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07 Aug 2026, 04:47
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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 \(\frac{1}{3}\) 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
A. 35
B. 36
C. 37
D. 38
E. 45
07 Aug 2026, 04:47
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Official Solution:
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 \(\frac{1}{3}\) 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
Start with the possible ways to get a score of 36.
• If an archer attempts 36 targets and hits all of them, the score is 36.
Targets attempted: 36
• If an archer hits 37 targets, the score is 1 point too high. To reduce the score by 1 point, the archer must miss 3 targets, since each miss reduces the score by \(\frac{1}{3}\) point.
Targets attempted: \(37 + 3 = 40\)
• If an archer hits 38 targets, the score is 2 points too high. To reduce the score by 2 points, the archer must miss 6 targets.
Targets attempted: \(38 + 6 = 44\)
• If an archer hits 39 targets, the score is 3 points too high. To reduce the score by 3 points, the archer must miss 9 targets.
Targets attempted: \(39 + 9 = 48\)
So the possible numbers of targets attempted are:
\(36, 40, 44, 48, \ldots\)
Each time the number of hits increases by 1, the number of misses must increase by 3, so the number of targets attempted increases by 4.
Therefore, the possible numbers of targets attempted to get a score of 36 are the multiples of 4 from 36 through 180, inclusive:
\(36, 40, 44, \ldots, 180\)
Since no two archers in the group shot at the same number of targets, each value in this list can correspond to at most one archer. Therefore, the maximum possible number of archers in the group is the number of multiples of 4 from 36 through 180, inclusive:
\(\frac{180 - 36}{4} + 1 = 37\)
Answer: C
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 \(\frac{1}{3}\) 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
Start with the possible ways to get a score of 36.
• If an archer attempts 36 targets and hits all of them, the score is 36.
Targets attempted: 36
• If an archer hits 37 targets, the score is 1 point too high. To reduce the score by 1 point, the archer must miss 3 targets, since each miss reduces the score by \(\frac{1}{3}\) point.
Targets attempted: \(37 + 3 = 40\)
• If an archer hits 38 targets, the score is 2 points too high. To reduce the score by 2 points, the archer must miss 6 targets.
Targets attempted: \(38 + 6 = 44\)
• If an archer hits 39 targets, the score is 3 points too high. To reduce the score by 3 points, the archer must miss 9 targets.
Targets attempted: \(39 + 9 = 48\)
So the possible numbers of targets attempted are:
\(36, 40, 44, 48, \ldots\)
Each time the number of hits increases by 1, the number of misses must increase by 3, so the number of targets attempted increases by 4.
Therefore, the possible numbers of targets attempted to get a score of 36 are the multiples of 4 from 36 through 180, inclusive:
\(36, 40, 44, \ldots, 180\)
Since no two archers in the group shot at the same number of targets, each value in this list can correspond to at most one archer. Therefore, the maximum possible number of archers in the group is the number of multiples of 4 from 36 through 180, inclusive:
\(\frac{180 - 36}{4} + 1 = 37\)
Answer: C
