PLEASE HELP  DIVISORS PROB
posted by failureinmath .
What is the probability that a randomly selected divisor of 720 is a multiple of 4? Express your answer as a common fraction.

since 720 = 4*180
and 180 = 2^2 3^2 5
180 has 18 factors:
1 2 4 6 12 18 36
10 20 30 60 90
3 9 15 45 180
I'd say 1/18, since all the factors that are multiples of 4 will be 4 times one of the factors in the list. 
thanks alot

3/5

15

Using the above logic, the answer is 3/5.
All the reasoning is correct (About the factors of 180). The only problem is that the probability of selecting one. Since the prime factorization of 720 is 2^4 x 3^2 x 5, then the number of factors that 720 has is (4+1)(2+1)(1+1)=30. So, the answer is 18/30=3/5 
The prime factorization of 720 is $2^4 \cdot 3^2 \cdot 5$, so every divisor of 720 is of the form $2^a \cdot 3^b \cdot 5^c$, where $a$, $b$, and $c$ are integers with $0 \le a \le 4$, $0 \le b \le 2$, and $0 \le c \le 1$.
This divisor is divisible by 4 if and only if $a \ge 2$, i.e. $a$ is equal to 2, 3, or 4. The total number of possible values of $a$ is 5, and they are all equally likely, so the probability that the divisor is divisible by 4 is $\boxed{3/5}$.
Note that we basically ignored what was happening with the primes other $2$. You can be reassured that this is ok through an alternate solution: By the factorization above, we see there are $(4+1)(2+1)(1+1)=5 \cdot 3 \cdot 2$ total divisors. If the divisor is a multiple of $4$, then the value of $a$ above must be $2$, $3$, or $4$. Combining this with the possibilities for the other primes besides 2, we have $3 \cdot 3 \cdot 2$ divisors that are multiples of $4$. When we take the ratio, the factors contributed from the primes $3$ and $5$ cancel out: $\dfrac{3 \cdot 3 \cdot 2}{5 \cdot 3 \cdot 2} = \boxed{\dfrac35}$.