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19 Jun 2010, 11:37
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
49% (02:03) correct
51%
(01:59)
wrong
based on 560
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If k is a positive integer, is k a prime number?
(1) No integers between 2 and \(\sqrt{k}\), inclusive divides k evenly.
(2) No integers between 2 and k/2 inclusive divides k evenly, and k is greater than 5.
(1) No integers between 2 and \(\sqrt{k}\), inclusive divides k evenly.
(2) No integers between 2 and k/2 inclusive divides k evenly, and k is greater than 5.
19 Jun 2010, 12:35
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If k is a positive integer, is k a prime number?
Given: \(k=integer>0\). Question: \(k=prime\)?
A prime number is a positive integer with exactly two distinct divisors: 1 and itself.
(1) No integer between 2 and \(\sqrt{k}\) inclusive divides k evenly --> let's assume \(k\) is not a prime, then there must be some integers \(a\) and \(b\) (\(1<a<k\) and \(1<a<k\)), a factors of \(k\), for which \(ab=k\). As given that \(k\) has no factor between 2 and \(\sqrt{k}\) inclusive, then both factors \(a\) and \(b\) must be more than \(\sqrt{k}\). But it's not possible, as the product of two positive integers more than \(\sqrt{k}\) will yield an integer more than \(k\) (\(ab>k\)). Hence our assumption that \(k\) is not a prime is not true --> \(k\) is a prime. Sufficient.
(2) No integers between 2 and \(\frac{k}{2}\) inclusive divides k evenly, and k is greater than 5 --> the same here : let's assume \(k\) is not a prime, then there must be some integers \(a\) and \(b\) (\(1<a<k\) and \(1<a<k\)), a factors of \(k\), for which \(ab=k\) --> \(k=ab\geq{\frac{k^2}{4}}\) (as both \(a\) and \(b\) are more than or equal to \(\frac{k}{2}\), then their product \(ab\), which is \(k\), must be more than or equal to \(\frac{k}{2}*\frac{k}{2}\)) --> \(4k\geq{k^2}\) --> \(k(4-k)\geq{0}\). But this inequality cannot be true as \(4-k\) will be negative (as given \(k>5\)) and \(k\) is positive so \(k(4-k)\) must be negative not positive or zero. Hence our assumption that \(k\) is not a prime is not true --> \(k\) is a prime. Sufficient.
Answer: D.
P.S. The first statement is basically the way of checking whether some # is a prime:
Verifying the primality (checking whether the number is a prime) of a given number \(n\) can be done by trial division, that is to say dividing \(n\) by all integer numbers (primes) smaller than \(\sqrt{n}\), thereby checking whether \(n\) is a multiple of \(m\leq \sqrt{n}\).
Example: Verifying the primality of \(161\): \(\sqrt{161}\) is little less than \(13\), from integers from \(2\) to \(13\), \(161\) is divisible by \(7\), hence \(161\) is not prime.
Hope it helps.
Given: \(k=integer>0\). Question: \(k=prime\)?
A prime number is a positive integer with exactly two distinct divisors: 1 and itself.
(1) No integer between 2 and \(\sqrt{k}\) inclusive divides k evenly --> let's assume \(k\) is not a prime, then there must be some integers \(a\) and \(b\) (\(1<a<k\) and \(1<a<k\)), a factors of \(k\), for which \(ab=k\). As given that \(k\) has no factor between 2 and \(\sqrt{k}\) inclusive, then both factors \(a\) and \(b\) must be more than \(\sqrt{k}\). But it's not possible, as the product of two positive integers more than \(\sqrt{k}\) will yield an integer more than \(k\) (\(ab>k\)). Hence our assumption that \(k\) is not a prime is not true --> \(k\) is a prime. Sufficient.
(2) No integers between 2 and \(\frac{k}{2}\) inclusive divides k evenly, and k is greater than 5 --> the same here : let's assume \(k\) is not a prime, then there must be some integers \(a\) and \(b\) (\(1<a<k\) and \(1<a<k\)), a factors of \(k\), for which \(ab=k\) --> \(k=ab\geq{\frac{k^2}{4}}\) (as both \(a\) and \(b\) are more than or equal to \(\frac{k}{2}\), then their product \(ab\), which is \(k\), must be more than or equal to \(\frac{k}{2}*\frac{k}{2}\)) --> \(4k\geq{k^2}\) --> \(k(4-k)\geq{0}\). But this inequality cannot be true as \(4-k\) will be negative (as given \(k>5\)) and \(k\) is positive so \(k(4-k)\) must be negative not positive or zero. Hence our assumption that \(k\) is not a prime is not true --> \(k\) is a prime. Sufficient.
Answer: D.
P.S. The first statement is basically the way of checking whether some # is a prime:
Verifying the primality (checking whether the number is a prime) of a given number \(n\) can be done by trial division, that is to say dividing \(n\) by all integer numbers (primes) smaller than \(\sqrt{n}\), thereby checking whether \(n\) is a multiple of \(m\leq \sqrt{n}\).
Example: Verifying the primality of \(161\): \(\sqrt{161}\) is little less than \(13\), from integers from \(2\) to \(13\), \(161\) is divisible by \(7\), hence \(161\) is not prime.
Hope it helps.
General Discussion
22 Jun 2010, 11:56
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Thanks Bunuel. You have been very helpful
+1
+1
