If m is the product of the first 99 prime numbers and n is the sum of

Question

If m is the product of the first 99 prime numbers and n is the sum of the first 99 prime numbers, then what is the value of  (-1)^m+(-1)^n ?

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

This is a  PS Butler Question

Check the links to other Butler Projects:

Data Sufficiency Butler  
  Critical Reasoning Butler  
  Reading Comprehension Butler  
  Data Insights Butler

ManifestDreamMBA · Answer

m is the product of the first 99 prime numbers. Given this includes 2, m will be even

n is the sum of the first 99 prime numbers. Given this includes 2 and 98 other odd primes, it will be even + even, making n even

(-1)^m+(-1)^n 
= (-1)^{even}+(-1)^{even} 
=1+1
=2

GMATinsight · Answer

m = product of the first 99 prime numbers = Even * (98 odd numbers) = even
n = sum of the first 99 prime numbers = one even + 98 Odd = Even

(-1)^m+(-1)^n = (-1)^{even}+(-1)^{even} = 1+1 = 2

Answer: Option E

Related Video:  
  https://www.youtube.com/watch?v=xWR1VpyCKP0

Missinga · Answer

m is the product of the first 99 prime numbers (only 2 is the even prime and rest are all odd primes. so, Even*(98 odd numbers)= Even

n is the sum of the first 99 prime numbers (1 even+ 98 odd numbers= even+even= even) 
 
(-1)^m+(-1)^n= (-1)^even+(-1)^even= 1+1= 2

Answer: E

Dereno · Answer

m is the product of first 99 prime numbers .  2 is the ONLY even and first prime number

Hence , (-1)^m = 1 (even power)

Now (-1)^n ==== ???

First 99 primes include 2,3,5,7,11,13...... and list goes on.  APART FROM 2 every other 98 numbers are odd.

Odd + odd = even (3+5 = 8 for example).

2 + 49 odd pairs

Even + even = even

2 + even = even

Hence (-1)^n will be 1

Sum of the two values will be 2  OPTION E

Prep Toolkit