SuicideFuel Problem

[trying to ascend]

A hexagon is divided into 3 equilateral triangles. In how many ways can the integers the from 1 to 6 be placed, without repetition, in a way that the sum of three adjacent triangles is always a multiple of 3?

 

[Pancakecel]

Who cares

 

[trying to ascend]

Who cares

Solution?

 

[InMinecraft]

Who cares

 

[NoBitches]

A hexagon is divided into 3 equilateral triangles. In how many ways can the integers the from 1 to 6 be placed, without repetition, in a way that the sum of three adjacent triangles is always a multiple of 3?

3-Not possible
6-Possible for 3 2 1 therfore giving 6 combinations possible[Im factoring in permutation eg(321) and(231) are distint]
9-6 21 531 432 so 18 possible ways
12- 651 642 543 so 18 possible ways
15-654 therefore only 6 possibilities
18-Not possible
6+18+18+6=48

 

[trying to ascend]

3-Not possible
6-Possible for 3 2 1 therfore giving 6 combinations possible[Im factoring in permutation eg(321) and(231) are distint]
9-6 21 531 432 so 18 possible ways
12- 651 642 543 so 18 possible ways
15-654 therefore only 6 possibilities
18-Not possible
6+18+18+6=48

Correct