Bunuel
In a league of 8 teams, each team played every other team 10 times. The number of wins of the 8 teams formed an arithmetic sequence. What is the least possible number of games won by the champion?
A. 41
B. 42
C. 43
D. 44
E. 45
Are You Up For the Challenge: 700 Level Questions: 700 Level QuestionsTotal matches if every team plays other team once = 8C2 = 8*7/2 = 28
So if every team plays other team 10 times, then no of matches = 280
Now, no. of wins needs to be in arithmetic sequence and we need to find least possible largest number in this sequence, which means given a sum, we need to maximize the smallest number in the sequence.
\(S = \frac{n}{2}*(2a + (n-1)*d)\)
\(280 = \frac{8}{2}*(2a + 7d)\)
\(70 = 2a + 7d\)
\(a = \frac{70 - 7d}{2}\)
We need to maximize a, so we need to minimize d, so we will start with d = 1 and see when we get an integer value of a.
d = 1, a = 63/2 (Not possible)
d = 2, a = 28
So last term will be => a + 7d => 28 + 14 => 42
IMO: B