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If n > 2, then the sum, S, of the integers from 1 through n [#permalink]

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07 Jun 2012, 04:53

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If n > 2, then the sum, S, of the integers from 1 through n can be calculated by the following formula: S = n(n + 1)/2. Which one of the following statements about S must be true?

A. S is always odd. B. S is always even. C. S must be a prime number D. S must not be a prime number E. S must be a perfect square

Can you please explain between B & D. Both need to be correct in order for the question to be valid right ?

(S needs to be even to be divisible by 2 & S shouldn't be a prime number)

If n > 2, then the sum, S, of the integers from 1 through n can be calculated by the following formula: S = n(n + 1)/2. Which one of the following statements about S must be true? A. S is always odd. B. S is always even. C. S must be a prime number D. S must not be a prime number E. S must be a perfect square

Notice that we are asked "which of the following MUST be true, not COULD be true. For such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

A. S is always odd --> not necessarily true if n=3 then 1+2+3=6=even. B. S is always even --> not necessarily true if n=5 then 1+2+3+4+5=15=odd. C. S must be a prime number --> not true if n=3 then 1+2+3=6=not prime. E. S must be a perfect square --> not necessarily true if n=3 then 1+2+3=6=not a perfect square.

Re: If n > 2, then the sum, S, of the integers from 1 through n [#permalink]

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06 Nov 2012, 04:43

Correct me if I am not right, but since n > 2, S is always even since odd * even = even and 2 is the only even prime number S can never be a prime number!

Correct me if I am not right, but since n > 2, S is always even since odd * even = even and 2 is the only even prime number S can never be a prime number!

No, that's not correct. If S is always even, then B must also be correct. But if n=5 then 1+2+3+4+5=15=odd. _________________

Re: If n > 2, then the sum, S, of the integers from 1 through n [#permalink]

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06 Jan 2013, 01:05

shreya717 wrote:

If n > 2, then the sum, S, of the integers from 1 through n can be calculated by the following formula: S = n(n + 1)/2. Which one of the following statements about S must be true?

A. S is always odd. B. S is always even. C. S must be a prime number D. S must not be a prime number E. S must be a perfect square

Can you please explain between B & D. Both need to be correct in order for the question to be valid right ?

(S needs to be even to be divisible by 2 & S shouldn't be a prime number)

Thanks, Shreya

S = [n(n+1)]/2 for n>2 S should be divisible by either n or n+1 (for n = odd S is divisible by n and for n=even S is divisible by n+1) so it cannot be a prime no. Answer: D

If n > 2, then the sum, S, of the integers from 1 through n can be calculated by the following formula: S = n(n + 1)/2. Which one of the following statements about S must be true? (A) S is always odd. (B) S is always even. (C) S must be a prime number. (D) S must not be a prime number. (E) S must be a perfect square.

If n > 2, then the sum, S, of the integers from 1 through n can be calculated by the following formula: S = n(n + 1)/2. Which one of the following statements about S must be true? (A) S is always odd. (B) S is always even. (C) S must be a prime number. (D) S must not be a prime number. (E) S must be a perfect square.

Though i agree that the OA is right but even option B should be correct.

Put n = 5

S = 5*6/2 = 15 S is not always even. It may be even, it may be odd. If the even integer (out of n and n+1) is not a multiple of 4, then S will be odd. _________________

If n > 2, then the sum, S, of the integers from 1 through n can be calculated by the following formula: S = n(n + 1)/2. Which one of the following statements about S must be true? (A) S is always odd. (B) S is always even. (C) S must be a prime number. (D) S must not be a prime number. (E) S must be a perfect square.

Re: If n > 2, then the sum, S, of the integers from 1 through n [#permalink]

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07 Jan 2014, 23:54

S= n (n+1) /2 , Either n or n+1 , is even & also n > 2, Thus after dividing by 2, S can be shown to be a product of two distinct numbers (not including 1) ----> S can never be prime . So D it is

Re: If n > 2, then the sum, S, of the integers from 1 through n [#permalink]

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16 Feb 2014, 09:43

Bunuel wrote:

If n > 2, then the sum, S, of the integers from 1 through n can be calculated by the following formula: S = n(n + 1)/2. Which one of the following statements about S must be true? A. S is always odd. B. S is always even. C. S must be a prime number D. S must not be a prime number E. S must be a perfect square

Notice that we are asked "which of the following MUST be true, not COULD be true. For such kind of questions if you can prove that a statement is NOT true for one particular set of numbers, it will mean that this statement is not always true and hence not a correct answer.

A. S is always odd --> not necessarily true if n=3 then 1+2+3=6=even. B. S is always even --> not necessarily true if n=5 then 1+2+3+4+5=15=odd. C. S must be a prime number --> not true if n=3 then 1+2+3=6=not prime. E. S must be a perfect square --> not necessarily true if n=3 then 1+2+3=6=not a perfect square.

Only choice D is left.

Answer: D.

Does anyone know why can't the sum be a prime number?

So I began trying to understand this. First since all prime numbers greater than 3 are of the form 6k+1 or 6k-1 Now then let's take 1+6k, that means that 2+3+4......+n cannot be a multiple of 6, but i'm trying to figure out why this can't be true?

Re: If n > 2, then the sum, S, of the integers from 1 through n [#permalink]

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05 Jul 2015, 19:42

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