6
$\begingroup$

The sum of the ages of six sisters known to me is 92. Though there is no single whole number greater than 1 that simultaneously divides the ages of any three of them, I did notice this morning, while they lined up for the ski lift, that the ages of any two of them standing one behind the other had a common divisor greater than 1. How old are the six sisters?

$\endgroup$
  • $\begingroup$ "simultaneously divides the ages of any three of them" - as in 'the sum of the ages of any three of them'? $\endgroup$ – rhsquared Jan 11 '18 at 14:20
  • $\begingroup$ @rhsquared Most likely not, rather as in that 2 divides 4, 6 and 8 simultaneously (though a consequence is that it also divides the sum, but dividing the sum does not imply the former) $\endgroup$ – Lolgast Jan 11 '18 at 14:22
  • 1
    $\begingroup$ after some hard work, 2 pages of my notebook, and a serious headache,I can assure you that if you misread and think that the ages of any two of them has a common divisor greater than 1, there is no solution ! $\endgroup$ – Florian Bourse Jan 11 '18 at 14:25
  • $\begingroup$ I don't understand that part: "any two of them standing one behind the other". $\endgroup$ – Alix Eisenhardt Jan 11 '18 at 14:26
  • $\begingroup$ Well, they are in a line, say they are order like this: A B C D, then A and B share a common divisor greater than one, B and C as well, and C and D too $\endgroup$ – Florian Bourse Jan 11 '18 at 14:29
8
$\begingroup$

Well ...

The common divisors must all be coprime, because otherwise we get three sisters whose ages have a common factor. Given that there are six sisters, that total age of 92 is rather small; let's begin by trying to make the total as small as we can and see what we get. So let's make the five common factors 2,3,5,7,11 in some order.

The ages will then be

(the numbers at the ends of our list and) products of adjacent pairs, so we'll make the total smallest by putting larger factors next to smaller ones -- so perhaps try the order 11,2,7,3,5. That gives us possible ages of 11,22,14,21,15,5. The total of these is 88, slightly too small. Let's try making a small change to the order and see what it does. How about 11,2,5,3,7? This gives 11,22,10,15,21,7 for a total of 86, actually a bit smaller than before. (Duh, of course, because we've moved a larger number to an end of the list; I should have put 7 at the end from the outset.) OK, let's try a reordering that increases the sum as little as we can: 11,2,3,5,7 gives 11,22,6,15,35,7. Total is 96, slightly too large.

So, what about

a different way of increasing the ages a little? Stick with 11,2,5,3,7 (giving the smallest total age so far) and then change 11 to 13. This gives ages 13,26,10,15,21,7; total is 92 as required.

(I thought it was worth "showing my working" to give some idea of how one can approach a problem of this kind. Though really I'd be more inclined to make a computer do the searching -- but that seemed kinda cheating in this case. I've reproduced the exact process, including one instance where I did something that in hindsight was clearly suboptimal.)

$\endgroup$
  • 1
    $\begingroup$ The answer is unique. Can any one prove this, preferably without recurring to computer verification? $\endgroup$ – Bernardo Recamán Santos Jan 11 '18 at 19:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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