What is the trick to solving the odd number camel puzzle?

[musicgold]

Homework Statement

I came across the puzzle shown below. I know how to solve it and the general structure behind it. But I wish to know how to express it algebraically.

Relevant Equations

A man leaves 19 camels in his will. He orders that 1/2 the camels to be given to his son, 1/4 to his daughter and 1/5 to the servant who took care of him. How should they distribute the camels?

A wise man adds his own camel to make the total 20. He gives 10 to the son, 5 to the daughter and 4 to the servant and take his own camel back. So the wise man enables the distribution of the remaining 1/5 th camel without cutting it.

I know that the puzzle is related to the LCM of the denominators of the fractions. I could create a few of my own puzzles too (for example, 23 camels to be divided in 1/2, 1/3, 1/8th). But I have some more questions about this puzzle.

1. What is the formal name for this solution method?

2. How would one go about this problem, if they did not have the help of the one camel of the wise man?

3. Is there a way to frame this situation in algebraic equations?

 

[PeroK]

What is $\frac 1 2 + \frac 1 4 + \frac 1 5$?

 

[BvU]

And what if the old man would have left 1/20 of his estate to his dear wife ? Huh !

 

[Frabjous]

https://en.wikipedia.org/wiki/Diophantine_equation

 

[Frabjous]

And what if the old man would have left 1/20 of his estate to his dear wife ? Huh !

Taxes.

 

[FactChecker]

I'm afraid that the true name of the "solution" is "sloppy arithmetic". One half of 20 is not one half of 19, etc.

 

[PeroK]

I'm afraid that the true name of the "solution" is "sloppy arithmetic". One half of 20 is not one half of 19, etc.

How would you have prevented a family feud?

 

[Frabjous]

x = # camels =19
y = # wise man camels

a = # son’s camels = (x+y)/2
b = # daughter’s camels = (x+y)/4
c = # servant’s camels = (x+y)/5

a+b+c=x
19(x+y)/20 = x
y=x/19 = 1

Without the wise man, normal rounding gets you to the same answer.
9.5 rounds to 10
4.75 rounds to 5
3.8 rounds to 4

I am assuming that only whole camels are dispersed and that all of them, not 95%, are dispersed

 

[BvU]

Without the wise man, normal rounding gets you to the same answer.
9.5 rounds to 10
4.75 rounds to 5
3.8 rounds to 4

So everybody is happy: they all get more than foreseen !
Now let's get rid of the national debt in the same way !