Smarter solution to Medium Quant Practice P Jan 17,2022

GRE Quant Problem Solving
1

I think I figured out a smarter way to do a problem covered from a lecture: Medium Quant Practice Problem Jan 17, 2022. The question is below:

In a certain competition, there are 8 players. In every stage, each player still in the game has to play every other player still in the game in a head to head match. When a stage is over, half of the players are eliminated from the game. How many head to head matchups are needed until a winner is declared.

Answer choices:

  1. 16
  2. 32
  3. 35
  4. 128
  5. 256

Using combination solves the problem. But if you look carefully at the answer choices, there’s an odd one out, LITERALLY (#3: 35). Since it’s a head to head game, there has to be only 1 game at the last stage.

We know that even+even=even, even+odd=odd, and odd+odd=even (not relevant in this question). We also know that all factorials are even, except 1!

#3 is the correct answer. I thought this approach is simpler and am pretty proud of spotting it. But does this way of thinking have any drawbacks? Are there cases where this wouldn’t work?

2

OK,

Maybe it’s something to do with the chosen method, but what does factorials have to do here? Your solution is correct when the number of players is a power of 2, and this can be done by deducing that you need an even number of matches to reduce the number of players by half.

3

Factorials are relevant because the combination formula uses them. I was trying to think of a case where the formula would spit out an odd number