[AccessD] OT: Friday Puzzles

Drew Wutka DWUTKA at Marlow.com
Mon Apr 26 10:07:08 CDT 2010


Which means you want to be the greatest 'even' power of 2.  ;)

Drew

-----Original Message-----
From: accessd-bounces at databaseadvisors.com
[mailto:accessd-bounces at databaseadvisors.com] On Behalf Of Tina Norris
Fields
Sent: Sunday, April 25, 2010 4:21 PM
To: Access Developers discussion and problem solving
Subject: Re: [AccessD] OT: Friday Puzzles

I'd just like to be number 2 in each round.
T

Drew Wutka wrote:
> Actually, it said everyone was shot until it got down to one person.
So
> if there were two people, the first person was shot, and the last
person
> would then be the lone survivor.
>
> But if you look at the pattern, the 'safe' person, for each round is a
> power of 2.  This makes sense, if you think about this problem in a
> binary fashion.
>
> You are getting rid of the first person, then every other person,
which
> means every odd number.  
>
> So let's take 64 people, the last person would be 00100000 and the
first
> person would be 00000001, and every combination in between (except,
the
> person in position 64 would be the only one with a 1 in the 64 column,
> and all zeros).  By removing all the numbers, you are removing any
> number with a 1 in the 1's position.  'renumbering' them, in binary,
is
> just removing the first digit (the 1's position).  Ie, instead of
> 00100000 you would now have 0010000, and every combination in between
> 10000 and 00001.  When you get to the second to last 'round', you will
> have one of two scenarios:
>
> 10 and 01
>
> Or 
>
> 11,10, and 01
>
> Either way, 01 is gone (as the first person) and in the case of three
> numbers left, the third person is shot too (cause it's an odd number,
> when renumbered).
>
> So the surviving person, no matter how many there are, is the highest
> multiple of 2 without going over.
>
> Ie, the 64th person would survive in cases where you had 64 people to
> 127 people.  128th position would survive from 128 people to 255
people.
> 512th person would survive with 512 people to 1023 people, and so on.
>
> Boy, this was a fun riddle, still haven't heard if I was right on the
> second one!
>
> Drew
>
> -----Original Message-----
> From: accessd-bounces at databaseadvisors.com
> [mailto:accessd-bounces at databaseadvisors.com] On Behalf Of Perry
Harold
> Sent: Friday, April 23, 2010 3:43 PM
> To: Access Developers discussion and problem solving
> Subject: Re: [AccessD] OT: Friday Puzzles
>
> If the first one is always shot it doesn't matter what position.  When
> it 
> gets down to only one standing if you start with the first that one is
> shot 
> as well.  Being in any other position would only prolong the angony.
>
> Perry
>
> ----- Original Message ----- 
> From: "Drew Wutka" <DWUTKA at marlow.com>
> To: "Access Developers discussion and problem solving" 
> <accessd at databaseadvisors.com>
> Sent: Friday, April 23, 2010 11:03 AM
> Subject: Re: [AccessD] OT: Friday Puzzles
>
>
>   
>> First round survivors:
>>
>>
>>     
>
2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48,50,5
>   
>
2,54,56,58,60,62,64,66,68,70,72,74,76,78,80,82,84,86,88,90,92,94,96,98,1
>   
>
00,102,104,106,108,110,112,114,116,118,120,122,124,126,128,130,132,134,1
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