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.
all factors of a number
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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!
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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To determine the number of positive factors of an integer: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
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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ODD * ODD = ODD.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
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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Another approach: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
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