# James Booth's 2 sets of 3 digit numbers

James Booth showed in a paper published in 1854 that a six figure number in which the first three places coincided with the final three places, for example, 376376, 459459 or 301301, is always divisible by 7, 11 an 13. He effectively uses modular arithmetic. However, it may be easier to see using the fact that
$abcabc = 1001.abc$

Since 1001 = 7.11.13 the result is clear.
Booth then writes:-
Like properties may be found for 17, 19, 23, but the periods are longer. The prime divisor being 2n + 1, it is manifest the number of places in the period cannot exceed; however it may fall short of n. Thus when the divisor is 17, the number of places in the period is eight.
It isn't clear what he means by this. If $2n + 1$ only refers to 17, 19, 23 then he is correct. The periods are 8, 9 and 11. However, these are exactly $n$ and they do not fall short. Clearly 2 sets of $n$ digits gives numbers divisible by the factors of $10^{n} + 1$ so these facts are easy to check. If, however, Booth means that the result hold for all other values of $n$ where $2n + 1$ is prime then he is incorrect. For example, it is false for $2n + 1 = 31$.

Now if $p$ is an odd prime, say $p = 2n + 1$, then 2 sets of $n$ digits is divisible by $p$ for $p = 7, 11, 17, 19, 23, 29$ but not for 31, 37, 41, 43. For 13 we get periods of $3 + 6k$ but not 6. For primes 47, 59, 61 again we get period $(p - 1)/2$ but not for 53, 67, 71. Again we get period $(p - 1)/2$ for 73, 97, 101, 103, 109, 113 but not 79, 83, 107. For 89 we get a period of 22 but not of 44.

Last Updated February 2016