The question gives us 2 constraints:
C1: A total score of exactly 36 point for each archer
C2: A total of 180 different targets
C3: No two archers shot the same number of targets
Now, lets assume the number of hits and misses be h and m respectively. Now according to the question
Eq1: h + m <= 180
Eq2: h - m/3 = 36
Now, lets say archer 1 shoots all on target. Then for archer 1, we get
Therefore: h = 36 and m = 0. Therefore Total = 36
and h + m = 36 (Valid)
Lets say Archer 2 misses 3 shots (Keep in mind that in order for h and m to be integers, we will need the misses to be in multiples of 3)
Therefore, from equation 2 we can get
h = 37 and m = 3. Therefore Total = 40
Now for Archer 3
m = 6
h = 38
=> Total shots = 44
This gives us a sequence of 36, 40, 44 .... 180 (As 180 is the total number of targets)
The number of terms in the sequence is the number of possible archers
In the AP we have a = 36, d = 4
For number of terms we can use the equation 36 + (n-1)4 = 180. Therefore n = 37
Hence option C
Bunuel
In an archery contest with 180 different targets, an archer may shoot at any number of these targets. For each target hit, the archer earns 1 point, and for each target missed, the archer loses 1/3 point. A group of archers participated in the contest, and no two archers in the group shot at the same number of targets. If every archer in the group finished with a score of exactly 36 points, what is the maximum possible number of archers in the group?
A. 35
B. 36
C. 37
D. 38
E. 45
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