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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) Thanks, Shreya
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Re: If n > 2, then the sum, S, of the integers [#permalink]
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07 Jun 2012, 05:41
S does not necessarily have to be even. For example, when n=5, you have:
S = [(5)(5+1)]/2 = 15, an odd number



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Re: If n > 2, then the sum, S, of the integers [#permalink]
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07 Jun 2012, 05:49
Hi,
S = n(n + 1)/2, n > 2 for ex, n = 3, S = 6 n=5, S = 15 = 3*5 and so on
Now, if n is even, then, n = 2k, S = k(2k+1)....(a)
if n is odd, then, n = 2k+1, S = (2k+1)(k+1)...(b)
Depending on value of k (integer), S can be odd/even But for n>2, it will always be a product of two numbers....(from (a) & (b))
Thus, answer is (D)
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Re: If n > 2, then the sum, S, of the integers from 1 through n [#permalink]
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07 Jun 2012, 14:26
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.
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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!



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Re: If n > 2, then the sum, S, of the integers from 1 through n [#permalink]
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06 Nov 2012, 04:56



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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



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Re: If n > 2, then the sum, [#permalink]
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06 Jan 2014, 21:49
akankshasoneja 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. 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.
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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



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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 6k1 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? Thanks Cheers J Bumpinggg



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