"The Legacy" Problem

zenith20

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Aug 20, 2010
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Another puzzle from Shakuntala Devi, i wouldnt understand the solution:

"The Legacy"
When my uncle in Madura died recently, he left a will, instructing his
executors to dividehis estate of
Rs 1,920,000 in this manner:
Every son sould receive three times as much as a daughter, and that every
daughter should get
twice as much as their mother. What is my aunt's share?

the answer is:
Rs. 49200(10/13)

-----------------------------------
my problem is i can't get the answer! cause there's not cleared the number
of each son/daughter.
could you please explain it a bit more to me?
 
zenith20 said:
Another puzzle from Shakuntala Devi, i wouldnt understand the solution:

"The Legacy"
When my uncle in Madura died recently, he left a will, instructing his
executors to dividehis estate of
Rs 1,920,000 in this manner:
Every son sould receive three times as much as a daughter, and that every
daughter should get
twice as much as their mother. What is my aunt's share?

the answer is:
Rs. 49200(10/13)

-----------------------------------
my problem is i can't get the answer! cause there's not cleared the number
of each son/daughter.
could you please explain it a bit more to me?

Does the answer look something like:

492001013\displaystyle 49200\frac{10}{13}

.
 
Subhotosh Khan said:
Does the answer look something like:

492001013\displaystyle 49200\frac{10}{13}

.

Yes, Subhotosh Khan, That's the answer, and i can't get it :(
 
Hello, zenith20!

Sorry, I don't get it either . . . Is there some missing information?


The Legacy: When my uncle in Madura died recently, he left a will
instructing his executors to divide his estate of $1,920,000 in this manner:

Every son should receive three times as much as a daughter,
and that every daughter should get twice as much as their mother.

What is my aunt's share?

The answer is: .492001013\displaystyle 49200\tfrac{10}{13} . . . . How?

Let: mother's share = x\displaystyle x dollars.

Then each daughter gets 2x\displaystyle 2x dollars.

And each son gets 6x\displaystyle 6x dollars.


If there are S\displaystyle S sons, they get a total of 6Sx\displaystyle 6Sx dollars.

If there are D\displaystyle D daughters, they get a total of 2Dx\displaystyle 2Dx dolars.


So we have: .6Sx+2Dx+x=1,920,000\displaystyle 6Sx + 2Dx + x \:=\:1,920,000

. . . That is: .(6S+2D+1)x  =  1,920,000\displaystyle (6S + 2D + 1)x \;=\;1,920,000 .[1]



At first, I assumed there is an integer solution,

On the left side of [1], note that (6S+2D+1)\displaystyle (6S+2D+1) is an odd integer.

On the right side: .1,920,000=210354\displaystyle 1,920,000 \:=\:2^{10}\cdot3\cdot5^4
. . Hence, the odd factors are: .1,3,5,15,25,75,125,375,625,1875\displaystyle 1, 3, 5, 15, 25, 75, 125, 375, 625, 1875


For most of the factors, there are multiple solutions.

For example: .6S+2D+1=156S+2D=143S+D=7\displaystyle 6S + 2D + 1 \:=\:15 \quad\Rightarrow\quad 6S + 2D \:=\:14 \quad\Rightarrow\quad 3S + D \:=\:7

. . . .Hence: .(S,D)  =  (0,7),  (1,4),  (2,1)\displaystyle (S,D) \;=\;(0,7),\;(1,4),\;(2,1)

So that approach is a dead end . . .



Then we are told that:   x  =  492001013  =  639,61013\displaystyle \text{Then we are told that: }\;x \;=\;49200\tfrac{10}{13} \;=\; \frac{639,610}{13} . impossible!


Substitute into [1]:

. . (6S+2D+1)639,61013  =  1,920,0006S+2D+1  =  13639,610(1,920,000)\displaystyle (6S + 2D+1)\cdot\frac{639,610}{13} \;=\;1,920,000 \quad\Rightarrow\quad 6S + 2D + 1 \;=\;\frac{13}{639,610}(1,920,000)


\(\displaystyle \text{And we have: }\;\underbrace{6S + 2D + 1}_{\text{This is an integer}} \;=\;\underbrace{39.02378012\hdots}_{\text{This is not!}}\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


Suppose there is a typo in the given answer.

Suppose it was actually:   492301013=640,00013\displaystyle \text{Suppose it was actually: }\;49230\tfrac{10}{13} \:=\:\frac{640,000}{13}


Substitute into [1]:

. . (6S+2D+1)640,00013  =  1,920,0006S+2D+1=39\displaystyle (6S + 2D+1)\cdot\frac{640,000}{13} \;=\;1,920,000 \quad\Rightarrow\quad 6S + 2D + 1 \:=\:39

Then we have: .6S+2D=383S+D=19\displaystyle 6S + 2D \:=\:38 \quad\Rightarrow\quad 3S + D \:=\:19


Then the number of sons and daughters would be:

. . \(\displaystyle (S,D) \;=\;(1,16),\:(2,13),\:(3,10),\:(4,7).\:(5,4),\:(6,1)\)


But why would we choose any of those?

 
Dear Soroban,

Thanks for your kindness, i also think there's something missing.

most of Shakuntala's solution to her Puzzles are in brief, and we cannot get if it's kinda misspelling or not!

thank you anyway :)
 
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