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Re: If S1 = {1, 2, 3, 4, ... , 23} and S2 [#permalink]
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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?
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Re: If S1 = {1, 2, 3, 4, ... , 23} and S2 [#permalink]
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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.
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Re: If S1 = {1, 2, 3, 4, ... , 23} and S2 [#permalink]
Is there any short way?
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Re: If S1 = {1, 2, 3, 4, ... , 23} and S2 [#permalink]
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The explanation above is pretty fast.

Not all the time there is.

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Re: If S1 = {1, 2, 3, 4, ... , 23} and S2 [#permalink]
Carcass,
do we really get this much time in GRE to multiply as see each set of nums?
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Re: If S1 = {1, 2, 3, 4, ... , 23} and S2 [#permalink]
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In the old GRE I saw a similar question but not exactly the same.

It is a matter of counting the prime numbers for which the numbers are divisible. There is not so many steps to perform. It is just a matter of counting.

Yes, it is reasonable as question but is rare to have in the new GRE a question about set of numbers
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Re: If S1 = {1, 2, 3, 4, ... , 23} and S2 [#permalink]
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