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 $45?$

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

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

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.

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