Every so often, a puzzle comes along and is just right for its time. Not so hard that you waste hours on it, but not so easy that it pops out straight away. I heard this from Simon at Big MathsJam last year and thought it’d be a good one to share and analyse. I’ve adopted (and slightly adapted) @colinthemathmo’s wording of it:
Apart from exactly one exception, the digital root ((The digital root of a number is the result of summing its digits, and summing the digits of that sum, repeatedly, until you reach a single number. For example, the digital root of 981 is 9: 9+8+1 = 18 and 1+8=9.)) of the product of twin primes is always 8. Why?
I’d recommend convincing yourself that it’s true, finding the exception, and having a go at a proof before reading on. Assuming you want to.
My proof is as follows: except for the pair 3 and 5, twin primes are always of the form $6n-1$ and $6n+1$, or else one of them would be divisible by 2, 3 or both.
The product $(6n-1)(6n+1) = 36n^2 - 1$ is one less than a multiple of 9. Taking the digital root of this gives the remainder modulo 9, which is 8.
With a tombstone.
In fact, it can be taken further: all of these twin prime products are congruent to 35 (modulo 36). It turns out that you can find a similar result in other bases than 10 - in fact, in any base that’s one more than a factor of 36 (except for base 2).
In base 3, the iterated digit sum also turns out to be 1; this is the remainder modulo 2, which tells you the product of twin primes is odd. Again, not exactly impressive.
Base 4 is where we start getting somewhere. The iterated digit sum here is two, meaning the product is one less than a multiple of 3.
For bases 3, 4, 5, 7, 10, 13, 19 and 37, the product of twin primes turns out to be two less than the base.
Isn’t that neat?
* Edited 2017-13-03 to include a definition of digital root, at @colinthemathmo’s suggestion. Thanks, Colin!
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