🚀 go-pugleaf

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