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