If x = -1 and if n is the sum of the first 404 prime numbers

Question

If x = -1 and if n is the sum of the first 404 prime numbers, then  x + x^n + x^{n+1} + x^{n+2}  =

A. -4
B. -2
C. 0
D. 1
E. 4

gmatophobia · Accepted Answer

Great question!

The first 404 prime numbers have one even prime and 403 odd primes.

The sum of 403 odd primes = odd.

Reason: Sum of an odd number of odd integers = odd

Sum of first 404 primes = odd + even = odd

x + x^N + x^{N+1} + x^{N+2}

x + x^{odd} + x^{even} + x^{odd}

-1 + (-1^{odd}) + (-1^{even}) + (-1^{odd})

-1 -1 +1 -1

-2

Option B

JeffTargetTestPrep · Answer

The first prime number is 2, which is the only even prime number.

N = 2 + the sum of 403 odd prime numbers = even + odd = odd

The expression in question can be written as:

(-1) + (-1)^odd+(-1)^even+(-1)^odd = (-1) + (-1) + 1 + (-1) = 1 – 3 = -2

Answer: B

Archit3110 · Answer

since N is sum of even count of first 404 prime numbers , we can say that 403 are odd and 1 is even so sum will be odd of N
for the given expression
 x + x^N + x^{N+1} + x^{N+2} 
-1+(-1)^O + (-1)^0 +(-1)^ O ; where O is odd sum value of N
-1-1+1-1
-2
option B

ifyouknowyouknow · Answer

Points to remember which would help in such questions.
a.  x^{any number} =1 where x=1. 
b.  x^{any even number} =1, where x=-1.
c.  x^{any odd number} =-1, where x=-1.
d. 2 is the only even prime number.

So the equation  x + x^N + x^{N+1} + x^{N+2} , 
where x=-1 can 
N = 2 + the sum of 403 odd prime numbers = even + odd = odd
N+1= 2 + the sum of 404 odd prime numbers = even + even = even
N+2 = 2 + the sum of 405 odd prime numbers = even + odd = odd 
Can be rewritten as,
(-1) + (-1)^odd+(-1)^even+(-1)^odd. = (-1) + (-1) + 1 + (-1) = 1 – 3 = -2
The correct answer is -2.

user1592 · Answer

When calculating the sum of the first 404 prime numbers, why aren't we able to use the sum = avg x count formula, which would make the sum an even number?

ex. sum= avg x 404 (anything times even = even)

Fish181 · Answer

You can't use it because that formula requires that the numbers are evenly spaced in the set and 2-3-5-7-11 have different distances between them.

You don't even need to calculate the sum though. This problem is a play on even and odd exponents and how it affects negative signs. If you recognize that only the first prime number "2" is even and every other prime number to infinity is odd, then you know that an even + odd = odd number.

Since you know that the exponent will be ODD you know that the negative 1 will carry. When you add +1 to an odd number, it becomes an even number.... in that case the -1 will be squared and it turns into +1.

With that info you plug it in and see that -1 + -1^(ODD) + -1^(EVEN) + -1^(ODD) = -1 - 1 + 1 - 1 = - 2

CapnSal · Answer

If I take the approach that 2 + (3+5+7+11...). The numbers inside the parenthesis are all odd numbers. So won't their sum also be even? and 2 + even = even

Using that I get the answer as 0.

Can someone help where I am going wrong?

mrfrantic · Answer

Ans (B)

Let's first check whether n, which is the sum of the first 404 prime numbers, is even or odd by figuring out the pattern in the first initial prime numbers.

Prime no. = 2
# of terms = 1
Sum = 2

Prime no. = 2,3
# of terms = 2
Sum = 5

Prime no. = 2,3,5
# of terms = 3
Sum = 10

Prime no. = 2,3,5,7
# of terms = 4
Sum = 17

So we notice that for odd no. of prime numbers (2,3,5), the sum is even (10) and vice versa.

Now applying this logic on the given equation, we have been given 404 prime numbers, which is even. Therefore, its sum N must be odd.

-> x + x^N + x^(N+1) + x^(N+2)
-> x + x^(odd) + x^(even) + x^(odd)
-> (-1) + (-1)^(odd) + (-1)^(even) + (-1)^(odd)
-> (-1) + (-1) + 1 + (-1)
-> -2

ahalail · Answer

The answer to this question is correct, however the logic is flawed. The sum of the first 402 prime numbers are even. 
So sum of 401 odd primes is actually even.

If the reason that the sum of an odd number of odd integers is always odd, then this should not be the case.

I got the question wrong in my practice because I assumed that sum of 404 primes is even. Based on the "pattern" of the sum of primes every even n'th sum is even as well, but the 404th prime sum goes against this pattern. Can someone explain how we are supposed to know this in a test setting where we have to make a quick choice?

tusharmurkute22 · Answer

Answer's to this questions are right but logic is flawed

I tried to solve this question the following way :-

Prime number are - 2,3,5,7,11,13,17,19 ...

Here only 2 is even prime number so let's add this in the end

3 (1st) +5(2nd) = 8 (E)
8(result)+7(3rd next prime number) = 15 (O)
15 (result) + 11 (4th next prime number) =26 (E)
26 + 13 = 39 (O) and so on

here we can see that addition of 2nd and 3rd prime number results in ODD number and addition of their result (i.e. 8) and 4th prime number results in EVEN number and the trend follows, thus we can say the addition of 403th 404th prime number will be an  ODD  number

Thus, 2 (EVEN) + addition of 2nd to 404th prime number (ODD) = ODD

=> -1 + (-1)^odd +(-1)^even + (-1)^odd
=> -1 + (-1) + 1 + (-1)
=> -2 
Option B

miag · Answer

Hi,

I wanted to clarify the reason behind why n = odd

N = (2) + (3 + 5 + 7..)
> The second component is the sum of odd number of odd primes which is why the sum there will be odd. But if these were even in number for e.g. 402 or 404 then this sum will be even

N = even + odd = odd

Let me know if my logic is correct here - thank you!

DanTheGMATMan · Answer

The sum of 404 prime numbers is meant to sound overwhelming. That's just one even and a bunch of odds, which will sum to an odd number. So that's just a way of saying N is odd:

https://www.youtube.com/watch?v=b7iSK6X6vCs

KarishmaB · Answer

On reading "prime numbers," one of the first things that should come to your mind is only one even number and all others odd. Many of the prime number questions are based on this construct.

goosens123 · Answer

You could even do it by just factoring out x^n:

x + x^n +x^n+1 +x^n+2

--> x + x^n(1 + x + x^2)

Substitute the x value in:

--> -1 + x^n(1-1+1)
--> -1 + -1^n

Sum of 404 prime is odd so:

-1+ (-1)
=-2

If x = -1 and if n is the sum of the first 404 prime numbers

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If x = -1 and if n is the sum of the first 404 prime numbers

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