Conference rusure::math

    re .0,
    
    Me think Ya friend's son got a bad math teacher for it has nothing to 
    do with 'em being primes...
    
                  2        
    (n-1)(n+1) = n  - 1
    
    As to the second part, I tried a few example, and you are right.  But I
    have no clue as to why they are so or if there are any counterexamples.
    I get nervous when people add primes...
    
    Eugene

    Re .1, .0:
    
    > As to the second part, I tried a few example, and you are right.  But I
    > have no clue as to why they are so or if there are any counterexamples.

    This one has a little to do with the fact that they are primes.
    
    What is the remainder of the first number when divided by 9?  It could
    be 0, 1, 2, 3, 4, 5, 6, 7, or 8 -- except that it couldn't be 0, 3, or
    6, because then it would be divisible by 3 and would not be prime
    (unless the first number were exactly 3, which gives us the sole
    exception -- the 3 and 5 pair).  So the remainder of the first number
    is 1, 2, 4, 5, 7, or 8.  The second number is two greater than the
    first, so it's corresponding remainders are 3, 4, 6, 7, 0, or 1. 
    Again, 3, 6, and 1 are not possible.  So the pair of numbers has
    remainders 2 and 4, 5 and 7, or 8 and 1.  And 2*4 = 8, 5*7 = 35, and
    8*1 = 8.  The remainder of 35 when divided by 9 is 8, so the product of
    each pair of twin primes, except 3*5, is 8 modulo 9.
    
    
    				-- edp

    re .*
    
    On the second part, it is well known that all primes are of the form
    6n�1 excepting 2 and 3. (The converse is not true.)
    
    Thus the product is (6n-1)(6n+1) = 36n�-1 == -1 (mod 9) == 8 (mod 9).
    QED.

RE: .2 and .3

    Implicit in the previous two responses is the well but not universally
    known fact that repeatedly adding the digits in a number gives the
    remainder of that number when divided by 9.  To see this take the
    number X:

			abcdef

    where a, b, c, d, e, and f are the 6 digits of the number.  X therefore
    equals:

	a*100000 + b*10000 + c*1000 + d*100 + e*10 + f =
	a*99999 + b*9999 + c*999 + d*99 + e*9 + a + b + c + d + e + f =
	(a*11111 + b*1111 + c*111 + d*11 + e)*9 + (a+b+c+d+e+f)

    Now the term on the left is 0 mod 9 so the term on the right -- the
    sum of the digits -- will have the same remainder as X.  Since it will
    be smaller than X unless X has one digit, repeating the process will
    reduce the value until it has 1 digit, which will be the remainder when
    divided by 9 unless the result is 9, in which case the remainder is 0.

    This is why casting out the 9's works -- it is a simple way to check
    that the right relationship holds between the remainders of the input
    terms and the output value of an arithmetic computation.

			    Topher

Topher, thanks for presenting casting-out-9's like that.  It's a good
way to see "why it works".

/Eric

    Hi...
    
    Many thanks to all who responded. I've passed on the information to
    those concerned.
    
    Cheers Lawrie