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TAGS: Director  Joined: 16 May 2014
Posts: 595
GRE 1: Q165 V161 Followers: 97

Kudos [?]: 477  , given: 64

If S1 = {1, 2, 3, 4, ... , 23} and S2 [#permalink]
1
KUDOS
Expert's post 00:00

Question Stats: 31% (01:25) correct 68% (02:35) wrong based on 19 sessions
If S1 = {1, 2, 3, 4, ... , 23} and S2 = {207, 208, 209, 210, 211, ... , 691}, how many elements of the set
S2 are divisible by at least four distinct prime numbers that are elements of the set S1?
(a) 9
(b) 8
(c) 11
(d) 12
(e) 7
[Reveal] Spoiler: OA

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If you find this post helpful, please press the kudos button to let me know !  Director  Joined: 16 May 2014
Posts: 595
GRE 1: Q165 V161 Followers: 97

Kudos [?]: 477  , given: 64

Re: If S1 = {1, 2, 3, 4, ... , 23} and S2 [#permalink]
2
KUDOS
Expert's post

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.
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If you find this post helpful, please press the kudos button to let me know ! Manager Joined: 02 Dec 2018
Posts: 74
Followers: 0

Kudos [?]: 7 , given: 60

Re: If S1 = {1, 2, 3, 4, ... , 23} and S2 [#permalink]
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.

Why 11? I see you listed 8 elements divisible by 4 or more primes in set S1.
Manager Joined: 02 Dec 2018
Posts: 74
Followers: 0

Kudos [?]: 7 , given: 60

Re: If S1 = {1, 2, 3, 4, ... , 23} and S2 [#permalink]
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? Manager  Joined: 01 Nov 2018
Posts: 87
Followers: 0

Kudos [?]: 45  , given: 22

Re: If S1 = {1, 2, 3, 4, ... , 23} and S2 [#permalink]
1
KUDOS
Expert's post
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.
Director Joined: 09 Nov 2018
Posts: 508
Followers: 0

Kudos [?]: 27 , given: 1

Re: If S1 = {1, 2, 3, 4, ... , 23} and S2 [#permalink]
Is there any short way?
Founder  Joined: 18 Apr 2015
Posts: 6931
Followers: 114

Kudos [?]: 1344 , given: 6333

Re: If S1 = {1, 2, 3, 4, ... , 23} and S2 [#permalink]
Expert's post
The explanation above is pretty fast.

Not all the time there is.

Regards
_________________ Re: If S1 = {1, 2, 3, 4, ... , 23} and S2   [#permalink] 12 Jan 2019, 10:24
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