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Re: How many prime numbers n exist such that 90 < n < 106 and [#permalink]
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OFFICIAL SOLUTION

C. This difficult factor problem requires you to do two things:

1) Determine the prime numbers between 90 and 106. Those numbers are 97, 101, and 103 (91 is divisible by 7; 93 is divisible by 3; 95 is divisible by 5; 99 is divisible by 3; 105 is divisible by 5, and all the even numbers are divisible by 2). This allows you to eliminate answer choice E, as there are not more than three primes in that range.

2) Recognize that \(99999919\) can be rewritten as \(108−81\)
, allowing you to use the Difference of Squares rule to factor out the number into:

\((104+9)(104−9)\)
And \((104−9)\)
allows you to use Difference of Squares again, creating:

\((104+9)(102+3)(102−3)\)
, which equals:

\((10009)(103)(97)\)

Here you have two of the primes in that range, 103 and 97, and you just have to test 10009 to see if it's divisible by 101. It is not, so the correct answer is 2.
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Re: How many prime numbers n exist such that 90 < n < 106 and [#permalink]
Thank you Bunuel !!! this is a very nice solution.

I would like to say that, even if you are not sure whether a number between 90 and 106 is a prime you can check whether it divides 99999919. Because if it doesn't, you don't need to search further if it's indeed a prime. This could save some time and also eliminates some possible answers.
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Re: How many prime numbers n exist such that 90 < n < 106 and [#permalink]
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Solution:

The number of prime numbers b/w 90-106 are 3,that is 97, 101 and 103

99999919 = 10^8-81 = 1009*(103)*97

Thus option (c)

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Re: How many prime numbers n exist such that 90 < n < 106 and [#permalink]
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Re: How many prime numbers n exist such that 90 < n < 106 and [#permalink]
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