Question 592549
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The fact that the original number is prime is just a red herring.  In fact, if you subtract the sum of the digits from any integer, the result is divisible by 9 and therefore not prime.


Proof for two-digit integers:


Let *[tex \LARGE x] represent the 10s digit and let *[tex \LARGE y] represent the 1s digit.


Then the original integer is *[tex \LARGE 10x\ +\ y] and the sum of the digits is *[tex \LARGE x\ +\ y].


Subtracting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 10x\ +\ y\ -\ (x\ +\ y)\ =\ 9x]


Which is divisible by 9.  So if ANY two digit integer less the sum of its digits is non-prime, any prime two digit integer must exhibit the same characteristic.


Q.E.D.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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