I have been aware for many years of a simple parlour trick that appears almost magical to most people. This involves the following steps.

1. Ask members of the audience to select a three digit number, eg 245.
2. Now repeat the three digits and making a 6 digit number eg 245245.
3. Ask everyone to divide the 6 digit number by 7.
4. Ask everyone to divide the result of 3. above by 11.
5. Ask everyone to divide the result of 4. above by 13.

Astound everyone by revealing the result is their original 3 digit number.

Alternatively, you could replace one of the above divisors by the original 3 digit number, and the result for everybody will be the divisor that was replaced.

For example, with the number 245 this becomes a 6 digit number 245245.
Divide 245245 by 7 giving 35035.
Divide 35035 by 11 giving 3185.
Divide 3185 by 13 giving 245, our original 3 digit number.
If you divided 3185 by our original number 245 we get 13.

Why does this work?

Well, if you start with a 3 digit number and repeat the digits to make a 6 digit number, this is the same as multiply our 3 digit number by 1001.
For our example 245 x 1001 = 245245.

But 1001 = 7x11x13
So any 3 digit number will work to give the required result.

This then got me thinking, does this work in any other base, such base 12.

Well, for base 12, 1001* = 1728 + 1 =1729
and 1729 = 7 x 13 x 19 or 7 x 11* x 17*

So then I thought, what about other bases.

So here is the table I came up with.

Now, are there any maths wizzes that know if there is a general proof that numbers of this form i.e (n^3 +1) are always non-prime?

n^3+1 = (n+1)(n^2-n+1)
or, in reverse notation:


From basic algebra
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
a^3 - b^3 = (a - b)(a^2 + ab + b^2)