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Difference between revisions of "2019 AMC 10A Problems/Problem 5"

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Though impractical, a proof of maximality can proceed as follows:  Let the desired sequence of consecutive integers be <math>a, a+1, \cdots, a+(N-1)</math>, where there are <math>N</math> terms, and we want to maximize <math>N</math>.  Then the sum of the terms in this sequence is <math>aN + \frac{(N-1)(N)}{2}=45</math>.  Rearranging and factoring, this reduces to <math>N(2a+N-1) = 90</math>.  Since <math>N</math> must divide <math>90</math>, and we know that <math>90</math> is an attainable value of the sum, <math>90</math> must be the maximum.
 
Though impractical, a proof of maximality can proceed as follows:  Let the desired sequence of consecutive integers be <math>a, a+1, \cdots, a+(N-1)</math>, where there are <math>N</math> terms, and we want to maximize <math>N</math>.  Then the sum of the terms in this sequence is <math>aN + \frac{(N-1)(N)}{2}=45</math>.  Rearranging and factoring, this reduces to <math>N(2a+N-1) = 90</math>.  Since <math>N</math> must divide <math>90</math>, and we know that <math>90</math> is an attainable value of the sum, <math>90</math> must be the maximum.
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==Solution 2==
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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>.
  
 
==See Also==
 
==See Also==

Revision as of 23:10, 9 February 2019

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 $45?$

$\textbf{(A) } 9 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 90 \qquad\textbf{(E) } 120$

Solution

Note that every term in the sequence $-44, -43, \cdots, 44, 45$ cancels out except $45$. Thus, the answer is, intuitively, $\boxed{\textbf{(D) } 90 }$ integers.

Though impractical, a proof of maximality can proceed as follows: Let the desired sequence of consecutive integers be $a, a+1, \cdots, a+(N-1)$, where there are $N$ terms, and we want to maximize $N$. Then the sum of the terms in this sequence is $aN + \frac{(N-1)(N)}{2}=45$. Rearranging and factoring, this reduces to $N(2a+N-1) = 90$. Since $N$ must divide $90$, and we know that $90$ is an attainable value of the sum, $90$ 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 $\frac12$ if the middle two numbers are $0$ and $1$, so the answer is $\frac{45}{\frac12}=\boxed{\textbf{(D) } 90 }$.

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

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