<s>[INST] A partition of 186 into five parts under divisibility constraints The sum of 5 positive natural numbers, not necessarily distinct, is 186. If placed appropriately on the vertices of the following graph, two of them will be joined by an edge if and only if they have a common divisor greater than 1 (that is, they are not relatively prime).

What, in non-decreasing order, are those 5 numbers? The answer is unique.
 [/INST] We cannot have at least three even numbers, for then there would be a triangle. So, we have four odd numbers and one even.
None of the numbers in the square can be a prime number, for else the two numbers it is connected two would have that prime number as a common divisor, and share an edge. 
So, each number in the square must share one prime factor with one neighbor, and a different prime factor with another, and since it does not share en edge with the third, those two numbers have different prime factors altogether, meaning that there are at least four odd prime factors involved among the four numbers in the square.
One of them could be multiplied by $2$, but maybe the singleton is the even number ... well, let's see what is possible:
With those four odd prime numbers being $3,5,7,11$, we could form:
$3\cdot 5$, $3 \cdot 7$, $5 \cdot 11$, and $7 \cdot 11$ ... sum is $168$, so leftover number is $18$, which has prime factor $3$ ... does not work. 
We can try to multiply one of the numbers by $2$, but not going over $186$ that can only be done for $3 \cdot 5$, meaning a sum of $183$ with a leftover of $5$ ... also does not work.
$3 \cdot 5$, $5 \cdot 7$, $3 \cdot 11$, $7 \cdot 11$ ... sum is $160$, so leftover is $26$ ... works!
Multiplying $3 \cdot 5$ by $2$ gives leftover $11$ .. does not work
$3 \cdot 7$, $3 \cdot 11$, $5 \cdot 7$, $5 \cdot 11$ ... sum is $144$ ... leftover $42$ ... does not work
Multiplying by $2$ gives leftovers $21$, $9$, and $7$ ... none work
Taking our four odd primes as $3,5,7,13$:
$3\cdot 5$, $3 \cdot 7$, $5 \cdot 13$, $7 \cdot 13$ ... sum is $193$ ... too high
$3 \cdot 5$, $3 \cdot 13$, $5\cdot 7$, $7 \cdot 13$ ... sum $180$ ... leftover $6$ ... does not work
$3 \cdot 7$, $3 \cdot 13$, $5 \cdot 7$, $5 \cdot 13$ ... sum $160$ ... leftover $26$ ... does not work
multiplying by $2$: only $3 \cdot 7$ possible, but gives leftover of $5$ does not work.
Any other combination of four odd primes gives sum above $186$.
So: only one combination possible: the square is $15, 33, 77, 35$, and the singleton is $26$
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