On each side of an equilateral triangle with side length $n$ units, where $n$ is an integer, 1 $\le$ $n$ $\le$ 100, consider $n$ – 1 points that divide the side into $n$ equal segments. Through these points, draw lines parallel to the sides of the triangle, obtaining $a$ net of equilateral triangles of side length one unit. On each of the vertices of these small triangles, place $a$ coin head up. Two coins are said to be adjacent if the distance between them is 1 unit. A move consists of flip$\text{ping}$ over any three mutually adjacent coins. Find the number of values of $n$ for which it is possible to turn all coins tail up after $a$ finite number of moves.

Step 1: Total number of vertices

Step 2: ( $n$ + 1)( $n$ + 2)

Step 3: Let Head be donate by ‘0’ and tail by ‘1’.

Step 4: => initial sum of all vertices = 0.

Step 5: After each move three ‘0’ becomes three ‘1’.

Step 6: Let after kth move the sum be f(k).

Step 7: => f(k + 1) = f(k) + 3

Step 8: ( $n$ + 1)( $n$ + 2)

Step 9: Finally number of tails sum =

Step 10: But initially it was zero.

Step 11: Hence 3 divides

Step 12: ( $n$ + 1)( $n$ + 2)

Step 13: => $n$ = 3k + 1 or 3k + 2.

Step 14: Number of such $n$ which are less then 100 are 67.