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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1410.0. "Twin Primes" by MEO78B::HANSON (Resistance is useless) Wed Apr 03 1991 02:01

    Hi...
    
    A friend's son had an interesting problem set as homework.
    
    First he was asked to calculate all twin primes between 0 and 100.
    No problems here (3 & 5, 5 & 7, 11 & 13, etc)
    
    Then he was asked to calculate the products and note anything
    interesting about the series of numbers produced (15, 35, 143, etc)
    
    The answer that he got back was that each product was the square of the
    mean of the twin primes minus 1.
    
    But one thing I observed was that with the exception of the first
    product (15), each of products digits if summed until a single digit is
    left, always results in 8. (3 + 5 = 8, 1 + 4 + 3 = 8, etc). Intrigued,
    I even checked it for the twin primes 1000000009649 & 1000000009651, 
    the product of which is 1000000019300000093122499. And yes, summing the 
    digits produces 53, which of course sums to 8.
    
    Is this a general rule for twin primes? Is there a proof?
    
    Thanks & Regards,
    
    Lawrie 
T.RTitleUserPersonal
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1410.1HPSTEK::XIAIn my beginning is my end.Wed Apr 03 1991 04:0013
    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
1410.2JARETH::EDPAlways mount a scratch monkey.Wed Apr 03 1991 09:5221
    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
1410.3ELIS::GARSONV+F = E+2Wed Apr 03 1991 10:077
1410.4"digital roots"CADSYS::COOPERTopher CooperWed Apr 03 1991 14:0827
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
1410.5thanks Topher, clear explanationHANNAH::OSMANsee HANNAH::IGLOO$:[OSMAN]ERIC.VT240Thu Apr 04 1991 16:107
Topher, thanks for presenting casting-out-9's like that.  It's a good
way to see "why it works".

/Eric


1410.6Thanks everyoneMEO78B::HANSONResistance is uselessThu Apr 04 1991 23:186
    Hi...
    
    Many thanks to all who responded. I've passed on the information to
    those concerned.
    
    Cheers Lawrie