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If a number is selected from the factors of 3^{3} 5^{2} 7^{2}
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Updated on: 06 Oct 2016, 12:28
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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
Re: If a number is selected from the factors of 3^{3} 5^{2} 7^{2}
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26 Nov 2017, 07:25
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
Re: If a number is selected from the factors of 3^{3} 5^{2} 7^{2}
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27 Nov 2017, 08:10
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}
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25 Mar 2019, 18:47
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36% (02:06) correct 
