Sara is an avid lottery player. In the certain game she plays, she must pick one number between 30 and 39, inclusive, one number between 40 and 49, inclusive, and one number between 50 and 59, inclusive. She believes that she will have the best chance of winning if her three numbers, as a set, has the greatest number of distinct prime factors possible. According to Sara's theory, which of the following sets of three numbers should she use?
A. 32 - 48 - 52
B. 33 - 42 - 56
C. 39 - 40 - 54
D. 38 - 49 - 51
E. 36 - 42 - 56
The OA is D.
I'm really confused with this PS question. Experts, any suggestion about how can I solve it? Thanks in advance.
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
Sara is an avid lottery player. In the certain game...
This topic has expert replies
-
BTGmoderatorLU
- Moderator
- Posts: 2207
- Joined: Sun Oct 15, 2017 1:50 pm
- Followed by:6 members
-
EconomistGMATTutor
- GMAT Instructor
- Posts: 555
- Joined: Wed Oct 04, 2017 4:18 pm
- Thanked: 180 times
- Followed by:12 members
Hello LUANDATO.
Let's take a look at your question.
We have to find all the prime factors of each option.
Option A: $$32=2^5\ \ \ -\ \ \ 48=2^4\cdot3\ \ \ -\ \ \ 52=2^2\cdot13.$$ Option B: $$33=3\cdot11\ \ \ -\ \ \ 42=2\cdot3\cdot7\ \ \ -\ \ \ 56=2^3\cdot7.$$ Option C: $$39=3\cdot13\ \ \ -\ \ \ 40=2^3\cdot5\ \ \ -\ \ \ 54=2\cdot3^3.$$ Option D: $$38=2\cdot19\ \ \ -\ \ \ 49=7^2\ \ \ -\ \ \ 51=3\cdot17.$$ Option E: $$36=2^2\cdot3^2\ \ \ -\ \ \ 42=2\cdot3\cdot7\ \ \ -\ \ \ 56=2^3\cdot7.$$ If we take a look of all distinct prime factors on each set, we can conclude that the option D has 5 distinct prime factors and the rest has 4 distinct prime factors or less.
Hence, the correct answer is the option D. I hope this explanation may help you.
Feel free to ask me again if you have a doubt.
Regards.
Let's take a look at your question.
We have to find all the prime factors of each option.
Option A: $$32=2^5\ \ \ -\ \ \ 48=2^4\cdot3\ \ \ -\ \ \ 52=2^2\cdot13.$$ Option B: $$33=3\cdot11\ \ \ -\ \ \ 42=2\cdot3\cdot7\ \ \ -\ \ \ 56=2^3\cdot7.$$ Option C: $$39=3\cdot13\ \ \ -\ \ \ 40=2^3\cdot5\ \ \ -\ \ \ 54=2\cdot3^3.$$ Option D: $$38=2\cdot19\ \ \ -\ \ \ 49=7^2\ \ \ -\ \ \ 51=3\cdot17.$$ Option E: $$36=2^2\cdot3^2\ \ \ -\ \ \ 42=2\cdot3\cdot7\ \ \ -\ \ \ 56=2^3\cdot7.$$ If we take a look of all distinct prime factors on each set, we can conclude that the option D has 5 distinct prime factors and the rest has 4 distinct prime factors or less.
Hence, the correct answer is the option D. I hope this explanation may help you.
Feel free to ask me again if you have a doubt.
Regards.
GMAT Prep From The Economist
We offer 70+ point score improvement money back guarantee.
Our average student improves 98 points.
We offer 70+ point score improvement money back guarantee.
Our average student improves 98 points.
