A certain number ends with the digit 2. Moving the 2 from the end of the number to its front doubles it. Can you find this number?

(Hint: it's quite large)

let x be the number WITHOUT the 2 at the end or the beginning. Then, if we append a 2 to the end, the value of this number is 10*x + 2. On the other hand, if a 2 is pre-appended to x, the value is 2*10^n + x, where n is the number of digits in x. The latter is double the former, 2*10^n + x = 2*(10*x + 2). Solving this, we get 2*10^n - 4 = 19*x. In other words, x is the quotient when a number of the form 199...996 is evenly divisible by 19. I obtained the solution by long division.

Posted by steve on 2003-05-22 03:09:30