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.
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Number of odd factors
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Mike@Magoosh
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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
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
Magoosh GMAT Instructor
https://gmat.magoosh.com/
https://gmat.magoosh.com/
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chieftang
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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
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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SnitchSeeker
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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
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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Brent@GMATPrepNow
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Nice comment, Janelle.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
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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Matt@VeritasPrep
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Take the GMAT now! You'll ace it, and you can always use it down the lineSnitchSeeker 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
(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.)
