Last visit was: 24 Apr 2024, 22:57 It is currently 24 Apr 2024, 22:57

Close

GRE Prep Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
User avatar
Director
Director
Joined: 16 May 2014
Posts: 592
Own Kudos [?]: 1965 [22]
Given Kudos: 0
GRE 1: Q165 V161
Send PM
Most Helpful Expert Reply
User avatar
Director
Director
Joined: 16 May 2014
Posts: 592
Own Kudos [?]: 1965 [5]
Given Kudos: 0
GRE 1: Q165 V161
Send PM
General Discussion
avatar
Manager
Manager
Joined: 02 Dec 2018
Posts: 74
Own Kudos [?]: 31 [1]
Given Kudos: 0
Send PM
avatar
Manager
Manager
Joined: 02 Dec 2018
Posts: 74
Own Kudos [?]: 31 [1]
Given Kudos: 0
Send PM
Re: If S1 = {1, 2, 3, 4, ... , 23} and S2 [#permalink]
1
soumya1989 wrote:

Explanation



If a number of the set S2 is divisible by at least four distinct prime numbers of the set S1, then it will be divisible by their product as well.
  • The number of numbers in S2 divisible by the product of 2, 3, 5 and 7, i.e. 210 = 3.
  • The number of numbers in S2 divisible by the product of 2, 3, 5 and 11, i.e. 330 = 2.
  • The number of numbers in S2 divisible by the product of 2, 3, 5 and 13, i.e. 390 = 1.
  • The number of numbers in S2 divisible by the product of 2, 3, 5 and 17, i.e. 510 = 1.
  • The number of numbers in S2 divisible by the product of 2, 3, 5 and 19, i.e. 570 = 1.
  • The number of numbers in S2 divisible by the product of 2, 3, 5 and 23, i.e. 690 = 1.
  • The number of numbers in S2 divisible by the product of 2, 3, 7 and 11, i.e. 462 = 1.
  • The number of numbers in S2 divisible by the product of 2, 3, 7 and 13, i.e. 546 = 1.
There is no other combination of four or more prime
numbers in set S1 that divides any of the elements of set S2.
Hence, the required number of elements = 11.


There are so many numbers (more than 400), how do you quickly find the numbers that are divisible by the product of the prime numbers?
User avatar
Manager
Manager
Joined: 01 Nov 2018
Posts: 87
Own Kudos [?]: 140 [4]
Given Kudos: 0
Send PM
Re: If S1 = {1, 2, 3, 4, ... , 23} and S2 [#permalink]
3
Expert Reply
1
Bookmarks
Think of this
we have these prime numbers :2,3,5,7,11,13,17,19,23
the range is from 207 to 691

the product of at least 4 distinct primes sounds scary, but we have the primes all listed out.
just multiply the first 4:
2*3*5*7= 210, and there three multiples of 210 within the given bounds, which are: 210,420,630
now try
2*3*5*11=330, two multiples of that are 330 and 660
2*3*5*13= 390
2*3*5*17= 510
2*3*5*19=570
2*3*5*23=690

2*3*7*11=462
2*3*7*13=546

thus we have 11 multiples within the given bounds with at least 4 distinct primes.
avatar
Director
Director
Joined: 09 Nov 2018
Posts: 505
Own Kudos [?]: 129 [0]
Given Kudos: 0
Send PM
Re: If S1 = {1, 2, 3, 4, ... , 23} and S2 [#permalink]
Is there any short way?
Verbal Expert
Joined: 18 Apr 2015
Posts: 28633
Own Kudos [?]: 33112 [0]
Given Kudos: 25173
Send PM
Re: If S1 = {1, 2, 3, 4, ... , 23} and S2 [#permalink]
Expert Reply
The explanation above is pretty fast.

Not all the time there is.

Regards
User avatar
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 4414
Own Kudos [?]: 68 [0]
Given Kudos: 0
Send PM
Re: If S1 = {1, 2, 3, 4, ... , 23} and S2 [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Prep Club for GRE Bot
[#permalink]
Moderators:
Moderator
1085 posts
GRE Instructor
218 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne