Difference between revisions of "2019 AMC 10A Problems/Problem 5"
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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 <math>\frac12</math> if the middle two numbers are <math>0</math> and <math>1</math>, so the answer is <math>\frac{45}{\frac12}=\boxed{\textbf{(D) } 90 }</math>. | 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 <math>\frac12</math> if the middle two numbers are <math>0</math> and <math>1</math>, so the answer is <math>\frac{45}{\frac12}=\boxed{\textbf{(D) } 90 }</math>. | ||
| − | ==Solution | + | ===Solution 2a (more detailed ver. of Solution 2)=== |
Note that the sum of <math>n</math> consecutive integers whose mean (median; the two are equal in this case) is <math>a</math> is equal to <math>an</math>. Here, we want to maximize <math>n</math> such that <math>an=45</math>. The mean/median of a set of consecutive integers is either an integer or half of an integer; clearly, <math>n</math> is maximized when <math>a=0.5</math> and <math>n=\boxed{\textbf{(D) }90}</math>. | Note that the sum of <math>n</math> consecutive integers whose mean (median; the two are equal in this case) is <math>a</math> is equal to <math>an</math>. Here, we want to maximize <math>n</math> such that <math>an=45</math>. The mean/median of a set of consecutive integers is either an integer or half of an integer; clearly, <math>n</math> is maximized when <math>a=0.5</math> and <math>n=\boxed{\textbf{(D) }90}</math>. | ||
Latest revision as of 00:12, 3 November 2024
- The following problem is from both the 2019 AMC 10A #5 and 2019 AMC 12A #4, so both problems redirect to this page.
Contents
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 
.
Solution 2a (more detailed ver. of Solution 2)
Note that the sum of 
consecutive integers whose mean (median; the two are equal in this case) is 
is equal to 
. Here, we want to maximize such that 
. The mean/median of a set of consecutive integers is either an integer or half of an integer; clearly, is maximized when 
and 
.
~Technodoggo
Video Solution 1
https://youtu.be/ZhAZ1oPe5Ds?t=665
~ pi_is_3.14
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. 
