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The sum of k consecutive integers is 41. If the least intege

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The sum of k consecutive integers is 41. If the least intege [#permalink]

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New post 11 Mar 2014, 02:20
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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

I have tried solving this problem by using formula sum of k consecutive integers = k(k+1)/2

Sum = 41
But I am struggling to understand what role does the least integer play over here/. Can you please help?
[Reveal] Spoiler: OA

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Last edited by Bunuel on 11 Mar 2014, 02:35, edited 1 time in total.
Renamed the topic and edited the question.
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Re: The sum of k consecutive integers is 41. If the least intege [#permalink]

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enigma123 wrote:
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

I have tried solving this problem by using formula sum of k consecutive integers = k(k+1)/2

Sum = 41
But I am struggling to understand what role does the least integer play over here/. Can you please help?


The sum of k consecutive integers is 41 and the least integer is -40.

So, we have the following sum: (-40) + (-39) + ... = 41. The sum to be positive you must have positive numbers in the sequence as well. Each positive number and its mirror negative number will give the sum of 0, for example, -1+1=0, -2+2=0, -3+3=0, -40+40=0. Since the sum is 41, then the greatest number must be 41:

(-40) + (-39) + ... + (-1) + 0 + 1 + ... + 39 + 40 + 41 = 41 (the sum of red part is 0 + 41 = 41).

How many terms do we have? 40 negative numbers, 0 and 41 positive numbers, so total of 82.

Answer: E.
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Re: The sum of k consecutive integers is 41. If the least intege [#permalink]

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New post 11 Mar 2014, 06:13
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enigma123 wrote:
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

I have tried solving this problem by using formula sum of k consecutive integers = k(k+1)/2

Sum = 41
But I am struggling to understand what role does the least integer play over here/. Can you please help?


I think bunuel has explain it very nice !

-40........0.........41......total 82 integers.. -40 to 1-----40+1(0)+41(1 to 41)=82 ...
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Re: The sum of k consecutive integers is 41. If the least intege [#permalink]

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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


Consecutive Integers ---The difference is consistent with each integers , therefore the series can be A.P .

Sum of A.P. = A + (N-1 ) D

A= First term
D= Difference between each integer
N=number of terms

Sum = A + (N - 1 ) D
41= -40 + N -1
82 = N

CONSIDER N=K

Answer = E
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Re: The sum of k consecutive integers is 41. If the least intege [#permalink]

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New post 07 Jul 2016, 21:10
Hello.
we can use direct formula = Sn =n/2(2a+(n-1)d
here sn =41, n =?( we have to find) a=-40 & d=1

after applying given value, we will get n=82
so correct answer sis E
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Re: The sum of k consecutive integers is 41. If the least intege [#permalink]

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New post 27 Oct 2016, 13:02
This is another good problem where it is helpful to visualize the number line:

We have -40 off in space on the left hand side of the line and we don't know how far out it goes on the right, but we do know that the sum is a positive number and its 41. Right away, you should think to yourself "okay, if I go out to 40 on the right hand side of the number line, everything on the left hand side will cancel it out." Therefore, if we add one more number, 41, on the right hand side we arrive at our value of 41 for the sum & the number of values on the line altogether is 41+40+1 = 82
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The sum of k consecutive integers is 41. If the least intege [#permalink]

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New post 22 Jan 2018, 19:32
enigma123 wrote:
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

I have tried solving this problem by using formula sum of k consecutive integers = k(k+1)/2

Sum = 41
But I am struggling to understand what role does the least integer play over here/. Can you please help?


The Sum of an AP Series is \((n/2)[2a+(n-1)d]\).. Given that \(d = 1\) & \(Sum = 41\), \(a = -40\)
Solving above we get, \(n(n-81) = 82\).. No need to solve further.. n = 82
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Re: The sum of k consecutive integers is 41. If the least intege [#permalink]

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New post 22 Jan 2018, 23:52
enigma123 wrote:
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

I have tried solving this problem by using formula sum of k consecutive integers = k(k+1)/2

Sum = 41
But I am struggling to understand what role does the least integer play over here/. Can you please help?



(-40 , -39 , -38.............-3 , -2 , - 1 , 0 , 1 , 2 , 3 ..............38 , 39 , 40 ) 41

The result of the highlighted part will be 0 , and we have 81 Numbers in this range (-40 to 40 including 0)

So, The answer will be (E) 82

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Re: The sum of k consecutive integers is 41. If the least intege [#permalink]

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New post 24 Jan 2018, 09:37
enigma123 wrote:
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


If the least (or first) integer is -40, we see that the sum of the consecutive integers from -40 to 40 will all cancel to zero, and if we include 41 as the final integer in the set, then the sum will be 41.

Thus, there are 41 - (-40) + 1 = 82 integers in total.

Answer: E
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Re: The sum of k consecutive integers is 41. If the least intege   [#permalink] 24 Jan 2018, 09:37
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