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k is less than or equal to 7

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14y ago

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What is the method used in simplifying '2k minus 3 brackets h minus 5kbrackets'?

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For n = 4, 4! = 4*3*2*1 = 24 and 24 = 2*2*2*2 = 16 So the statement is true for n = 4. Suppose it is true for n = k, that is, k! > 2k Then (k+1)! = (k+1)*k! > (k+1)*2k (since k! > 2k) > 2*2k (since k >= 4 > 2) = 2k+1 So if the statement is true for n = k then it must be true for n = k+1. Therefore, since it is true for n = 4 it must be true for all n > 4.


Why is the sum of two prime numbers greater than two never prime?

Consider prime numbers p1, p2 greater than 2. Since p1 and p2 are prime and greater than 2, they are both necessarily odd. Hence, they are of the form: p1 = 2k+1, p2 = 2j+1, where j and k are positive integers. Their sum is then: p1+p2 = (2k+1)+(2j+1) = 2k+2j+2 = 2 (k+j+1). So 2 divides their sum, hence the sum can't be prime


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