## Gmat loves factors

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### Gmat loves factors

by bblast » Sun Jan 09, 2011 4:27 am
In the number 36 :

1>How many total factors ?
2>How many Odd factors ?
3>How many even factors ?
4>How many prime factors ?

This is a self devised question, so no OA.

[spoiler]1>9
2>3
3>6
4>2[/spoiler]

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by [email protected] » Sun Jan 09, 2011 4:39 am
bblast wrote:In the number 36 :

1>How many total factors ?
2>How many Odd factors ?
3>How many even factors ?
4>How many prime factors ?
36 = (2^2)*(3^2)

1. Number of total factors = (Number of ways to select any number of 2's out of 2)*(Number of ways to select any number of 3's out of 2) = (2 + 1)*(2 + 1) = 9

2. Number of total odd factors = (Number of ways to select no 2)*(Number of ways to select any number of 3's out of 2) = 1*(2 + 1) = 3

3. Number of even factors = (Number of ways to select at least one 2 out of 2)*(Number of ways to select any number of 3's out of 2) = (2)*(2 + 1) = 6

4. Number of prime factors = 2
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by aleph777 » Mon Jan 10, 2011 11:12 am
[email protected],

I'm not familiar with your method of breaking down total factors. Can you explain in a bit more detail?

Thanks!

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by GMATGuruNY » Mon Jan 10, 2011 11:46 am
In the number 36 :

1>How many total factors ?
2>How many Odd factors ?
3>How many even factors ?
4>How many prime factors ?

This is a self devised question, so no OA.

[spoiler]1>9
2>3
3>6
4>2[/spoiler]
To determine the number of positive factors of an integer:

1) Prime-factorize the integer
2) Add 1 to each exponent
3) Multiply

36 = 2^2 * 3^2. Adding 1 to each exponent and multiplying, we get (2+1)*(2+1) = 9 factors.

Here's the reasoning. To determine how many factors can be created from 36 = 2^2 * 3^2, we need to determine the number of choices we have of each prime factor:

For 2, we can use 2^0, 2^1, or 2^2, giving us 3 choices.
For 3, we can use 3^0, 3^1, or 3^2, giving us 3 choices.

Multiplying, we get 3*3 = 9 possible factors.

Another example: How many positive factors does 882 have?

882 = 2 * 3^2 * 7^2. Adding 1 to each exponent and multiplying, we get 2*3*3 = 18 factors.

To determine the number of odd positive factors of an integer:

1) Prime-factorize the integer
2) Add 1 to the exponent of each odd prime factor
3) Multiply

36 = 2^2 * 3^2. The only odd prime factor is 3, with an exponent of 2. Adding 1 to the exponent, we get 2+1 = 3 odd factors.

Number of even positive factors = Total possible factors - Odd factors = 9-3 = 6.
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by vk_vinayak » Mon Sep 24, 2012 3:36 am
So, to calculate EVEN POSITIVE FACTORS, we must find the total factors and subtract ODD POSITIVE FACTORS from it?

From 36 = 2^2 * 3^2. Why can't we say that only EVEN prime factor is 2, with an exponent of 2 and adding 1 to the exponent, we get 2+1 = 3 EVEN factors ?
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by [email protected] » Mon Sep 24, 2012 6:31 am
aleph777 wrote: I'm not familiar with your method of breaking down total factors. Can you explain in a bit more detail?
If N = (p^a)(q^b)(r^c)..., where p, q, r,...(etc.) are prime numbers, then the total number of positive divisors of N is equal to (a+1)(b+1)(c+1)...

Example: 14000 = (2^4)(5^3)(7^1)
So, the number of positive divisors of 14000 = (4+1)(3+1)(1+1) = 5x4x2=40

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by GMATGuruNY » Mon Sep 24, 2012 11:49 am
vk_vinayak wrote:So, to calculate EVEN POSITIVE FACTORS, we must find the total factors and subtract ODD POSITIVE FACTORS from it?

From 36 = 2^2 * 3^2. Why can't we say that only EVEN prime factor is 2, with an exponent of 2 and adding 1 to the exponent, we get 2+1 = 3 EVEN factors ?
This approach counts one combination that is NOT even (2â�°) but omits many combinations that ARE even (2*3, 2*3Â², etc.).
A factor will be EVEN if its prime-factorization includes AT LEAST ONE 2.
To directly count the EVEN positive factors of a positive integer, we could do the following:

1. Prime-factorize the integer.
2. Add 1 to every exponent OTHER THAN 2's exponent.
3. Multiply the results by 2's exponent.

To illustrate:
720 = 2â�´ * 3Â² * 5Â¹
The total number of EVEN factors = (4)(2+1)(1+1) = 24.

The reason that we DON'T add 1 to 2's exponent is that an EVEN factor must include AT LEAST ONE 2, so 2â�° is not an option.
An even factor of 720 must include either 2Â¹, 2Â², 2Â³, or 2â�´.
Thus, the total number of options with regard to 2â�´ is 4 -- the value of 2's exponent.
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by Ian Stewart » Mon Sep 24, 2012 8:45 pm
If you have the prime factorization of an even number, and it looks like this:

(2^k) * some odd primes

then the ratio of the number of even factors to the number of odd factors is k to 1.

So if you take a number like:

120 = (2^3)(3)(5)

then the ratio of even to odd divisors is 3 to 1, and so 3/4 of the factors of 120 will be even, and 1/4 of the factors of 120 will be odd.

As a consequence, every even number has at least as many even divisors as odd divisors, and any multiple of 2^2 = 4 has at least twice as many even divisors as odd divisors.
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