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Number of odd factors

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Number of odd factors

by chieftang » Mon Jan 02, 2012 10:37 am
How many factors of 1200 are odd integers?

(A) 6
(B) 8
(C) 12
(D) 22
(E) 24

Just curious to see various approaches. The key to these is time management!

FYI: This comes from a MOEMS problem set, not a GMAT problem set.
Last edited by chieftang on Mon Jan 02, 2012 12:30 pm, edited 10 times in total.
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by Mike@Magoosh » Mon Jan 02, 2012 11:25 am
Hi there! I'm happy to help. :)

To simplify this problem, remember that you have an almost "magic" understanding of the factors of a number when you know its prime factorization.

1200 = (12)(100)
= (4)(3)(10)(10)
= (2)(2)(3)(2)(5)(2)(5)
= (2)(2)(2)(2)(3)(5)(5) (regroup)

That's the prime factorization of 1200. Every factor of 1200 is the product of some subset of these numbers. We want ODD factors of 1200. We know that (even) x (odd) = (even), so the only odd factors will be the product of some subset of the odd prime factors: 3, 5, and 5.

The possibilities are
a) 1
b) 3
c) 5
d) 15 = 3 x 5
e) 25 = 5 x 5
f) 75 = 3 x 5 x 5

Six odd factors. Therefore, Answer = C

Does that make sense? Let me know if you have any questions.

Here's another example of a question involving factors.

https://gmat.magoosh.com/questions/308

Mike :-)
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https://gmat.magoosh.com/
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by chieftang » Mon Jan 02, 2012 12:00 pm
Good stuff, Mike.

Obviously it's critical to not solve this one by determining all 30 factors and then reducing that set to the odd values. (Exercise: if you don't know how to quickly determine that there are 30 factors of 1200, do some drills!)

So, as you say, the key to these is deriving a complete set of prime factors.

Here's my approach:

Get the prime factors of 1200:

1200 = 12*100 = 2*2*3*2*5*2*5 = 2^4 * 3^1 * 5^2

Now, utilize the multiplication rules:

odd*odd = odd
even*odd = even

So, the number of odd factors can be determined by calculating the number of factors when x=0 for 2^x. Why? Because then we are left with only odd*odd.

In this case that means we want the number of factors of 2^0 * 3^1 * 5^2, which is (0+1)(1+1)(2+1) = 1*2*3 = 6.

Answer A

Also, consider this approach:

Given a prime factorization of 2^x * 3^y * 5^z, then the number of odd factors will be:
(total number of factors)/(x+1)

Why is it so? (It's just another format of the first approach)


Try it in this problem:

(4+1)(1+1)(2+1)/(4+1) = 6

Answer A
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by SnitchSeeker » Tue Nov 01, 2016 1:36 pm
Hi! I saw this and I wanted to suggest a simpler version!

I'm in a sixth grade pre-algebra class. Recently my teacher showed us a very simple and quick way of doing this type of problem.

Example:

First prime factorize the original number - 1200 = 2*2*2*2*3*5*5 = 2^4*3^1*5^2

Then use the plus 1 method on the odd prime integers - (1+1)(2+1) = 6 odd factors

To find even factors, just find the total number of factors - in this case (4+1)(1+1)(2+1) = 30 total factors - and then subtract the number of odd factors 30 total - 6 odd = 24 even factors.


Hope this helped!
Janelle
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by Brent@GMATPrepNow » Tue Nov 01, 2016 2:57 pm
SnitchSeeker wrote:Hi! I saw this and I wanted to suggest a simpler version!

I'm in a sixth grade pre-algebra class. Recently my teacher showed us a very simple and quick way of doing this type of problem.

Example:

First prime factorize the original number - 1200 = 2*2*2*2*3*5*5 = 2^4*3^1*5^2

Then use the plus 1 method on the odd prime integers - (1+1)(2+1) = 6 odd factors

To find even factors, just find the total number of factors - in this case (4+1)(1+1)(2+1) = 30 total factors - and then subtract the number of odd factors 30 total - 6 odd = 24 even factors.


Hope this helped!
Janelle
Nice comment, Janelle.
If anyone is interested in learning more, here's a video that explains how/why Janelle's strategy works: -

Counting the divisors of numbers: https://www.gmatprepnow.com/module/gmat ... /video/828

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Matt@VeritasPrep » Fri Nov 11, 2016 4:28 pm
SnitchSeeker wrote:Hi! I saw this and I wanted to suggest a simpler version!

I'm in a sixth grade pre-algebra class. Recently my teacher showed us a very simple and quick way of doing this type of problem.

Example:

First prime factorize the original number - 1200 = 2*2*2*2*3*5*5 = 2^4*3^1*5^2

Then use the plus 1 method on the odd prime integers - (1+1)(2+1) = 6 odd factors

To find even factors, just find the total number of factors - in this case (4+1)(1+1)(2+1) = 30 total factors - and then subtract the number of odd factors 30 total - 6 odd = 24 even factors.


Hope this helped!
Janelle
Take the GMAT now! You'll ace it, and you can always use it down the line :)

(Also, kudos to your teacher. If I had learned this sort of nifty arithmetic at age 12, my whole life would've turned out differently.)
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