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Bunuel
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Randomly choosing an element from the set of positive divisors of 60, 16 Jan 2021, 11:02
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62% (01:22) correct 38% (01:22) wrong based on 162 sessions

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Randomly choosing an element from the set of positive divisors of 60, the probability that it will be a prime number is:

(A) 1/2
(B) 1/3
(C) 1/4
(D) 1/5
(E) 1/6
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C
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Randomly choosing an element from the set of positive divisors of 60, 16 Jan 2021, 21:58
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Bunuel
Randomly choosing an element from the set of positive divisors of 60, the probability that it will be a prime number is:

(A) 1/2
(B) 1/3
(C) 1/4
(D) 1/5
(E) 1/6
Show more

\(60=2^2*3*5\)
number of positive factors = \((2+1)(1+1)(1+1)=12\)
number of prime factors out of these 12 = 3

\(P = \frac{3}{12}=\frac{1}{4}\)

C
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Randomly choosing an element from the set of positive divisors of 60, 19 Jan 2021, 11:28
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chetan2u
Bunuel
Randomly choosing an element from the set of positive divisors of 60, the probability that it will be a prime number is:

(A) 1/2
(B) 1/3
(C) 1/4
(D) 1/5
(E) 1/6

\(60=2^2*3*5\)
number of positive factors = \((2+1)(1+1)(1+1)=12\)
number of prime factors out of these 12 = 3

\(P = \frac{3}{12}=\frac{1}{4}\)

C
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Don't you think we should count both the 2s in the prime factorisation of 60? In that case the answer turns out to be 1/3, or option (B).
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Randomly choosing an element from the set of positive divisors of 60, 19 Jan 2021, 19:18
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accannate
chetan2u
Bunuel
Randomly choosing an element from the set of positive divisors of 60, the probability that it will be a prime number is:

(A) 1/2
(B) 1/3
(C) 1/4
(D) 1/5
(E) 1/6

\(60=2^2*3*5\)
number of positive factors = \((2+1)(1+1)(1+1)=12\)
number of prime factors out of these 12 = 3

\(P = \frac{3}{12}=\frac{1}{4}\)

C

Don't you think we should count both the 2s in the prime factorisation of 60? In that case the answer turns out to be 1/3, or option (B).
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The set of positive divisors would be: {1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60}(12 items that you can pair up).
It would not be: {2, 2, 3, 5}.
The above method is a handy way of figuring out how many factors there are, but is not a reflection of the actual positive divisors of 60.
Prime factorization is a useful method of breaking down problems but wouldn't necessarily lead you to the correct one here!
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Randomly choosing an element from the set of positive divisors of 60, 19 Jan 2021, 22:08
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\(60 = 2^2 * 3 * 5\)

Total divisors: (2 + 1) * (1 + 1) * (1 + 1) = 3 * 2 * 2 = 12

Total prime divisors: 3 [2,3,5]

Probability: \(\frac{3}{12} = \frac{1}{4}\)

Answer C
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Randomly choosing an element from the set of positive divisors of 60, 19 Feb 2021, 05:30
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chetan2u
Bunuel
Randomly choosing an element from the set of positive divisors of 60, the probability that it will be a prime number is:

(A) 1/2
(B) 1/3
(C) 1/4
(D) 1/5
(E) 1/6

\(60=2^2*3*5\)
number of positive factors = \((2+1)(1+1)(1+1)=12\)
number of prime factors out of these 12 = 3

\(P = \frac{3}{12}=\frac{1}{4}\)

C
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Why isn't 1 being considered as a divisor of 60?
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Randomly choosing an element from the set of positive divisors of 60, 19 Feb 2021, 06:10
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BillHill
chetan2u
Bunuel
Randomly choosing an element from the set of positive divisors of 60, the probability that it will be a prime number is:

(A) 1/2
(B) 1/3
(C) 1/4
(D) 1/5
(E) 1/6

\(60=2^2*3*5\)
number of positive factors = \((2+1)(1+1)(1+1)=12\)
number of prime factors out of these 12 = 3

\(P = \frac{3}{12}=\frac{1}{4}\)

C

Why isn't 1 being considered as a divisor of 60?
Show more

The formula is to get the number to prime factor and then add 1 to each power.

The number of positive factors that you, thus, get it includes 1 too.
The 12 factors of 60 are : 1,2,3,4,5,6,10,12,15,20,30 and 60.
So it includes 1 and the number itself.
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Randomly choosing an element from the set of positive divisors of 60, Updated on: 19 Feb 2021, 06:26
Originally posted by BillHill on 19 Feb 2021, 06:20.
Last edited by BillHill on 19 Feb 2021, 06:26, edited 1 time in total.
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\(60=2^2*3*5\)
number of positive factors = \((2+1)(1+1)(1+1)=12\)
number of prime factors out of these 12 = 3

\(P = \frac{3}{12}=\frac{1}{4}\)

C[/quote]

Why isn't 1 being considered as a divisor of 60?[/quote]

The formula is to get the number to prime factor and then add 1 to each power.

The number of positive factors that you, thus, get it includes 1 too.
The 12 factors of 60 are : 1,2,3,4,5,6,10,12,15,20,30 and 60.
So it includes 1 and the number itself.[/quote]

Thank you for your reply.
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Randomly choosing an element from the set of positive divisors of 60, 19 Feb 2021, 06:26
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BillHill
\(60=2^2*3*5\)
number of positive factors = \((2+1)(1+1)(1+1)=12\)
number of prime factors out of these 12 = 3

\(P = \frac{3}{12}=\frac{1}{4}\)

C
Why isn't 1 being considered as a divisor of 60?
The formula is to get the number to prime factor and then add 1 to each power.

The number of positive factors that you, thus, get it includes 1 too.
The 12 factors of 60 are : 1,2,3,4,5,6,10,12,15,20,30 and 60.
So it includes 1 and the number itself.
Thank you for your reply. I suppose my real question is: if we are considering 1 as a divisor of 60, then there are 4 prime numbers that are divisors of 60 (1,2,3 & 5). Therefore 4/12 = 1/3 and the answer is "B."

I'm not disputing the answer, I just don't understand where my logic breaks down.
Show more

1 is not a prime number. That is why we just take 2, 3 and 5 only.

Even if someone takes 1. What is the power of 1.
1*2^2*3*5=1^(100)*2^2*3*5
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Randomly choosing an element from the set of positive divisors of 60, 19 Feb 2021, 06:53
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chetan2u
BillHill
\(60=2^2*3*5\)
number of positive factors = \((2+1)(1+1)(1+1)=12\)
number of prime factors out of these 12 = 3

\(P = \frac{3}{12}=\frac{1}{4}\)

C
Why isn't 1 being considered as a divisor of 60?
The formula is to get the number to prime factor and then add 1 to each power.

The number of positive factors that you, thus, get it includes 1 too.
The 12 factors of 60 are : 1,2,3,4,5,6,10,12,15,20,30 and 60.
So it includes 1 and the number itself.
Thank you for your reply. I suppose my real question is: if we are considering 1 as a divisor of 60, then there are 4 prime numbers that are divisors of 60 (1,2,3 & 5). Therefore 4/12 = 1/3 and the answer is "B."

I'm not disputing the answer, I just don't understand where my logic breaks down.

1 is not a prime number. That is why we just take 2, 3 and 5 only.

Even if someone takes 1. What is the power of 1.
1*2^2*3*5=1^(100)*2^2*3*5
Show more

Probability that I'm an idiot = 1

Thanks!
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