground floor,
b) No cell could be unoccupied,
c) There must always be eleven prisoners in the six cells on each of the four sides.
One night, nine prisoners escaped. Yet the next morning, when the warden made his rounds, nothing seemed amiss.
How many prisoners were there to begin with? How did the remaining prisoners rearrange themselves to conceal the escape?

Originally posted by Yshania:
<STRONG>@Xandax - I was scribbling on bits of paper in the kitchen while cooking pizza! lol! I bet I had the answer before you! (not!)
</STRONG>

Well I was just looking at it and couldn't figure it out - then I eate dinner, and presto - the idea appeare to me .
So you properly knew it before me

First guess...

Considering there are six cells along each four walls that would give you twenty cells (corner cells counting twice)

In order to have minimum 11 prisoners along each wall I calculated ground floor 2 to each cell, with two cells having just one occupant therefore 18 cells x 2 inmates and 2 cells x 1 inmate

= 18x2 = 36
& 2x1 = 2

therefore ground floor = 38 prisoners.

First floor therefore = 76 prisoners (twice ground floor occupancy)

with 18 cells having 4 occupants and 2 cells having 2 occupants.

= 18x4 = 72
& 2x2 = 4

Total prisoners to start = 38+76 = 114.

9 inmates escape leaving 105.

To redistribute themselves ensuring all cells are occupied they would need 35 on the ground floor and 70 (35x2) on the first floor. Totalling 105.

35 on the ground floor would be 15 cells with 2 inmates and 5 cells with one inmate.
(20 cells, 35 inmates)

70 on the first floor would be 15 cells with 4 inmates and 5 cells with 2 inmates.
(20 cells, 70 inmates)

Each cell on each floor has a minimum of one inmate.

Phew...I hope this is right lol!

[ 07-24-2001: Message edited by: Yshania ]

Veh where are you?

He's probably sleeping right now (he's on the same timeline as I am ) (good smiley though )

[ 07-24-2001: Message edited by: Brink ]

Doh! I will have to wait until tomorrow now to see if I can match Xandax...

@Xandax - were you nicking my kids pizza? I thought they had eaten a bit last night! lol!
I don't know, pinching my kids dinner and reading my notes...

[ 07-24-2001: Message edited by: Yshania ]

Awww.... can I cheer you up with my pillow adven...erm, nevermind

Is it a feather...I think we should stop spamming this thread and get back to the empty board thread b4 Veh hits us with a big stick...

I agree

Ok, I'm awake now

Yshania... where'd you get the idea of 20 cells?

There should only be 16 cells.

8 on each floor.

_______
|1|2|3|
|4| |5|
|6|7|8|
-------

Don't count the corner cells twice It's only one cell.

You should be on the right track though.

Originally posted by Yshania:
<STRONG>
@Xandax - were you nicking my kids pizza? I thought they had eaten a bit last night! lol!
I don't know, pinching my kids dinner and reading my notes...
</STRONG>

I didn't do it - nobody saw me do it

Oh Veh!!!

Originally posted by Veh

c) There must always be eleven prisoners in the six cells on each of the four sides.

6 cells along each of the four sides.

= | 1| 2| 3| 4| 5| 6|
| 7|-----------| 8|
| 9|-----------|10|
|11|-----------|12|
|13|-----------|14|
|15|16|17|18|19|20|

You did not specify that 6 cells along each wall was three downstairs and three upstairs

OK. Recalculating...

[ 07-25-2001: Message edited by: Yshania ]

I think I might have it now - though my first answer is not really wrong as it did work out, just a different number of cells.

Original no of inmates = 36 =
12 on the ground floor and 24 on the first.

Ground floor cells = nos
....i..ii..iii

a) 1 - 5 - 1
b) 1 - - - 1
c) 1 - 1 - 1 = 12

First floor cells = nos

...iv...v...vi

d) 1 - 2 - 1
e) 6 - - - 6
f) 1 - 6 - 1 = 24

considering row

a is above d total inmates = 11
c is above f total = 11

considering column

i) is above iv) total = 11
iii) is above vi) total = 11

9 escape. 36 - 9 = 27 redistributed as follows:

9 on ground floor, 18 on first floor:

Ground floor cells = nos

....i..ii..iii

a) 1 - 1 - 1
b) 1 - - - 1
c) 2 - 1 - 1 = 9

First floor cells = nos

...iv...v...vi

d) 3 - 2 - 3
e) 2 - - - 1
f) 2 - 1 - 4 = 18

considering row

a is above d total inmates = 11
c is above f total = 11

considering column

i) is above iv) total = 11
iii) is above vi) total = 11

I am not doing this one again!!!!!

I think I deserve 2 points now hehe!