# Ages of mathematician's five children

Two mathematicians meet at their school reunion.

A: Hey old friend, I heard you have 5 children. How old are they?
B: The sum of their ages is a cube number,
A: But, I still don't know their ages.
B: The product of their ages is a cube number too.
A: OK, I know now.

Find the most reasonable answer. How old are B's children?

Note, with most reasonable it is meant that you must look for an answer such that:
The age of the children is a positive integer number.
The children all have a unique age.
Of all valid possibilities, the one where the oldest child is as young as possible.

• I assume that having 1 day old quintuplets does not count as a reasonable answer: $0+0+0+0+0 = 0\times0\times0\times0\times0 = 0^3$ – Marius Oct 7 '16 at 9:13
• Just restricting the ages to positive integers or even unique ages (you don't even need to be a twin to have the same age as your 11 months older sibling) does not create a 'most reasonable' answer. Unless you find an age difference of 10 years perfectly reasonable but 13 years a clear oddity. – Hans Janssen Oct 7 '16 at 13:01
• @aretxabaleta : Thank you 4 correction – Jamal Senjaya Oct 7 '16 at 13:18
• What kind of a friend answers like this?! :P – user541686 Oct 7 '16 at 19:36
• People who participate on this site – bleh Oct 8 '16 at 16:25

$2, 3, 4, 6, 12$

Like Lynn's answer, the sum is

$27=3^3$

and the product is

$1728=12^3$

In this answer all the children have different ages, so there are no twins or triplets.

considering the product has to be a cube. This means that every prime factor used must occur 3 or 6 times, maybe even 9 times but I did not go that far and then the ages would probably get a bit large. With 5 ages, their sum will probably exceed 8, so a sum of 27 seems likely. Maybe a sum of 64 is possible too, but then they're more teenagers than children. If you use too few prime factors, you tend to get many children of the same age, or have to pack many of the factors in one or two of the ages to reach a sum of 27. Choosing six 2s and three 3s is the fewest where you begin to get enough factors to build distinct ages that are not too far apart. Moving the primes amongst the five ages, it is fairly easy to work towards a sum of 27.

The children could be possibly:

1, 6, 6, 6 and 8.

because

$1+6+6+6+8=27=3^3$

and

$1\cdot 6\cdot 6\cdot 6\cdot 8=1728=12^3$

One important thing to notice is that, since the product of the ages must be a cube, the prime factors of all ages can only appear 3 times, 6 times, 9 times, etc.

I first tried with the product $2^3\cdot3^3=216$, but couldn't manage to distribute the factors such that the sum would also be a cube. Then I experimented with $2^6+3^3$ and I had more luck.

• This is right, but having triplets is very rare. – Jamal Senjaya Oct 7 '16 at 9:16
• +1 because that was a really speedy answer, and technically OP asked for reasonable, not probable. Kudos to Lynn! – Xenocacia Oct 7 '16 at 9:20

I think

$4, 16, 35, 35, 35$

works as well - even though it is not a common, but still a possible solution:

sum is $125=5^3$, product is $2744000=140^3$

Other - not yet mentioned solutions are:

$2, 4, 7, 7, 7$ with $\sum=3^3, \prod=14^3$
$1, 1, 2, 2, 2$ with $\sum=2^3, \prod=2^3$
$0, 2, 2, 2, 2$ with $\sum=2^3, \prod=0^3$
$0, 2, 5, 9, 11$ with $\sum=3^3, \prod=0^3$

• My solution was the second one 1,1,2,2,2. Although this means once twins and once triplets. – FrenkyB Oct 7 '16 at 9:51
• Reminds me of the parents who, having once recovered from having quads, tried for one more and got triplets. A baseball team in two pregnancies. – Pieter Geerkens Oct 8 '16 at 6:15

I found the following answers for children <= 10:

1,1,2,2,2, S=8, P=8
1,6,6,6,8, S=27, P=1728
2,2,6,8,9, S=27, P=1728
2,4,7,7,7, S=27, P=2744
3,5,5,5,9, S=27, P=3375

This gives a product of $12^3$ and a sum of $4^3$.
Or: 25, 12, 10, 9, 8 (with product $60^3$, sum $4^3$).