



 
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 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 10^{4} = 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 10^{i} = 19h + 2 for some positive integer i. Thus, 10^{i}  2 divides 19. or in other words, 10^{i} 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 10^{17} mod 19 = 2. So i = 17, then p = 10^{17}. Then h = (10^{17}  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. Cheers,  


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