difficult questions

[sukh]

Is the product abcd even? (1)a^2 +b^2 +c^2 + d^2 = 0 (2) a = b = c = d
plz expain in detail

If N is a positive integer, is N ! divisible by 66 ? (1)N! is divisible by 11 (2)N! is divisible by 12

[shankar.ashwin]

Statement 1:

a^2+b^2+c^2+d^2 = 0.

Since all the terms have an even power, none of them can be -ve. And sum of 4 +ve numbers cannot be 0.
So all the terms, a,b,c and d should be 0. Their product would be =0 (even) Sufficient.

Statement 2:

You can have 4 even numbers, even product
or 4 odd numbers, odd product. A IMO

[shankar.ashwin]

If N is a positive integer, is N ! divisible by 66 ?
(1)N! is divisible by 11
(2)N! is divisible by 12

Statement 1:
Since 11 is a prime number N should be >11, for it to be divisible by 11. N! would contain (6*11) Hence would be divisible by 66. Sufficient.

Statement 2:

N! is divisible by 12.

N! could be 4! (or) 12!. One case its divisible by 66 and in the other its not. Insuff.

A IMO

[neelgandham]

Is the product abcd even?

(1)a^2 +b^2 +c^2 + d^2 = 0 square of a number is always >0. So, only one case exists - a=b=c=d=0 ! - Sufficient!

(2)a = b = c = d If a = b = c = d = Even, then product abcd is Even; If a = b = c = d = Odd, then product abcd is Odd - Insufficient

IMO A

__________________________________________

If N is a positive integer, is N ! divisible by 66 ? Rephrasing the question, Is N > 11 ?

(1)N! is divisible by 11 => N > 11 - Sufficient
(2)N! is divisible by 12 => N > 4 - Insufficient

IMO A

[vaibhavgupta]

Is the product abcd even? (1)a^2 +b^2 +c^2 + d^2 = 0 (2) a = b = c = d
plz expain in detail

If N is a positive integer, is N ! divisible by 66 ? (1)N! is divisible by 11 (2)N! is divisible by 12

One of the most basic rules that u keep in ur head is that a square of any non-imaginary i.e. without iota, will always be non negative.

[saketk]

Vaibhav is correct. Anyways, GMAT never tests imaginary numbers so you are good to go.