View source for 2019 AMC 10A Problems/Problem 5

{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #5]] and [[2019 AMC 12A Problems|2019 AMC 12A #4]]}}

==Problem==
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>

==Solution 1==
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.

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.

==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 <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 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>. 

~Technodoggo

==Solution 3==

If we want the answer to have as much terms as possible we need to make as much terms cancel out as much as we can.Because 45 - 1 < <math>\frac12</math>60 so we can eliminate E. Now lets make every integer cancel to zero except for 45 and then we account for 0 and 45 to get 44*2+1+1=<math>\boxed{\textbf{(D) }90}</math>. Feel free to edit and clarify.


~Twinotter


== Video Solution 1 ==
https://youtu.be/ZhAZ1oPe5Ds?t=665

~ pi_is_3.14

==Video Solution 2==
https://youtu.be/ivYp-eNOIZA

~savannahsolver

==See Also==

{{AMC10 box|year=2019|ab=A|num-b=4|num-a=6}}
{{AMC12 box|year=2019|ab=A|num-b=3|num-a=5}}
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