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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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11 Aug 2019, 04:42
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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

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Director
Joined: 08 Aug 2017
Posts: 732
Re: The sum of five consecutive integers is 35. How many of the five conse  [#permalink]

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11 Aug 2019, 04:47
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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11 Aug 2019, 05:10
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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11 Aug 2019, 05:16
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.
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Re: The sum of five consecutive integers is 35. How many of the five conse  [#permalink]

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11 Aug 2019, 06:38
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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11 Aug 2019, 09:39
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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Posts: 1253
The sum of five consecutive integers is 35. How many of the five conse  [#permalink]

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11 Aug 2019, 11:16
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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11 Aug 2019, 11:39
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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14 Aug 2019, 16:46
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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.

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Re: The sum of five consecutive integers is 35. How many of the five conse   [#permalink] 14 Aug 2019, 16:46

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

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