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

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==Problem 5==
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{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #5]] and [[2019 AMC 12A Problems|2019 AMC 12A #4]]}}
What is the greatest number of consecutive integers whose sum is <math>45 ?</math>
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==Problem==
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What is the greatest number of consecutive integers whose sum is <math>45?</math>
  
 
<math>\textbf{(A) } 9 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 90 \qquad\textbf{(E) } 120</math>
 
<math>\textbf{(A) } 9 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 90 \qquad\textbf{(E) } 120</math>
  
==Solution==
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==Solution 1==
Note that every term in the sequence <math>-44, -43..., 44, 45</math> cancels out except <math>45</math>. This results in <math>\boxed{\textbf{(D) } 90 }</math> integers.
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We might at first think that the answer would be <math>9</math>, because <math>1+2+3 \dots +n = 45</math> when <math>n = 9</math>. But note that the problem says that they can be integers, not necessarily positive. Observe also that every term in the sequence <math>-44, -43, \cdots, 44, 45</math> cancels out except <math>45</math>. Thus, the answer is, intuitively, <math>\boxed{\textbf{(D) } 90 }</math> integers.
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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.
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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>.
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== Video Solution 1 ==
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https://youtu.be/ZhAZ1oPe5Ds?t=665
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~ pi_is_3.14
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==Video Solution 2==
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https://youtu.be/ivYp-eNOIZA
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~savannahsolver
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==See Also==
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{{AMC10 box|year=2019|ab=A|num-b=4|num-a=6}}
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{{AMC12 box|year=2019|ab=A|num-b=3|num-a=5}}
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{{MAA Notice}}

Revision as of 22:11, 17 January 2021

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 1

We might at first think that the answer would be $9$, because $1+2+3 \dots +n = 45$ when $n = 9$. But note that the problem says that they can be integers, not necessarily positive. Observe also 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 $\frac{45}{\frac12}=\boxed{\textbf{(D) } 90 }$.

Video Solution 1

https://youtu.be/ZhAZ1oPe5Ds?t=665

~ pi_is_3.14

Video Solution 2

https://youtu.be/ivYp-eNOIZA

~savannahsolver

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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