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The sum of the ages of a father, his son, and his two grandsons is 181. The father´s age has a common divisor greater than 1 with the ages of each of his three descendants, but not two of the latter have ages with a common divisor greater than 1.

How old is the father?

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  • $\begingroup$ Is the solution unique and is there a clever way to see why? $\endgroup$ – Arnaud Mortier Apr 30 at 12:43
  • $\begingroup$ @ArnaudMortier The solution is unique and there is a clever way to see why. $\endgroup$ – Bernardo Recamán Santos Apr 30 at 12:44
  • $\begingroup$ Then have my upvote! $\endgroup$ – Arnaud Mortier Apr 30 at 12:49
  • $\begingroup$ @ArnaudMortier Enjoy! $\endgroup$ – Bernardo Recamán Santos Apr 30 at 12:52
  • $\begingroup$ For those interested, this is were it all began:youtube.com/watch?v=6Giujua_s80&t=18s – $\endgroup$ – Bernardo Recamán Santos May 1 at 12:07
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The father is

$105$

Proof.

Let us denote by $a$ the age of the father, $b$ his son, and $c,d$ the ages of his grandsons.

Considering

The equation $a+b+c+d=181$ modulo $2$ shows that the amount of even unknowns is either $1$ or $3$. But $3$ is impossible as $b$, $c$, $d$ are pairwise coprime.

Therefore

Exactly one unknown is even, and the age of the father is at least a multiple of $3\times 5\times 7$ (the three least possible pairwise coprime factors after $2$ which cannot appear more than once) which is already 105. The smallest possible age for the father, after $105$, would be $3\times 5\times 11=165$ which is unrealistic.

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  • 1
    $\begingroup$ The next smallest age is not just unrealistic: Subtract the resulting age from the 181 (to get the sum of the other 3 ages) and you can see that it is only possible make from sums of multiples of the 3 prime factors if one of the ages is 0. So, Methuselah is disqualified too $\endgroup$ – Chronocidal Apr 30 at 13:45
  • $\begingroup$ I found it straightforward enough to find what the father's age has to be and to find a solution for the others, but as yet I don't see any neat way to see why the whole thing is unique... $\endgroup$ – Gareth McCaughan Apr 30 at 14:19
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The descendants' ages are:

57, 14, and 5

Because

They must add up to 76 (181 - 105), and exactly one each must have 3, 5, and 7 as factors (though having multiples of the same factor would be okay), and exactly one of them must be even. They may have other prime factors of course, as these do not affect the result.

So:

57 = 19 x 3, 14 = 7 x 2, and 5 is by itself. I ran a test and these are the only numbers which satisfy all the conditions and add up to 76.

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