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N is the sum of five consecutive integers, and when these five integer

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N is the sum of five consecutive integers, and when these five integer  [#permalink]

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New post 14 Aug 2018, 01:18
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N is the sum of five consecutive integers, and when these five integers are arranged in ascending order, the sum of first three integers is 0. What is the product of last three integers, when these five integers are arranged in descending order, from highest to lowest?

A. -1
B. 0
C. 1
D. 2
E. 6
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Re: N is the sum of five consecutive integers, and when these five integer  [#permalink]

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New post 14 Aug 2018, 03:07
amanvermagmat wrote:
N is the sum of five consecutive integers, and when these five integers are arranged in ascending order, the sum of first three integers is 0. What is the product of last three integers, when these five integers are arranged in descending order, from highest to lowest?

A. -1
B. 0
C. 1
D. 2
E. 6


Let the 5 consecutive integers be x-2,x-1,x,x+1, and x+2 , in ascending order.

Given, x-2+x-1+x=0
Or, 3x-3=0
Or, x=1

So, the five consecutive integers are -1,0,1,2, and 3.

Arranging them in descending order: 3,2,1,0, and -1.

Now, product of last 3 integers=1*0*(-1)=0.

Ans. (B)
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N is the sum of five consecutive integers, and when these five integer  [#permalink]

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New post 14 Aug 2018, 03:13
amanvermagmat wrote:
N is the sum of five consecutive integers, and when these five integers are arranged in ascending order, the sum of first three integers is 0. What is the product of last three integers, when these five integers are arranged in descending order, from highest to lowest?

A. -1
B. 0
C. 1
D. 2
E. 6


Given: The sum of the first 3 integers is zero.

The 3 integers are 0 and 2 integers which are equal and have an opposite magnitude(-1 and 1)

Since we are talking about 5 such consecutive integers, the numbers in the sequence are -1 0 1 2 3.

Therefore, the product of the last three integers(when the integers are arranged in descending order) are -1*0*1 or 0(Option B)
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Re: N is the sum of five consecutive integers, and when these five integer  [#permalink]

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New post 14 Aug 2018, 03:29
amanvermagmat wrote:
N is the sum of five consecutive integers, and when these five integers are arranged in ascending order, the sum of first three integers is 0. What is the product of last three integers, when these five integers are arranged in descending order, from highest to lowest?

A. -1
B. 0
C. 1
D. 2
E. 6


Since there are 5 consecutive integers and sum of first three integers is 0 when arranged in ascending order. Hence, 1st three integers must be -1,0,1
So, five integers in ascending order are -1,0,1,2,3
Five integers in descending order are 3,2,1,0,-1

Product of last 3 integers = -1*0*1 = 0

Answer B
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Re: N is the sum of five consecutive integers, and when these five integer  [#permalink]

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New post 14 Aug 2018, 07:16
amanvermagmat wrote:
N is the sum of five consecutive integers, and when these five integers are arranged in ascending order, the sum of first three integers is 0. What is the product of last three integers, when these five integers are arranged in descending order, from highest to lowest?

A. -1
B. 0
C. 1
D. 2
E. 6


\((a - 1) + a + ( a + 1 ) = 0\)

So, \(3a = 0\)

Or, \(a = 0\)

N = { ( a - 1 ), a , ( a + 1) , ( a + 2) , ( a + 3)}

Sum of first \(3\) term is \(0\) and \(a = 0\) so, Sum of the numbers are \(-1 , 0 , 1 , 2 , 3\)

Now, the product of last three integers, when these five integers are arranged in descending order, from highest to lowest is -1*0*1 = 0 ; Answer must be (B)
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Re: N is the sum of five consecutive integers, and when these five integer   [#permalink] 14 Aug 2018, 07:16
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