How many factors of 330 are odd numbers greater than 1?

[swerve]

How many factors of 330 are odd numbers greater than 1?

A. 3
B. 4
C. 5
D. 6
E. 7

The OA is E.

Primes factors of 330 are 2^1, 3^1, 5^1, and 11^1.

Total divisors = (power if a prime factor+1)

Total number of odd factors (3, 5, 11) = (1+1)(1+1)(1+1)=8

Since we need odd divisors other than 1 => 8-1 = 7 odd divisors.

Please, can anyone explain another way to solve this PS question? Thanks.

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Hi All,

We're asked for the number of ODD factors of 330 that are greater than 1. Looking at the answer choices, we know that there are at least 3, but no more than 7, ODD factors, so a bit of 'brute force' and prime factorization could get us the correct answer without too much trouble.

330 =
(33)(10) =
(3)(11)(2)(5)

Since we're looking for the ODD factors of 330, we have three that immediately fit what we're looking for: 3, 5 and 11.
Next, any combination of those ODD factors will lead to another ODD factor:
(3)(5) = 15
(3)(11) = 33
(5)(11) = 55
(3)(5)(11) = 165

At this point, we have 7 ODD factors - and that's the largest answer available, so we can stop working.

Final Answer: E

GMAT assassins aren't born, they're made,
Rich

[Scott@TargetTestPrep]

How many factors of 330 are odd numbers greater than 1?

A. 3
B. 4
C. 5
D. 6
E. 7

Let's break 330 into primes:

330 = 33 x 10 = 2 x 3 x 5 x 11

We want to determine the odd factors of 330 greater than 1. Thus, we cannot use 2 at all, since 2 and any of its multiples are even. Thus, we will use 3, 5, and 11, and any multiples that can be obtained from them.

3; 5; 11; 3 x 5 = 15; 3 x 11 = 33; 5 x 11 = 55; 3 x 5 x 11 = 165

So we have 7 factors of 330 that are odd numbers greater than 1.

Answer: E