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.
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
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.
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