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It appears that the product of any pair of twin primes (excluding the first pair 3 and 5) yields a semi prime whose digital root is equal to $8$.

Example: $$ 17 \cdot 19 = 323 $$ The digital root of $323$ is $8$.

I've tested the first twenty and a bunch of random large ones such as $$ 8231 \cdot 8233 = 67765823 $$ Its digital root is also $8$.

As an amateur in advanced mathematics I'm curious to know how I could prove or disprove this conjecture. All tips are welcomed and much appreciated. Thanks.

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  • $\begingroup$ Hint: apart from $\{3,5\}$, all twin primes are of the form $\{6k-1,6k+1\}$. $\endgroup$ – Théophile Jun 22 '15 at 19:38
  • $\begingroup$ @Théophile is that why they all equal 8? $\endgroup$ – Tony Jun 22 '15 at 19:40
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Your conjectured fact is true.

Of any $3$ consecutive integers, one is divisible by $3$. If $2$ of the integers are a pair of twin primes neither of which is $3$, then neither of the "end" numbers is divisible by $3$. So the "middle" number must be divisible by $3$.

So the twin primes are $3k-1$ and $3k+1$ for some integer $k$. It follows that their product is $9k^2-1=9k^2-9+8$. We conclude that the remainder when $(3k-1)(3k+1)$ is divided by $9$ is $8$.

But the remainder when a number $n$ is divided by $9$ is the same as the remainder when the sum of the decimal digits of $n$ is divided by $9$. Thus the digital root of our product is $8$.

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  • $\begingroup$ Very nice and detailed, +1 $\endgroup$ – Atvin Jun 22 '15 at 19:45
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My approach is similar but less elegant. Still, if it's of any use to you, I'm happy.

First of all, you're right to suspect there's something special about $3 \times 5$. This is because any number for which the digital root is equal to $3$, $6$ or $9$ is divisible by $3$ and must therefore be composite, unless that number is $3$ itself (or $-3$, but for this topic we don't gain much by considering negative numbers, so I'll ignore them from this point on). And we don't consider $1$ prime anymore (though I'm so old I was taught otherwise), so $3 \times 5$ is the only case for which one of the primes has a digital root equal to $3$.

This leaves three cases to consider:

  • $2 \times 4 = 8$, e.g., $11 \times 13 = 143$
  • $5 \times 7 = 8$, e.g., $5 \times 7 = 35$
  • $8 \times 1 = 8$, e.g., $17 \times 19 = 323$.

(Remember that you can multiply digital roots directly).

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The digital root of $n$ repeats with period $9$, viz. $1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, \ldots$ so the digital roots of a twin prime pair must be $(1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8), (7, 9)$, or $(8, 1)$. However, given that the digital root of a prime number greater than 3 cannot be $3, 6$ or $9$, that leaves $(2, 4), (5, 7)$, or $(8, 1)$. (adapted from a posting by David Radcliffe to a LinkedIn Number Theory group discussion)

And here's a link to a proof that contextualizes the digital root sequencing of twin primes: http://www.primesdemystified.com/twinprimesdigitalrootproof

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