Been beating my head against a wall on this question.
There exist an integer n such that 6n(to the power of two) + 27 is prime.
I realize this is false but can't prove it QED. Don't really care about a home work grade (for sure this is the only one wrong). But I've a test coming up and no time to review this after hand-in. Any help or hint would be greatly appreciated.
51 is not prime... divisible by 3 as are the rest of the solutions you would come up with Discrete Mathematics::
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no because if
A divides B and A divides C , then A divides the sum B + C.
3 divides (6n)squared because 3 divides 6
and 3 divides 27
therefore 3 divides (6n)^2 + 27
thus, it is not prime!
=]
You need to specify whether the equation is (6n)^2+27 or if it is 6(n^2)+27.
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File Format: PDF/Adobe Acrobat - View as HTMLthis time r is chosen at random from the set of all prime numbers less than . Suppose that the poker site from HW6 Question 1 catches on to your tricks
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I am going to assume that it is the latter, because parenthesis are not in the equation in the way that you presented it (which I believe to be the most likely posing of the equation).
First of all, how do you know that it is false? if n=2, then the statement is correct. 6(2^2)+27=51. Last time I checked, 51 is prime.
Let me know if I am a complete idiot, because I think you are looking too deeply into the problem (which, as math people, you and I do much too frequently).
Hope this helps.
Since 3 divides both 6 and 27, the sum you describe is divisible by 3. Am I missing something?
The method for proving this particular statement is false is relatively simple. The idea is that 3 divides evenly into 6*(any integer) + 27. Since 3 divides the value, clearly it is not prime.
If you need to make the answer more formal, factor out the 3:
p = 6 * n^2 + 27 =
p = 3 * (2 * n^2 + 9)
and here you would say p is of the form 3n which cannot be prime.
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Discrete Mathmatics Prime number question?