If n 2, then the sum, S, of the integers from 1 through n can be cal

Question

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

Swoosh617 · Answer

S does not necessarily have to be even. For example, when n=5, you have:

S =  /2 = 15, an odd number

cyberjadugar · Answer

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)

Regards,

Bunuel · Answer

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.

KevinBrink · Answer

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!

Bunuel · Answer

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.

geneticsgene · Answer

S =  /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

KarishmaB · Answer

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.

joe26219 · Answer

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 :-D

jlgdr · Answer

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?

Thanks
Cheers
J

Bumpinggg

jainil7 · Answer

A. S is always odd:  False. (When $n=3$, $S=6$, which is even).
  B. S is always even:  False. (When $n=5$, $S=15$, which is odd).
  C. S must be a prime number:  False. (6, 10, 15, and 21 are all composite).
  E. S must be a perfect square:  False. (None of our test results are perfect squares).

If n > 2, then the sum, S, of the integers from 1 through n can be cal

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