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# If a number is selected from the factors of 3^{3} 5^{2} 7^{2}

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Manager
Joined: 03 Oct 2016
Posts: 75
Concentration: Technology, General Management
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If a number is selected from the factors of 3^{3} 5^{2} 7^{2}  [#permalink]

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06 Oct 2016, 09:24
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29
00:00

Difficulty:

95% (hard)

Question Stats:

36% (02:05) correct 64% (01:58) wrong based on 281 sessions

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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

Keep the Kudos coming in and let the questions go out !!!!
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Joined: 11 Sep 2015
Posts: 4350
If a number is selected from the factors of 3^{3} 5^{2} 7^{2}  [#permalink]

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Updated on: 06 Oct 2016, 11:28
14
Top Contributor
16
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

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Originally posted by GMATPrepNow on 06 Oct 2016, 10:44.
Last edited by GMATPrepNow on 06 Oct 2016, 11:28, edited 1 time in total.
##### General Discussion
Manager
Joined: 26 Jan 2016
Posts: 93
Location: United States
GPA: 3.37
Re: If a number is selected from the factors of 3^{3} 5^{2} 7^{2}  [#permalink]

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06 Oct 2016, 11:24
1
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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Joined: 11 Sep 2015
Posts: 4350
Re: If a number is selected from the factors of 3^{3} 5^{2} 7^{2}  [#permalink]

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06 Oct 2016, 11:29
Top Contributor
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 )
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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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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

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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]

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27 Nov 2017, 07:10
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]

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30 Jan 2018, 01:00
1
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]

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25 Mar 2019, 17:47
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Re: If a number is selected from the factors of 3^{3} 5^{2} 7^{2}   [#permalink] 25 Mar 2019, 17:47
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