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The sum of five consecutive integers is 35. How many of the five conse

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The sum of five consecutive integers is 35. How many of the five conse  [#permalink]

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New post 11 Aug 2019, 05:42
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Re: The sum of five consecutive integers is 35. How many of the five conse  [#permalink]

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New post 11 Aug 2019, 05:47
n+n+1+n+2+n+3+n+4= 35
n=5.
Two prime numbers out of 5.
C is correct.
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Re: The sum of five consecutive integers is 35. How many of the five conse  [#permalink]

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New post 11 Aug 2019, 06:10
gvij2017 wrote:
n+n+1+n+2+n+3+n+4= 35
n=5.
Two prime numbers out of 5.
C is correct.


n is not 5 but 7.

Let the numbers be, n, n+1, n-1, n+2 and n-2; adding them we get=> n*5=35, so n is 7,

Numbers: 7,8,6,9,5; Primes: 5 and 7.
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Re: The sum of five consecutive integers is 35. How many of the five conse  [#permalink]

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New post 11 Aug 2019, 06:16
Bunuel wrote:
The sum of five consecutive integers is 35. How many of the five consecutive integers are prime numbers?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4



Let the five consecutive integers be \((n-2)\), \((n-1)\), \(n\), \((n+1)\), \((n+2)\).

Sum of the five consecutive integers is 35
\(\implies (n-2) + (n-1) + n + (n+1) + (n+2) = 35\)
\(\implies n = 7\)
\(\implies\) The five consecutive numbers are \(5\), \(6\), \(7\), \(8\), \(9\).

There are two prime numbers, \(5\) and \(7\), among them.
Answer is C.
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Re: The sum of five consecutive integers is 35. How many of the five conse  [#permalink]

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New post 11 Aug 2019, 07:38
LeoN88 wrote:
gvij2017 wrote:
n+n+1+n+2+n+3+n+4= 35
n=5.
Two prime numbers out of 5.
C is correct.


n is not 5 but 7.

Let the numbers be, n, n+1, n-1, n+2 and n-2; adding them we get=> n*5=35, so n is 7,

Numbers: 7,8,6,9,5; Primes: 5 and 7.


Both approaches are correct and yield the same result

First Approach: -
If we assume numbers as n, n+1, n+2, n+3 and n+4 then
n+n+1+n+2+n+3+n+4 = 5n+10 = 35
n=25/5 = 5

Numbers are 5,6,7,8 & 9
5 & 7 are prime

IMO C

Second Approach: -
If we assume numbers as n, n+1, n+2, n-1 and n-2 then
n+n+1+n+2+n-1+n-2 = 5n = 35
n=35/5 = 7

Numbers are 5,6,7,8 & 9
5 & 7 are prime

IMO C

But the end result of both approaches is the same.

It is better to assume 5 consecutive integers as n-2 ,n-1, n, n+1, n+2 since it eases the calculations in most cases.
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Re: The sum of five consecutive integers is 35. How many of the five conse  [#permalink]

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New post 11 Aug 2019, 10:39
Bunuel wrote:
The sum of five consecutive integers is 35. How many of the five consecutive integers are prime numbers?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


given
5x+10=35
x=5
so series ; 5,6,7,8,9
2 prime
IMO C
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The sum of five consecutive integers is 35. How many of the five conse  [#permalink]

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New post 11 Aug 2019, 12:16
Bunuel wrote:
The sum of five consecutive integers is 35. How many of the five consecutive integers are prime numbers?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


mean=35/5=7
sequence=5,6,7,8,9
2
C
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Re: The sum of five consecutive integers is 35. How many of the five conse  [#permalink]

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New post 11 Aug 2019, 12:39
Bunuel wrote:
The sum of five consecutive integers is 35. How many of the five consecutive integers are prime numbers?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


5+6+7+8+9 is 35. 2 numbers are prime in this.
IMO C.
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Re: The sum of five consecutive integers is 35. How many of the five conse  [#permalink]

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New post 14 Aug 2019, 17:46
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Bunuel wrote:
The sum of five consecutive integers is 35. How many of the five consecutive integers are prime numbers?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4


The sum of 5, 6, 7, 8, and 9 is 35. Of these five numbers, 5 and 7 are prime numbers.

Answer: C
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Re: The sum of five consecutive integers is 35. How many of the five conse   [#permalink] 14 Aug 2019, 17:46
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