2019 AMC 10A Problems/Problem 5
- The following problem is from both the 2019 AMC 10A #5 and 2019 AMC 12A #4, so both problems redirect to this page.
Problem
What is the greatest number of consecutive integers whose sum is
Solution 1
We might at first think that the answer would be , because when . But note that the problem says that they can be integers, not necessarily positive. Observe also that every term in the sequence cancels out except . Thus, the answer is, intuitively, integers.
Though impractical, a proof of maximality can proceed as follows: Let the desired sequence of consecutive integers be , where there are terms, and we want to maximize . Then the sum of the terms in this sequence is . Rearranging and factoring, this reduces to . Since must divide , and we know that is an attainable value of the sum, must be the maximum.
Solution 2
To maximize the number of integers, we need to make the average of them as low as possible while still being positive. The average can be if the middle two numbers are and , so the answer is .
Video Solution 2
~savannahsolver
See Also
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 3 |
Followed by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.