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 $9$.
A: We have the following:

Taking a single elimination torunament with $m$ partecipants, with $2^n < m \leq 2^{n+1}$ for some $n\in \mathbb N$, the number of rounds is exactly $n+1$. 

By induction: if $1 < m \leq 2$ this is obvious.
Let's take $m$ such that $2^{n+1}<m\leq 2^{n+2}$. After the first round we have two cases: 


*

*$m$ is even, then it remains $\frac{m}{2}$ players and $2^n<\frac{m}{2}\leq 2^{n+1}$ and by indution hypotesis we are done. 

*$m$ is odd, then it remains $\frac{m+1}{2}$ players. Since $m$ is odd, then is still valid $m+1\leq 2^{n+2}$ and then $2^{n}<\frac{m+1}{2}\leq 2^{n+1}$ and again by induction hypotesis we are done.
So in the case $351$ players, you get $2^8<351\leq 2^9$ so you have $9$ rounds.
A: Technically, in a single elimination tournament, you provide byes or passes to a certain number of players before round 1 who will directly compete in round 2.
let there be N no. of participants, then the no. of byes will be the next power of two bigger than N minus the no. of participants.
In this case, the no. of participants is $351$ and the next bigger power if 2 would be $512$. Hence bye will be given to $161$ people which means $190$ people would participate in round 1.
This would mean that the no. of people participating in round 2 would be $95$ from round 1 and $161$ from byes which would add up to $256$, hence a perfect power of two.
Therefore the total number of rounds will be $9$.
A: An alternative perspective is to consider "ghosts" who increase the # of participants up to $\;2^9 = 512.\;$  Then, instead of anyone ever getting a bye, if there is no human to play, they play one of the ghosts.  
Anyone who plays a ghost always wins.  It's easy to see that when the # of participants is $\;2^n,\;$ then after a single round is played the # of participants is reduced to $\;2^{(n-1)}.$
$\underline{\textbf{Addendum}}$
I agree with joriki's reaction to my answer.  Also, I just realized that I should have considered another question.  It is clear from the analysis in my answer that if the # of participants is <= $2^n$ then $n$ rounds are sufficient.  However, it still needs to be proven that if the # of participants is > $2^n,$ then $n$ rounds are not sufficient.
This is easily demonstrated by induction. 
If there are more than 2 players, then 1 round is not sufficient.
Inductively assume that if there are more than $2^N$ players, then $N$ rounds are not sufficient.
Suppose that there are more than $2^{(N+1)}$ players.  Then after 1 round, there will still be more than $2^N$ players.  Based on the inductive assumption, that means that the initial round + an additional $N$ rounds will be insufficient.
