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The sum of \(k\) consecutive integers is 41. If the least integer is -40, then \(k =\)

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by M7MBA » Sun Jan 24, 2021 1:02 am

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The sum of \(k\) consecutive integers is 41. If the least integer is -40, then \(k =\)

A. 40
B. 41
C. 80
D. 81
E. 82

Answer: E

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Re: The sum of \(k\) consecutive integers is 41. If the least integer is -40, then \(k =\)

by Scott@TargetTestPrep » Fri Feb 05, 2021 4:28 pm
M7MBA wrote: ↑
Sun Jan 24, 2021 1:02 am
 The sum of \(k\) consecutive integers is 41. If the least integer is -40, then \(k =\)

A. 40
B. 41
C. 80
D. 81
E. 82

Answer: E

Solution:

Since the least integer is -40 and the set is consecutive integers, we must have all the negative integers from -40 to -1 (inclusive), 0, and if we have all the positive integers from 1 to 40 (inclusive), the sum of all the integers would now be 0. Since the sum is 41, we must also have the integer 41. That is, we must have all the integers from -40 to 41, inclusive. Therefore, there are 41 - (-40) + 1 = 82 integers in the set.

Answer: E

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