I'm trying to understand this GMAT question.
I've tried looking at the following questions (Determine the Number of Multiples of Given Numbers $\le$ 1000 and How many multiples of 3 are between 10 and 100? (SAT math question)) to gain insights on how to tackle this but they aren't 100% applicable.
"D is the set of all the multiples of 3 between 20 and 100. E is the set of all the factors of 465. Set D and Set E have how many numbers in common?
I know that all members of Set D are multiples of 3 between 2 and 100. That means each of these multiples will have the prime number of 3.
I know that Set E is divisble by 3 (because 465 sum to 9 which is divisible by 3).
I know the number of multiples of 3 between 20 and 100 can be calculated by doing the following:
For 20
a. 3^1=3 --> 20/3 = 6r2
b. 3^2=9 --> 20/9 = 2r2
Total Number of Multiples of 3 in 20: 6+2= 8 Multiples of 3
For 100
a. 3^1=3 --> 100/3 =33r1
b. 3^2=9 --> 100/9 = 11r1
c. 3^3=27 --> 100/27= 3r19
d. 3^4=81 --> 100/81= 1r19
Total Multiples of 3 in 100 = 33+11+3+1=48 Multiples
Total Number of Multiples of 3 between 20 and 100= 48 - 8= 40 Multiples of 3
Prime Factorization of 465 is: 3, 5, 31
From here though, I'm stuck as I can't determine what is next do.
$$465=1*3*5*31$$ The divisors are the product of one or two of these:
So: $$1*3=3$$ $$1*5=5$$$$ 1*31=31$$ $$3*5=15$$ $$3*31=93$$ $$5*31=155$$
Of these, the only multiples of $3$ are $3,15,93$ of which only $93$ is in the range ${(20,100)}$