[Guest]

A very rich man known for his diamond collection just died.

He donated all his possessions to charities and he left his most prized possession, 41 identical 24-karat diamond rings, said to be worth well over $10 million a ring, to his three sons. In his will, he left 1/2 of the rings to the eldest, 1/3 to the second, and 1/7 to the youngest. How many rings does each son get? (we are assuming that the rings will lose their values when cut).

[Guest]

Is it 20 to first person, 7 to second and 1 to the third one?

[Guest]

If you do the math and round up, you have the eldest receiving 21 rings, the second receiving 14 and the youngest receiving 6 which adds up to 41 rings. Is this correct?

[Guest]

20,14 and 7

[Guest]

The fairest way is to give 21 to the first, 14 to the second and 6 to the third. The trick is that the three fractions total less than 1, so you can add 1 imaginary ring, do the maths, and then take back your imaginary ring!

[Guest]

[spoiler=The Simplest solution method: Don't look if you know how to work out already ]

Make the denominators equal:

For this case: 1/2, 1/3, 1/7 => 21/42, 14/42, 6/42

You could clearly see that 21+14+6 = 41.

[Guest]

[spoiler=The Simplest solution method: Don't look if you know how to work out already ]

Make the denominators equal:

For this case: 1/2, 1/3, 1/7 => 21/42, 14/42, 6/42

You could clearly see that 21+14+6 = 41.

I agree that this seems to be the right solution, however... since they do not add up to 1, there is an inherent logical flaw here...

[Guest]

All three get 0. "He donated all his possessions to charities...". Assuming he did that before he died, the distribution of rings in his will is irrelevant as he no longer owns any.

[Guest]

It is 21 for eldest, 14 for second and 6 for the youngest son.

With above, all will get more than their indicated share of total 41 but in same ratio & without cutting even one !!!

[Guest]

If you can cut the rings its

1: 20.5

2: 13.666666667

3: 5.85714

Else you would have to round off to 21, 14 and 6

I am sure they will be so rich they won't care about the devaluation of the rings 'if you cut them we are assuming that the rings will lose their values when cut' You never said we can't cut them.

[Guest]

A very rich man known for his diamond collection just died.

He donated all his possessions to charities and he left his most prized possession, 41 identical 24-karat diamond rings, said to be worth well over $10 million a ring, to his three sons. In his will, he left 1/2 of the rings to the eldest, 1/3 to the second, and 1/7 to the youngest. How many rings does each son get? (we are assuming that the rings will lose their values when cut).

If he indeed donated all his possessions to charities, the answer is zero zero and zero

[bonanova]

The executor of the will has to do something.

[Guest]

It's a simple math question and many of you have already answered it correctly.

As someone pointed out, 1/2 + 1/3 + 1/7 = 41/42. So you add one imaginary ring, and give 1/2 (21 rings), 1/3 (14 rings), and 1/7 (6 rings) to the sons.

[Guest]

A very rich man known for his diamond collection just died.

He donated all his possessions to charities and he left his most prized possession, 41 identical 24-karat diamond rings, said to be worth well over $10 million a ring, to his three sons. In his will, he left 1/2 of the rings to the eldest, 1/3 to the second, and 1/7 to the youngest. How many rings does each son get? (we are assuming that the rings will lose their values when cut).

21 will go to eldest, 14 will go to second, 6 will go to youngest.

Add one diamond to 41 to make it divisible by 2,3 and 7. Then divide 42 by 2, 3, 7 u will get 21, 14 and 6 respectively. 21+14+6=41. So then the diamond which we added can be returned.