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If a number is selected from the factors of 3^{3} 5^{2} 7^{2}
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idontknowwhy94
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If a number is selected from the factors of 3^{3} 5^{2} 7^{2} [#permalink] New post  06 Oct 2016, 09:24
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If a number is selected from the factors of \(3^{3} 5^{2} 7^{2}\) , what is the probability that the selected number is divisible by 21?

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

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BrentGMATPrepNow
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If a number is selected from the factors of 3^{3} 5^{2} 7^{2} [#permalink] New post  Updated on: 06 Oct 2016, 11:28
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idontknowwhy94 wrote:
If a number is selected from the factors of \(3^{3} 5^{2} 7^{2}\) , what is the probability that the selected number is divisible by 21?

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



Great question!

We'll use the following formula:

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) = (5)(4)(2) = 40
--------------------------------------

Okay, now let's answer the question!

P(selected number is divisible by 21) = (# of divisors that are divisible by 21)/(TOTAL # of divisors)


As always, we'll begin with the....

DENOMINATOR
We're given the number: (3³)(5²)(7²)
So, the number of divisors = (3+1)(2+1)(2+1) = (4)(3)(3) = 36

NUMERATOR - i.e., # of divisors that are divisible by 21
Let's factor 21 out of the given number.
(3³)(5²)(7²) = (3¹)(7¹)[(3²)(5²)(7¹)]
= 21[(3²)(5²)(7¹)]
At this point, we can see that, if we multiply 21 by any factor of [(3²)(5²)(7¹)], the product will definitely be divisible by 21
So, the question becomes "how many divisors does [(3²)(5²)(7¹)] have?"
Once again, we'll apply our nice formula.
We have (3²)(5²)(7¹)
So, the number of divisors = (2+1)(2+1)(1+1) = (3)(3)(2) = 18
-------------------

So, P(selected number is divisible by 21) = 18/36
= 1/2

Answer:
Show Spoiler
E


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Originally posted by BrentGMATPrepNow on 06 Oct 2016, 10:44.
Last edited by BrentGMATPrepNow on 06 Oct 2016, 11:28, edited 1 time in total.
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Re: If a number is selected from the factors of 3^{3} 5^{2} 7^{2} [#permalink] New post  06 Oct 2016, 11:24
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GMATPrepNow

why where did the other 5 go when you factored out the 21? Shouldn't it be 21[3²*5²5^1)
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BrentGMATPrepNow
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Re: If a number is selected from the factors of 3^{3} 5^{2} 7^{2} [#permalink] New post  06 Oct 2016, 11:29
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joannaecohen wrote:
GMATPrepNow

why where did the other 5 go when you factored out the 21? Shouldn't it be 21[3²*5²5^1)


Good catch! I've edited my response accordingly (the answer is still E though :-D )
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Re: If a number is selected from the factors of 3^{3} 5^{2} 7^{2} [#permalink] New post  26 Nov 2017, 06:25
GMATPrepNow wrote:
idontknowwhy94 wrote:
If a number is selected from the factors of \(3^{3} 5^{2} 7^{2}\) , what is the probability that the selected number is divisible by 21?

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



Great question!

We'll use the following formula:

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) = (5)(4)(2) = 40
--------------------------------------

Okay, now let's answer the question!

P(selected number is divisible by 21) = (# of divisors that are divisible by 21)/(TOTAL # of divisors)


As always, we'll begin with the....

DENOMINATOR
We're given the number: (3³)(5²)(7²)
So, the number of divisors = (3+1)(2+1)(2+1) = (4)(3)(3) = 36

NUMERATOR - i.e., # of divisors that are divisible by 21
Let's factor 21 out of the given number.
(3³)(5²)(7²) = (3¹)(7¹)[(3²)(5²)(7¹)]
= 21[(3²)(5²)(7¹)]
At this point, we can see that, if we multiply 21 by any factor of [(3²)(5²)(7¹)], the product will definitely be divisible by 21
So, the question becomes "how many divisors does [(3²)(5²)(7¹)] have?"
Once again, we'll apply our nice formula.
We have (3²)(5²)(7¹)
So, the number of divisors = (2+1)(2+1)(1+1) = (3)(3)(2) = 18
-------------------

