Guest Posted June 24, 2009 Report Share Posted June 24, 2009 Any prime number greater than or equal to 5 can be expressed in the form 6p+1 or 6p-1, where p is a positive integer. Can you prove it? The proof is very simple..... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 24, 2009 Report Share Posted June 24, 2009 The problem should read like this.... either 6p+1 or 6p-1.. e.g. 41 is in the form 6p-1 (where p is 7) and 13 is in the form 6p+1 (where p is 2).... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 24, 2009 Report Share Posted June 24, 2009 Its easy one...... Any number can be written as 6P or 6P+1 or 6P+2 or 6P+3 or 6P+4 or 6P+5......... Now, 6P, 6P+2 and 6P+4 are always divisible by 2. 6P+3 is always divisible by 3. so we are left with 6P+1 and 6P+5 for prime numbers...... And, 6P+5 is same as 6P-1..... proved. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted June 24, 2009 Report Share Posted June 24, 2009 Any prime number greater than or equal to 5 can be expressed in the form 6p+1 or 6p-1, where p is a positive integer. Can you prove it? The proof is very simple..... any number can be expressed as a factor of 6 with r remainder: x = 6m + r where m is the multiple and r >= 0 and r <=5 . if r = 0 , 2 ,3, 4 the number cannot be prime since it will be divisible by 2 or 3 therfore for a number to be prime r must be 1 or 5. Therefore x = 6m + 1 or: x = 6m +5 = 6(m+1) - 1 = 6(new m) -1 Thus proved (Note all prime numbers will be contained in this... but not all such numbers will be prime.... this is a superset of prime numbers ) Quote Link to comment Share on other sites More sharing options...
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Any prime number greater than or equal to 5 can be expressed in the form 6p+1 or 6p-1, where p is a positive integer. Can you prove it?
The proof is very simple.....
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