Double-Or-Quits-Dept:
(The Endgame was: "A number ends with the digit 2. If we move this 2 from the last place to the first, the new number is twice the original. What's the number" -- MS)
The answer is 105,263,157,894,736,842. Since the last digit '2' is going to be made the first digit of the doubled number, the original number must start with '1'. Moreover, as the last digit is '2', the last but one digit has to be '4' (since we are doubling the original number). Similarly, the last but two digit has to be '8' (double of last but one digit '4'). In this fashion, keep moving from right to left, and continue to derive the subsequent digits by doubling the previous digit, keeping into account the 'carry forwards' too, till you arrive at '1'. On reaching the digit '1' test the number for the required condition. You'll notice that it doesn't meet the requirements of the question on reaching the first '1'. Second time it does.
Sameer Bhargava, SameerB@mashreqbank.com
The number we are looking for is (hold your breath) 105,263,157,894,736,842. I derived it this way by writing an equation: 2x/2 = x2 (assuming the number to be two-digit). I found that whatever value x possessed, the last digit on the LHS would be 4. So I rearranged the equation as: 2x42 = x42. Here we see that whatever value x possesses, the second-last digit on LHS will be 8. So rearranging again, we get: 2x84/2 = x842 (note again that irrespective of value of x, the third-last digit on LHS is 6). Do I need to continue further and state that . . . (No you don't need to do any further damage -- MS)
Saifuddin Khomosi, Dubai,UAE
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