digit trap

Stmt I

r can be 0 or 2 (0! =1 1!=1)

INSUFF

Stmt II

I just cant think what non negative(o or positive) number would have 0 as its factorial.

Oops those are not equal to not factorial in either case...

!= means not equal to

I think answer is E.

N can be 2, in which case r=4 - Non Prime
N can be 1.5, in which case r=3 - Prime

Same for Stmt II

same here.
E for me too
Stmt 1 : If n is not equal to 1.
then r can have any even value( as multiplied by 2) and in that case...No prime no.

But if n = 0 , r =0
n = 3/2, r = 3 prime
Stmt 2 : same

[quote="maihuna"]Is the non-negative number r prime?

(1) r = 2n, where n != 1
(2) r = 2m, where m != 0[/quote]


What if we take another angle:

It is a Yes/No question-right?

State 1:
Let n=0, then r=2*0=0 (Not prime)
If n=2, r=4 (not prime)
If n=3, 4=6 (not prime)


Stete 2: We get mixed answers
Let m=1, then r=2 (prime)
If m =2, r=4 (not prime).
Therefore not sufficient

So my question is...what does '0' come under...Its not prime, it is not negative either...its not even...but doesn't the Statement 1 actually answers this Yes/No question that " the non-negative number r is not prime".

Please explain.

thanks a lot

Hi,
How would this look like......
Stmt 1.
r=2n and n not equal to 1
So r is either 0 or multiple by 2, hence r is not prime

Stmt 2.
r=2m and m not equal to 0
So r is either 2 or multiple of 2, hence stmt is not suff. because 2 is prime and rest are not
So should not the answer be A..... Please provide update if I'm thinking it wrong

How would this look like......
Stmt 1.
r=2n and n not equal to 1
So r is either 0 or multiple by 2, hence r is not prime

n = 1.5

2*1.5 = 3

so r could be prime.

We have no info that n and m are integers.

Hope this helps!