Q:

How many rounds (not matches) must there be in a fair single elimination tournament with 351 participants?

Is it possible to give a sketch/outline of how this problem is approached and not just the answer?
The way I attempted to solve this problem was by matching up the number of participants with each other for a match. So for round 1: 350/2 = 175 matches with 1 person getting a free pass. Since there are 175 matches that means there are 175 losers since they have a 1-1 correspondence, leaving 176 remaining contestants (351-175). I basically repeated the same logic from here on out.
Round 2: 176/2 = 88 matches with no people passing free.
round 3: 88/2 = 44 matches
round 4: 44/2 = 22 matches
round 5: 22/2 = 11 matches
round 6: 10/2 = 5 with 1 person getting a pass
round 7: 4/2 = 2 matches with 1 person getting a pass.
round 8: 2/2 = 1 match with no passes. This leaves the 3 people that got passes left, meaning there are an additional 2 rounds? Adding up to a total of 10 rounds?
I don't think this is the right solution and it also feels pretty brute force, any help would be appreciated! 

A:

Normally one gives byes only in the first round.  You give the right number of byes to reduce to the next power of $2$, so with $351$ competitors you would have $95$ matches in the first round and $161$ byes, reducing the count to $256$.  Then eight more rounds for a total of $9$.  
In your computation, there are six people entering round $7$ so you have three entering round $8$ and two for round $9$.

