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

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by Max@Math Revolution » Wed Sep 07, 2016 9:40 pm
How many factors of 330 are odd numbers greater than 1?
A. 3
B. 4
C. 5
D. 6
E. 7

*An answer will be posted in 2 days.
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by dustystormy » Wed Sep 07, 2016 10:14 pm
IMO E

330 = 2*3*5*11

for factors to be odd we don't need multiple of 2
therefore number of odd factors = 1*2*2*2 = 8
8 is total number of odd factors but it also has 1 as a factor so subtract 1 from 8
8-1=7

Hence total number of odd factors except 1 = 7
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by Cthulu » Thu Sep 08, 2016 1:39 am
330 is a small number to factorize...330 - 1*2*3*5*11....so you have 3,5 and 11...multiply with each other and all together : 15,55,15, 165..so total 7
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by Brent@GMATPrepNow » Thu Sep 08, 2016 6:41 am
Max@Math Revolution wrote:How many factors of 330 are odd numbers greater than 1?
A. 3
B. 4
C. 5
D. 6
E. 7

*An answer will be posted in 2 days.
330 is a pretty small #, so it wouldn't take long to LIST all of the odd factors.
We'll use the fact that 330 = (2)(3)(5)(11) AND the fact that ODD x ODD = ODD

So, the ODD factors of 330 are:
- 3
- 5
- 11
- (3)(5) [we need not evaluate this since we'll just going to count the factors in our list]
- (3)(11)
- (5)(11)
- (3)(5)(11)

Done!

Answer: E
Brent Hanneson - Creator of GMATPrepNow.com
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by Max@Math Revolution » Sat Sep 10, 2016 8:32 pm
If we calculate, we get 330=2*3*5*11. We are trying to find odd factors. So we have to find 3*5*11. From (1+1)(1+1)(1+1)=8, we only have to get rid of 1. The answer is 7. Hence, the correct answer is E.
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