You can
put this solution on YOUR website!this problem is somewhat analogous to counting in binary
decimal-binary __ 0-0, 1-1, 2-10, 3-11, 4-100, 5-101, 6-110, 7-111, 8-1000
if there are 2 snaps after the starting snap, it takes 3 sec
__ if there are 3 snaps after the starting snap, it takes 7 sec
if n is the number of snaps after the starting snap, then 2^n-1 is the number of seconds
2^n-1=31536000 __ adding 1 __ 2^n=31536001 __ taking log __ n[log(2)]=log(31536001)
n=[log(31536001)]/[log(2)] __ n=24.91 (approx)
this is rounded down (truncated) to 24 because 25 would take more than a year
__ remembering to add in the starting snap, gives 25 snaps in a year
