Hi Jim. Here's how I would approach the problem.

A number (N) ends in a 2. Therefore, there exists a number n, which you can multiply by 10 and add 2 to get N:

N = 10n + 2

Now if you move the 2 from the back to the front, shifting all the other digits one place towards the decimal point, you get twice the original, which mean 2N.

So
2N = 2(10n + 2) = 20n + 4.

but also, 2N = 2p + n, where p is a power of 10 where the power is the number of digits in n.

For example, if N = 65432, then n is 6543. p is 10⁴ = 10000, so 2p + n is 20000 + 6543 = 26543 (clearly this is not equal to 2N, but it was just an example after all).

Since both expressions equal 2N, they equal each other:

20n + 4 = 2p + n
19n + 4 = 2p
p = (19/2)n + 2

Since we know n and p are a positive integer, n must be even. Thus there exists a positive integer h = n/2.

p = 19h + 2.

Now remember that p is a power of 10, so

10ⁱ = 19h + 2

for some positive integer i. Thus,

10ⁱ - 2 divides 19.

or in other words,

10ⁱ mod 19 = 2.

Since this will cycle in 19 steps, it is easy to just check them using a calculator or spreadsheet. I find that 10¹⁷ mod 19 = 2.

So i = 17, then p = 10¹⁷.

Then h = (10¹⁷ - 2 ) / 19 = 5263157894736842.

And n is twice that: 10526315789473684.

Thus N is 10n + 2 = 105263157894736842.

So if we move the 2 from the back to the front we have 210526315789473684.

And is that 2N? Yes it is.
I hope you are convinced that we found a number that works, even if it is less clear why this is the lowest such number.

Cheers,
Stephen La Rocque.