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

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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