/sci/ - Science & Math

>>14860639
we’re talking about 7 numbers a1…a7 of which three are 0 mod 3, two are 1 mod 3 and two are 2 mod 3. If it is a solution then we also know [math]a_n\equiv a_{n+4}[/math] for n=1,2,3. So a_4 has to be 0 mod 3. Now the other three in any run of four have to be one 0, one 1, and one 2. In other words the whole sequence mod 3 is uniquely determined by a1,a2,a3 mod 3 and you can see any such arrangement will give rise to a solution. Therefore the answer is

( # of those a1 a2 a3 arrangements mod 3) x ( # of arrangements of the 0s) x (# arrangements of the 1s) x (# arrangements of the 2s)
[math]= 3!3!2!2! = 144[/math]