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Two mathematicians meet at their school reunion.

A: Hey old friend, I heard you have 4 children. How old are they?

B: The sum of the multiplicative inverse of their ages is 1, and the sum of their ages is your sister's age.

A: But, I still don't know their ages.

B: The 2 oldest are twins.

A: OK, I know now.

How old are B's children ?

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    $\begingroup$ I'm surprised this is not a duplicate, this riddle is pretty famous Just changed a bit $\endgroup$
    – Jason_
    Aug 16, 2016 at 8:17
  • $\begingroup$ I thought I'd seen it's like on this site a couple months ago, but I'm not finding it. $\endgroup$
    – dcfyj
    Aug 16, 2016 at 12:27
  • $\begingroup$ @dcfyi I created this puzzle, and have posted it at www.brainden.com/forum more than a month ago. $\endgroup$
    – Jamal Senjaya
    Aug 16, 2016 at 13:12
  • $\begingroup$ @Jason, yes I get the idea from the similar puzzle, but this one is completely different. $\endgroup$
    – Jamal Senjaya
    Aug 16, 2016 at 13:12
  • $\begingroup$ This is basically the Ages of Three Children(or daughters) puzzle, you just made there be different math. If this really is an original puzzle than it is really similar. $\endgroup$
    – Jason_
    Aug 16, 2016 at 16:06

2 Answers 2

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Answer:

$2$, $4$, $8$, and $8$

The sum of their multiplicative inverse ages is 1 and the two eldest are twins implies

$\frac1a+\frac1b+\frac2c=1, 1\leq a\leq b\leq c$ (can't have a $0$ year old as $\frac10$ is undefined)

Which is satisfied for

a  b  c  a+b+c+c
2  3 12    29
2  4  8    22
2  6  6    20
3  3  6    18
4  4  4    16

So

The ages are $(a,b,c,c)$ for whichever one of those has his sister's age on the right.

Since

He did not know the ages before being told the oldest two were of the same age the correct ages must be possible with the two eldest not being equal, this is true of $(2,4,8,8)$ since $(2,5,5,10)$ or $(3,3,4,12)$ also sum to $22$.

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  • $\begingroup$ Good start, the question seems incomplete, we do not know A's sister age. But we can deduce the ages from there. $\endgroup$
    – Jamal Senjaya
    Aug 16, 2016 at 3:33
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    $\begingroup$ But we don't need to: A knows his sister's age, otherwise he would still not know the children's ages. $\endgroup$
    – Jonathan Allan
    Aug 16, 2016 at 3:55
  • $\begingroup$ OK I added what I think the intended (flawed) solution is :/ $\endgroup$
    – Jonathan Allan
    Aug 16, 2016 at 4:08
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    $\begingroup$ pretend that you are A, than calculate the ages from first information. At first, you do not know two of them are twin. $\endgroup$
    – Jamal Senjaya
    Aug 16, 2016 at 4:55
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    $\begingroup$ Still multiple possibilities - for example $(3,3,6,6)$ has a unique sum and it's possible to have two twins of six and two non-twins of three (gestation is only nine months). $\endgroup$
    – Jonathan Allan
    Aug 16, 2016 at 5:01
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Jonathan Allan have answered correctly.
To get better understanding why the answer is 8,8,4,2
I give my version

1st : Sum of their multiplicative inverse is 1

All age possibility are

2 , 3 , 7 , 42 sums 54
2 , 3 , 8 , 24 sums 37
2 , 3 , 9 , 18 sums 32
2 , 3 , 10 , 15 sums 30
2 , 3 , 12 , 12 sums 29
2 , 4 , 5 , 20 sums 31
2 , 4 , 6 , 12 sums 24
2 , 4 , 8 , 8 sums 22
2 , 5 , 5 , 10 sums 22
2 , 6 , 6 , 6 sums 20
3 , 3 , 4 , 12 sums 22
3 , 3 , 6 , 6 sums 18
3 , 4 , 4 , 6 sums 17
4 , 4 , 4 , 4 sums 16

2nd : Sum of their age is your sister age

Because A said that knowing the total (from the age of her sister) did not help, we know that knowing the sum of the ages does not give a definitive answer; thus, there must be more than one solution with the same total.

Only three sets of possible ages add up to the same totals:

2 , 4 , 8 , 8 sums 22
2 , 5 , 5 , 10 sums 22
3 , 3 , 4 , 12 sums 22

3rd clue : 2 oldest are twin

A concludes that the correct solution is 8,8,4,2.

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