So, P(selected number is divisible by 21) = 18/36
= 1/2

Answer:
Show Spoiler
E


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wont the factors include 1 as well? shouldn't that be excluded for this question since the q stem asks for numbers divisble by 21?
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Re: If a number is selected from the factors of 3^{3} 5^{2} 7^{2} [#permalink] New post  27 Nov 2017, 07:10
1
Kudos
Total number of factors = (3+1)*(2+1)*(2+1) = 4*3*3
Dividing the number by 21 , n = 3^2*5^2*7^1, Favorable number of factors =(2+1)*(2+1)*(1+1) = 3*3*2

Required probability = 3*3*2/(4*3*3) = 1/2
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Re: If a number is selected from the factors of 3^{3} 5^{2} 7^{2} [#permalink] New post  30 Jan 2018, 01:00
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Total number of factors of \(3^3*5^2*7^2\) = \((3+1) * (2+1) * (2+1)\) = 36

to find number of multiples of 21, we find number of factors = \(3^3*5^2*7^2\) / \(21\) = \(3^2 * 5^2 * 7\) = (2+1) * (2+1) * (1+1) = 18

probability of getting number divisible by 21 = 18/36 = 1/2
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Re: If a number is selected from the factors of 3^{3} 5^{2} 7^{2} [#permalink] New post  13 Apr 2020, 19:47
1
Kudos
Total number of factors = 4×3×3=36
Factors which are multiple of 21
= 3×3×2 =18
Required Probability =1/2
Option E is the answer

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Re: If a number is selected from the factors of 3^{3} 5^{2} 7^{2} [#permalink] New post  13 Apr 2020, 20:38
Determining the sample space/all the possible numbers that could be made from those factors

1. There are 12 ways to choose different numbers from the factors of 3,5,7 (for instance 3x5x7, 3x5x7^2, 3x5^2x7, and so on).
2. There are 6 ways to choose different numbers from the factors of 3,5 (3x5, 3x5^2, and so on).
3. There are 4 ways to choose different numbers from the factors of 5, 7 (5x7, 5x7^2, and so on).
4. There are 6 ways to choose different numbers from the factors of 3,7 (3x7, 3x7^2, 3^2 x 7, 3^2 x 7^2, and so on)
5. There is a total of 7 numbers if we only account for 1 number from the factors (3, 3^2,3^3, 5, 5^2) and so on)
6. And indeed 1 is a factor of any number.

The event that we desire is only from numbers 1 and 4.

Pr(the number is divisible by 21) = 18/36 = 1/2
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Re: If a number is selected from the factors of 3^{3} 5^{2} 7^{2} [#permalink] New post  16 Apr 2020, 22:38
BrentGMATPrepNow wrote:
idontknowwhy94 wrote:
If a number is selected from the factors of \(3^{3} 5^{2} 7^{2}\) , what is the probability that the selected number is divisible by 21?

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



Great question!

We'll use the following formula:

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) = (5)(4)(2) = 40
--------------------------------------

Okay, now let's answer the question!

P(selected number is divisible by 21) = (# of divisors that are divisible by 21)/(TOTAL # of divisors)


As always, we'll begin with the....

DENOMINATOR
We're given the number: (3³)(5²)(7²)
So, the number of divisors = (3+1)(2+1)(2+1) = (4)(3)(3) = 36

NUMERATOR - i.e., # of divisors that are divisible by 21
Let's factor 21 out of the given number.
(3³)(5²)(7²) = (3¹)(7¹)[(3²)(5²)(7¹)]
= 21[(3²)(5²)(7¹)]
At this point, we can see that, if we multiply 21 by any factor of [(3²)(5²)(7¹)], the product will definitely be divisible by 21
So, the question becomes "how many divisors does [(3²)(5²)(7¹)] have?"
Once again, we'll apply our nice formula.
We have (3²)(5²)(7¹)
So, the number of divisors = (2+1)(2+1)(1+1) = (3)(3)(2) = 18
-------------------

So, P(selected number is divisible by 21) = 18/36
= 1/2

Answer:
Show Spoiler
E


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Why are we factoring out that 21 before calculating the number of factors? [Highlighted Part]
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Re: If a number is selected from the factors of 3^{3} 5^{2} 7^{2} [#permalink] New post  18 Apr 2021, 03:32
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Re: If a number is selected from the factors of 3^{3} 5^{2} 7^{2} [#permalink]
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