GMATPrep Q1

Question

please discuss

netskater · Answer

I would go with E

consider 1: n divisible by 3
n can be odd (3,9,15) or even (6,12,18) - INSUFF

consider 2: 2n has double the factors of n
not true for n = 6 (which has 4 factors), 2n = 12 (has 6 factors)
true of n = 5 (which has 2 factors) 2n = 10 (has 4 factors)
but again not true for n = 15 (which has 4 factors) but 2n=30 (has 7 factors)

 - INSUFF

yezz · Answer

I think the answer should be B

Even numbers n = ( 2,4,6) 2n(4,8,12) factorize and do not forget one as a divisor and the number itself.

Ie: 4 is divisible by 4,2,1 2 is divisible by 2,1

 8 is divis by 8,4,2,1, 4 is divisa by 4,2,1 and so on 

Try for odd Intigers

N = (1,3,5,9) 2n(2,6,10,18)

1 is div by 1 2 is div by 2,1

3 is div by 3,1 6 is divi by 6,3,2,1,

9 is div by 9,3,1 18 is divi by 18,9,2,6,1,3

Thus b is enough

kevincan · Answer

Actually, 30 has 8 factors: 1,2,3,5,6,10,15,30

Remember that only  for a number to have a prime number of factors, it must have only one prime factor . How do we know?

In general, we can count the number of factors a number n has by writing it as  a product of prime numbers  n=p1^k1*p2^k2..., where p1, p2,... are  distinct  prime numbers. The number of factors, including 1 and n is (k1+1)*(k2+1)...

For example 30=2^1*3^1*5^1 has (1+1)*(1+1)*(1+1)= 8 factors 

200= 2^3*5^2 has  12 factors . (3+1)*(2+1)

If n is even, it can be written as n= 2^k1 *p2^k2..., so it will have (k1+1)(k2+1)... factors where k1 is  at least 1 

2n will have (k1+ 2 )(k2+1)... factors

Notice that the ratio of the number of factors of 2n to that of n is  (k1+2)/(k1+1) , which will always be less than 2. Thus  an even number n will have more than half as many factors as 2n . Thus  (2) means that n is odd .

I know this is a bit dense, and picking numbers would work well for this question, but a bit of theory can make questions much easier.

gmatornot · Answer

I would go with B here... 

ST 1 is clearly INSUFF

ST 2 says if n has x factors then 2n should have 2x factors

This seems only possible for odd integers

1 has (1) factor, 2 has 2 (1,2)
3 has (1,3) and 6 has (1,2,3,6)
5 has (1,5) and 10 has (1,2,5,10)

for even numbers 

2 has (1,2) and 4 has (1,2,4)
4 has (1,2,4) and 8 has (1,2,4,8)

SUFF

sgrover · Answer

Would have marked (B) onthe exam day:

I. Insufficient.

II. Itappears that the only class of no.s that fall in this category are odd prime no.s Hence sufficient. the no. must be odd.

Every odd prime no. n has two factors: 1 and n
2n will have factors: 1, 2, n, 2n

rkatl · Answer

OA: B

Thanks for all the explainations..

kevincan · Answer

should read  odd numbers

heman · Answer

(1) n = 6 Even
 n =9 Odd

BCE

(2) n= 2 (div by 1&2 which are 2 +ve integers)
 2n =4 ( div by 1,2,4 whic are 3 + ve integers)
Not possible case since we need 2n to be divisible by twice as many + integ

 n = 3 ( div by two +ve integers 1&3)
 2n = 6 ( div by 1,2,3,6 Four + integers)

Hence n has to be odd

Hence B

Heman

GMATPrep Q1

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