Four mathematicians and their ages

Four mathematicians, none yet a centenarian, meet for coffee. The graph-theorist among them noticed that the common divisor graph of their ages (that is, the graph whose vertices are their ages, two of which are joined by an edge if, and only if, they have a common divisor greater than 1) is the simple square (left below).

After some thought, the number-theorist observed that if the Collatz recipe (multiply by 3, and add 1 if odd; divide by 2 if even) were applied to each of their ages, the resulting common divisor graph of the four numbers would be the star (right).

How old is Freddy, the youngest of the mathematicians? • Great puzzle! Fun and challenging. Feb 23 '20 at 16:32

Final solution

Freddy is

51

and the four mathematicians' ages are

51, 66, 70, 85 ($$51=3\times17$$, $$66=2\times3\times11$$, $$70=2\times5\times7$$, $$85=5\times17$$), with their Collatz-transformed ages being respectively

154, 33, 35, 256 ($$154=2\times7\times11$$, $$33=3\times11$$, $$35=5\times7$$, $$256=2^8$$).

Step-by-step deduction

This is essentially a tricky exercise in modular arithmetic and prime factorisations.

• There are at most two even numbers in each of the two graphs (three even numbers would give a triangle). But the Collatz transform sends every odd number to an even number, so there are exactly two odd and two even numbers in each graph.

• The number in the centre of the post-Collatz graph must be even, since the two even numbers are obviously linked. Every even post-Collatz number must be congruent to 1 mod 3, therefore congruent to 4 mod 6 by the Chinese remainder theorem.

So that central number in the post-Collatz graph must be

congruent to 4 mod 6, having at least three distinct prime factors (to be connected separately to the three others), and at most 300.

Let's consider all possible numbers satisfying these criteria:

$$70,130,154,190,220,238,280,286$$ (yes, I did find all these by hand!)

Most of these don't work because

either they give a prime number in the pre-Collatz graph ($$70=3(23)+1$$, $$130=3(43)+1$$, $$220=3(73)+1$$, $$238=3(79)+1$$) or they give a prime factor which is too large ($$280=3(3\times31)+1$$, something else in the pre-Collatz graph must be a multiple of $$31$$ which is impossible).

The smallest one which might work is

$$154=3(51)+1$$. This is the Collatz transform of $$51=3\times17$$, so we need a multiple of $$17$$ in the original square. Fortunately, $$5\times17=85$$ has Collatz transform $$3(85)+1=256$$, which is a power of 2.

Then for the post-Collatz graph we have

$$154=2\times7\times11$$ in the middle, $$256$$, a multiple of 7, and a multiple of 11,

corresponding in the pre-Collatz graph to

$$51=3\times17$$, $$85=5\times17$$, a multiple of 3, and a multiple of 5.

Putting the

3 with 11 and 5 with 7,

We had three remaining possibilities that might work as the central number in the post-Collatz graph. Now we know

154 does work; what about 190 and 286?

• $$286=2\times11\times13$$ is the Collatz transform of $$95=5\times19$$. The other multiple of 19 in the pre-Collatz graph must be $$3\times19=57$$, which Collatz transforms to $$172=2^2\times43$$. Then the post-Collatz graph is $$286$$, $$172$$, a multiple of 11, and a multiple of 13. But one of those last two must also be a multiple of 5, and both $$5\times11=55$$ and $$5\times13=65$$ are too big to be Collatz transforms of even numbers less than 100. Contradiction.

• $$190=2\times5\times19$$ is the Collatz transform of $$63=3^2\times7$$. The post-Collatz graph then must contain the number $$19$$, corresponding to $$38=2\times19$$ in the pre-Collatz graph. One of the remaining numbers in the pre-Collatz graph must be $$3\times19=57$$, which Collatz transforms to $$172=2^2\times43$$. Then the final number must be $$2\times5\times7=70$$ pre-Collatz and $$5\times7=35$$ post-Collatz.

So there is a second possibility, namely Freddy is

38

and the four mathematicians' ages are

38, 57, 63, 70 ($$38=2\times19$$, $$57=3\times19$$, $$63=3^2\times7$$, $$70=2\times5\times7$$)

with their Collatz-transformed ages being respectively

19, 172, 190, 35 ($$19=19$$, $$172=2^2\times43$$, $$190=2\times5\times19$$, $$35=5\times7$$).

• This is correct. However, I have a different solution which I believed was unique. In it Freddy is younger (which in fact he is!). Feb 23 '20 at 16:34
• @Bernardo Interesting. I'm still in the process of checking the remaining possibilities, but I'd assumed that this solution would be unique and everything else would be excludable somehow. Feb 23 '20 at 16:35
• @Bernardo OK, got it. There are exactly two solutions, nothing else works. Feb 23 '20 at 16:55