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the number which Collatz transforms to them is prime, thus can't be part of that square. ($70=3(23)+1$, $130=3(43)+1$, $220=3(73)+1$, $238=3(79)+1$, etc.)

we get the final answer.

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).

we get the final answer.

How about uniqueness?

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

Final solution

Freddy is

51

and the four mathematicians' ages are

51, 66, 70, 85,

with their Collatz-transformed ages being

33, 35, 154, 256.

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

the number which Collatz transforms to them is prime, thus can't be part of that square. ($70=3(23)+1$, $130=3(43)+1$, etc.)

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 get the final answer.

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

the number which Collatz transforms to them is prime, thus can't be part of that square. ($70=3(23)+1$, $130=3(43)+1$, $220=3(73)+1$, $238=3(79)+1$, etc.)

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 get the final answer.

Freddy is

51

and the four mathematicians' ages are

51, 66, 70, 85,

with their Collatz-transformed ages being

33, 35, 154, 256.

Full explanation coming ...

Final solution

Freddy is

51

and the four mathematicians' ages are

51, 66, 70, 85,

with their Collatz-transformed ages being

33, 35, 154, 256.

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

the number which Collatz transforms to them is prime, thus can't be part of that square. ($70=3(23)+1$, $130=3(43)+1$, etc.)

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 get the final answer.