idontknowwhy94
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....
DENOMINATORWe're given the number: (3
³)(5
²)(7
²)
So, the number of divisors = (
3+1)(
2+1)(
2+1) = (4)(3)(3) =
36NUMERATOR - 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 (3
²)(5
²)(7
¹)
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:
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