11
$\begingroup$

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.

Improve this question
$\endgroup$
5
  • 8
    $\begingroup$ I assume that having 1 day old quintuplets does not count as a reasonable answer: $0+0+0+0+0 = 0\times0\times0\times0\times0 = 0^3$ $\endgroup$
    – Marius
    Oct 7, 2016 at 9:13
  • $\begingroup$ 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. $\endgroup$
    – user29705
    Oct 7, 2016 at 13:01
  • $\begingroup$ @aretxabaleta : Thank you 4 correction $\endgroup$
    – Jamal Senjaya
    Oct 7, 2016 at 13:18
  • $\begingroup$ What kind of a friend answers like this?! :P $\endgroup$
    – user541686
    Oct 7, 2016 at 19:36
  • $\begingroup$ People who participate on this site $\endgroup$
    – bleh
    Oct 8, 2016 at 16:25

5 Answers 5

Reset to default
18
$\begingroup$

My answer is

$2, 3, 4, 6, 12$

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.

I found the answer by

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.

Improve this answer
$\endgroup$
0
12
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ This is right, but having triplets is very rare. $\endgroup$
    – Jamal Senjaya
    Oct 7, 2016 at 9:16
  • 1
    $\begingroup$ +1 because that was a really speedy answer, and technically OP asked for reasonable, not probable. Kudos to Lynn! $\endgroup$
    – Xenocacia
    Oct 7, 2016 at 9:20
5
$\begingroup$

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 $\sum=3^3, \prod=0^3$

Improve this answer
$\endgroup$
2
  • $\begingroup$ My solution was the second one 1,1,2,2,2. Although this means once twins and once triplets. $\endgroup$
    – FrenkyB
    Oct 7, 2016 at 9:51
  • 1
    $\begingroup$ 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. $\endgroup$
    – Pieter Geerkens
    Oct 8, 2016 at 6:15
3
$\begingroup$

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

Improve this answer
$\endgroup$
2
$\begingroup$

Next to the accepted answer another reasonable answer would be:

18, 16, 15, 10, 5.
This gives a product of $12^3$ and a sum of $4^3$.
Or: 25, 12, 10, 9, 8 (with product $60^3$, sum $4^3$).

The problem is, if we allow twins and triplet, an enormous amount of answers are possible...

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.