Question from GPREP

StmI)

Clearly insufficient

n=9 yes
n=5 no

INSUFF

Stmt II

2n-> even (but n could be odd or even since even*odd=even and even*even = even)

Lets take n as even and see if the statement is true

n=4

4 has 3 factors 1,2,4 whereas 8 has 4 factors clealry the statement doesnt hold good

n=6 6 has 1,2,3,6 as factors 12 has 6 factors again not true

We can see a pattern here that its not going to hold good for n is even


n=5 5 has 2 factors 1,5 2n=10 has 4 factors 1,2,5,10 TRUE

n=9 9 has 3 factors 1,3,9 2n=18 has 6 factors TRUE

N=15 15 has 4 factors 2n=30 has 8 factors TRUE
Pattern here also

n has to be odd

SUFF

B)

Another way to look at it conceptually:

If n was even then 2 would already be a factor of it so by multpliying by 2 each of the original factors clearly the factors cant double

n=10(even) factors are 1,2,5,10 (4 factors)

2n = 2* 10 = 20 factors are 1,2,5,10,4,20(6 factors)


Only if n was odd 2 would already not be a factor so each original factor when multiplied by 2 gives a new factor thereby doubling it

n=15(odd)

15 has 4 factors 1,3,5,15

Multiply by 2 so u are essentially getting 4 new factors

1*2,3*2,5*2,15*2 making it 8 factors (1,3,5,15,2,6,10,30)

Hope this helps!