"Suppose that Ingrid has $63$ chocolates. In every step, she meets two friends and divides her pile three ways between herself and her two friends. She divides the chocolates as evenly as possible, but in the event that the chocolates can not be divided perfectly evenly into three piles, she takes the larger pile for herself. How many steps until she is down to one chocolate?"
My logic :
1) Between which powers of 3 is 63 located
2) Low: $3^3 = 27$
3) High: $3^4 = 81$
So for me, I would pick $27 (3$ steeps$)$ instead of $81 (4$ steeps$)$ because she only has $63$ chocolates. $27$ and the leftover $(63-27)$ is for Ingrid.
However, the answer is $4$, anyone willing to explain me? thank you!
After three steps, she hasn't gotten down to one chocolate (she's got 21 after the first, 7 after the second, and 3 after the third), so it takes a fourth for her to get there.