2019 AMC 12A Problem 4
Attempt Problem 4 of the 2019 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2019 AMC 12A solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
4.
What is the greatest number of consecutive integers whose sum is
Answer: D
Small hint:
The integers are allowed to be negative
Big hint:
Adding contributes so extend the run just past
Solution:
Negative integers are allowed. The integers from to sum to so the integers from to sum to
This run has integers. Conversely, if consecutive integers have sum then twice their sum is times an integer, so is a multiple of Hence proving that this run is longest.
Thus, the correct answer is D.
Problem 4 in other years
- 1950 AMC 12
- 1951 AMC 12
- 1952 AMC 12
- 1953 AMC 12
- 1954 AMC 12
- 1955 AMC 12
- 1956 AMC 12
- 1957 AMC 12
- 1958 AMC 12
- 1959 AMC 12
- 1960 AMC 12
- 1961 AMC 12
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- 1963 AMC 12
- 1964 AMC 12
- 1965 AMC 12
- 1966 AMC 12
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- 1977 AMC 12
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