Legacy puzzle - elegant approach?

[musicgold]

Hi,
This is not homework. I am struggling with a puzzle from this book (page 6, puzzle #13). I know the answer of the puzzle but I can't seem to figure out a good approach.

1. Homework Statement
A man left legacies to his three sons and to a hospital, amounting in all to $1,320.00. If he had left the hospital legacy also to his first son, that son would have received as much as the other two sons together. If he had left it to his second son, he would have received twice as much as the other two sons together. If he had left the hospital legacy to his third son, he would have received then thrice as much as the first son and second son together. Find the amount of each legacy.

Homework Equations

$a+b+c+h =1320 ~ ....(1)\\ h= b+c-a ~ ....(2)\\ h=2a +2c-b ~ ....(3)\\ h=3a+3b-c ~ ....(4)$

The Attempt at a Solution

Using equation (1), I got $c =660 -b$
Also I equated equations (2) and (3) to get $2b = 3a +c$
I substituted the value of c in this equation to get $b =220 +a$
Using these values of c and b in equation (1) I got $b+h=880$
I am not sure what to do after this.

Also, is there an elegant way to solve this problem?

Thanks

 

[PeroK]

Hi,
This is not homework. I am struggling with a puzzle from this book (page 6, puzzle #13). I know the answer of the puzzle but I can't seem to figure out a good approach.

1. Homework Statement
A man left legacies to his three sons and to a hospital, amounting in all to $1,320.00. If he had left the hospital legacy also to his first son, that son would have received as much as the other two sons together. If he had left it to his second son, he would have received twice as much as the other two sons together. If he had left the hospital legacy to his third son, he would have received then thrice as much as the first son and second son together. Find the amount of each legacy.

Homework Equations

$a+b+c+h =1320 ~ ....(1)\\ h= b+c-a ~ ....(2)\\ h=2a +2c-b ~ ....(3)\\ h=3a+3b-c ~ ....(4)$

The Attempt at a Solution

Using equation (1), I got $c =660 -b$
Also I equated equations (2) and (3) to get $2b = 3a +c$
I substituted the value of c in this equation to get $b =220 +a$
Using these values of c and b in equation (1) I got $b+h=880$
I am not sure what to do after this.

Also, is there an elegant way to solve this problem?

Thanks

You just need to keep working on those equations.

 

[YoungPhysicist]

A quick hint after @PeroK :

$h=3a+3b−c ....(4)$

c = 660-b
b = 220+a
h=?

Edit:And also, check out Guassian elimination.

 

[PeroK]

I'm not sure if it's particularly elegant, but it's clear from the puzzle that the first son ($a$) must have got the least. So, you could forget about the total for a bit and solve for $b, c, h$ in terms of $a$. That would be using your equations (2), (3) and (4).

 

[kuruman]

I would rewrite (2) in the form $h+a=b+c$ and substitute in (1) to get
$2b+2c=1320~\rightarrow~b+c=660~~ (5)$.
By similar rewriting and substitution back to (1) of equations (3) and (4), you get
$a+c=\cdots~~(6)$
$a+b=\cdots~~(7)$
Now if you add (5)+(6)+(7), the right side of this equation is twice the sum $a+b+c$. Knowing that sum you can find the value of $h$. At his point you can subtract (6) from (5) to get an equation for the difference $b-a$. Add that to (7) to get an equation for $2a$, and so on and so forth.

 

[haruspex]

is there an elegant way to solve this problem?

Your best bet of finding an elegant solution is to generalise it first.
Let the r^th son get a_r, r=1..n. Write the equations using Σa_r terms.

 

[haruspex]

Your best bet of finding an elegant solution is to generalise it first.
Let the r^th son get a_r, r=1..n. Write the equations using Σa_r terms.

Since there's been no response, let me fill in the details.
Let S be the sum of the legacies to the N sons.
Total legacy L= h+S.
h+a₁=S-a₁
h+a₂=2S-2a₂
etc.
h+(r+1)a_r=rS
$a_r=\frac{rS-h}{r+1}=S-\frac{S+h}{r+1}$
Summing
$S=NS-L\Sigma\frac 1{r+1}$
$S=\frac L{N-1}\Sigma\frac 1{r+1}$