# Happy birthday Ramanujan!

On December 22 2019, Ramanujan would have been 132 years old. In his memory here are two puzzles around 132.

In the six vertices of each of these graphs place six positive integers that add up to 132, and such that two vertices are joined by an edge if, and only if, they have a common divisor greater than 1 (that is, they are not relatively prime). • Can we prove that the construction is impossible? specifically the left one? – Omega Krypton Dec 26 '19 at 2:58
• @OmegaKrypton Why not? – Bernardo Recamán Santos Dec 26 '19 at 2:59
• in fact both are impossible !?? – Omega Krypton Dec 26 '19 at 3:03
• It's if and only if, so if two integers share a common divisor greater than 1, they should have an edge. – justhalf Dec 26 '19 at 9:32
• Don't give up friends! Both puzzles have (unique) solutions as confirmed by Freddy Barrera, their creator. – Bernardo Recamán Santos Dec 26 '19 at 11:41

For the first one, With A starting from the top, B and C in the second.

A = 9, B = 15, C = 21, D = 10, E = 70, F = 7

    A
B   C
D   E   F For the second one with A, C, E in the top row

A = 9, B = 21, C = 42, D = 35, E = 20, F = 5

A   C   E

B   D   F


dot files generated and verified with this and the problem was solved manually based on choosing the factors that connected nodes will share.

On both figures, the numbers are shown on the vertices. • gcd(4,16)=4>1, but there's no edge. – JMP Dec 27 '19 at 4:25
• @JMP. The question asked for a common factor greater than 1. – Vassilis Parassidis Dec 27 '19 at 4:44
• two vertices are joined by an edge if, and only if, they have a common divisor greater than 1 – JMP Dec 27 '19 at 4:52