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To test whether an integer, \(n\), is prime, it is enough to be sure that none of the primes less than or equal to the square root of \(n\) divide \(n\). If you want to check that a number between 900 and 950 is prime with this rule, what is the largest prime divisor you need to test?

 

I don't get what this problem is asking for. Thanks in advance!

Guest Sep 3, 2018
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Well, suppose we want to check and see if 923 is a prime number or not. You take the square root of 923 and you get about 30.38. So, if you want to check 923 to see if it is prime or not, all you have to do is to check only prime divisors under 30.38, which means dividing 923 by all prime numbers from 2 to 30.38. So, 923/2, 923/3, 923/5........all the way to 923/29.

29 is the last prime below 30.38 to check. If 923 doesn't divide any of these primes EVENLY, then it is a prime number. As it happens, this number of 923 is NOT prime because it is divisible by 13.

So, 923/13 =71.

Guest Sep 3, 2018
edited by Guest  Sep 3, 2018
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I'm sorry, but I am still confused. What about other numbers in the range, like say 947?

Guest Sep 3, 2018
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Never mind, I found it

Guest Sep 3, 2018

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