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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?

  1. 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.

  2. I know that Set E is divisble by 3 (because 465 sum to 9 which is divisible by 3).

  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.

Omar
  • 155
  • Hint : There are four divisors of $465$ which are divisible by $3$. How many of them are between $20$ and $100$ ? – Peter May 06 '18 at 16:51
  • @Peter thanks for the hint. So you leveraged the prime factorization of 465 to determine all possible combinations of multiples of 3 and then mentally listed them out - four divisors of 465 (3,9,15, 95). Is there a more efficient way so that it can be applicable in other scenarios or is that the only way. Thank you so much! – Omar May 06 '18 at 16:58

1 Answers1

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$$465=1*3*5*31$$ 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 ${(20,100)}$

Rhys Hughes
  • 12,842
  • Basically correct, but $1$ is usually omitted in the prime factorization. And it is a matter of taste whether $1$ and $465$ are considered to be factors, which does not matter in the particular exercise anyway. – Peter May 06 '18 at 17:11
  • Agreed. I omitted the $465$ from my examples as it is obviously out of range. – Rhys Hughes May 06 '18 at 17:15