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all factors of a number

by topspin360 » Fri Jun 29, 2012 7:02 pm
Is it possible to use prime factorization and then combination formula to find out all the different factors of the following number:

How many different positive integers are factors of 225 ?
(A) 4
(B) 6
(C) 7
(D) 9
(E) 11


Also, how would we set up the combinations formula for the following problem?

If x is to be chosen at random from the set {1, 2, 3} and y is to
be chosen at random from the set {4, 5, 6, 7}, what is the
probability that xy will be odd?
(A) 1/6
(B) 1/3
(C) 1/2
(D) 2/3
(E) 5/6


Thanks.
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by shantanu86 » Fri Jun 29, 2012 7:12 pm
IMO [spoiler]9 and 1/6[/spoiler]
If you feel like it, hit thanks :)
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by spartacus1412 » Fri Jun 29, 2012 7:14 pm
Q.1 if any number can be written as N = X^a Y^b Z^c (X, Y, Z being prime)
then number of factors is given by (a+1) (b+1)(c+1)

here, 225 = 3^2 5^2
total number of factors of 225 == (2+1)(2+1) = 9.


Q.2 set A (1,2, 3)
set B (4, 5, 6,7 )


prob = 1/3(2/4+2/4) = 1/3

Hope this helps!
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by topspin360 » Fri Jun 29, 2012 8:30 pm
can you please explain 1/3(2/4 + 2/4)?
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by GMATGuruNY » Fri Jun 29, 2012 9:04 pm
topspin360 wrote:Is it possible to use prime factorization and then combination formula to find out all the different factors of the following number:

How many different positive integers are factors of 225 ?
(A) 4
(B) 6
(C) 7
(D) 9
(E) 11
To determine the number of positive factors of an integer:

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

For example:
72 = 2³ * 3².
Adding 1 to each exponent and multiplying, we get (3+1)*(2+1) = 12 factors.

Here's why:
To determine how many factors can be created from 72 = 2³ * 3², we need to determine the number of choices we have of each prime factor and to count the number of ways these choices can be combined:

For 2, we can use 2�, 2¹, 2², or 2³, giving us 4 choices.
For 3, we can use 3�, 3¹, or 3², giving us 3 choices.

Multiplying the number of choices we have of each factor, we get 4*3 = 12 possible factors.

As for the problem at hand:
225 = 3² * 5².
Adding 1 to each exponent and multiplying, we get:
Number of factors = (2+1)(2+1) = 9.

The correct answer is D.
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by GMATGuruNY » Fri Jun 29, 2012 9:20 pm
topspin360 wrote: If x is to be chosen at random from the set {1, 2, 3} and y is to
be chosen at random from the set {4, 5, 6, 7}, what is the
probability that xy will be odd?
(A) 1/6
(B) 1/3
(C) 1/2
(D) 2/3
(E) 5/6
ODD * ODD = ODD.
Thus, for the product of X and Y to be odd, both X and Y must be odd.

Of the 3 numbers in the top set, 2 are odd.
P(X is odd) = 2/3.

Of the 4 numbers in the bottom set, 2 are odd.
P(Y is odd) = 2/4.

Since we want both probabilities to happen, we multiply the fractions:
2/3 * 2/4 = 1/3.

The correct answer is B.
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by topspin360 » Sat Jun 30, 2012 9:34 am
understood. thanks all!
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by Brent@GMATPrepNow » Sat Oct 19, 2019 2:05 pm
topspin360 wrote:Is it possible to use prime factorization and then combination formula to find out all the different factors of the following number:

How many different positive integers are factors of 225 ?
(A) 4
(B) 6
(C) 7
(D) 9
(E) 11
Another approach:

When we scan the answer choices (always scan the answer choices before getting to work!!), we see that all 5 answer choices are pretty small.
So, we could just LIST all of the factors.
We'll do so in PAIRS of values that have a product of 225
We get:
1 and 225
3 and 75
5 and 45
9 and 25
15 and 15

So, the factors are: {1, 3, 5, 9, 15, 25, 45, 75 and 225}

Answer: D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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