BMAD Posted April 23, 2019 Report Share Posted April 23, 2019 (edited) I have in mind a number which, when you remove the units digit and place it at the front, gives the same result as multiplying the original number by 2. Am I telling the truth? Edited April 23, 2019 by BMAD Quote Link to comment Share on other sites More sharing options...

1 plasmid Posted May 8, 2019 Report Share Posted May 8, 2019 Spoiler Call the original number N and the new number M. Define N_{1} as the units digit of N, N_{2} as the tens digit of N, etc. If you use the same numbering system for M then the fact that M = 2N means that M_{1} = 2N_{1} mod 10, M_{2} = 2N_{2} mod 10 + int(N_{1}/5), M_{3} = 2N_{3} mod 10 + int(N_{2}/5), etc. And the fact that you can move the units digit of N to the front to also generate M means that M_{1} = N_{2}, M_{2} = N_{3}, … M_{max} = N_{1}. Combining the facts that M_{1} = 2N_{1} mod 10 and M_{1} = N_{2} means that N_{2} = 2N_{1} mod 10 M_{2} = 2N_{2} mod 10 + int(N_{1}/5) and M_{2} = N_{3} means N_{3} = 2N_{2} mod 10 + int(N_{1}/5) M_{3} = 2N_{3} mod 10 + int(N_{2}/5) … Well, I could keep writing all that stuff out, but you get the idea. If you’re given the ones digit, you can solve for the rest of the digits. So we can just go through all the possible values for the ones digit and find out if we can generate a number that fits the description of the problem. Note that we can skip the cases where N_{1} = 0 or N_{1} = 1 because in those cases it would be impossible for M (where N_{1} is now the lead digit) to be 2N. I wrote a spreadsheet to do those calculations for the various potential starting values N_{1} and will paste the results. Now we’re looking for a number where the original N_{1} digit (which becomes the M_{max} digit) would be 2N_{max} + int(N_{max-1}/5) which is what you know must be true if M = 2N. In English: if N_{1} is even we’re looking for a number with an N_{max} of N_{1}/2 and N_{max-1} less than 5, and if N_{1} is odd then we’re looking for an N_{max} of N_{1}/2 rounded down and N_{max-1} of 5 or greater. So taking the case of N_{1} = 2, we can look down the list of subsequent digits until we find a potential N_{max} where N_{max} = 1 and N_{max-1} < 5, which occurs at digit 18. That gives the number 105263157894736842. And manually checking, we see that in fact moving the ones digit to the front gives the same value as multiplying the original number by 2, which is 210526315789473684. A similar thing can be done for the other potential starting units digits to find more numbers that would be a solution to the problem. Quote Link to comment Share on other sites More sharing options...

0 rocdocmac Posted April 24, 2019 Report Share Posted April 24, 2019 (edited) Spoiler Answer withdrawn Edited April 24, 2019 by rocdocmac Quote Link to comment Share on other sites More sharing options...

0 rocdocmac Posted April 24, 2019 Report Share Posted April 24, 2019 Spoiler For 11 up to 9999 it's a lie! After that, maybe too. Some figures get close, however, e.g. 25, 37, 449, 4499, one or two off the mark.. Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted April 25, 2019 Author Report Share Posted April 25, 2019 On 4/24/2019 at 6:59 AM, rocdocmac said: Hide contents For 11 up to 9999 it's a lie! After that, maybe too. Some figures get close, however, e.g. 25, 37, 449, 4499, one or two off the mark.. you have not proven nor disproven this Quote Link to comment Share on other sites More sharing options...

0 rocdocmac Posted April 26, 2019 Report Share Posted April 26, 2019 (edited) I know, but somebody will perhaps! Spoiler Used trial and error up to 50 000. Don't know how to prove it mathematically (yet). I believe the statement will only be true at infinity. Edited April 26, 2019 by rocdocmac Quote Link to comment Share on other sites More sharing options...

0 janakiraman Posted April 26, 2019 Report Share Posted April 26, 2019 the answer will be '00'...lol Quote Link to comment Share on other sites More sharing options...

0 rocdocmac Posted April 26, 2019 Report Share Posted April 26, 2019 The way I see it. Spoiler Quote Link to comment Share on other sites More sharing options...

0 rocdocmac Posted April 26, 2019 Report Share Posted April 26, 2019 ... this ... (not thus)! Quote Link to comment Share on other sites More sharing options...

0 BMAD Posted April 27, 2019 Author Report Share Posted April 27, 2019 I think you are on the right track rocdocmac Quote Link to comment Share on other sites More sharing options...

0 Wilson Posted April 28, 2019 Report Share Posted April 28, 2019 Off base probably...... Spoiler Binary 01.......10 ? Quote Link to comment Share on other sites More sharing options...

0 rocdocmac Posted April 28, 2019 Report Share Posted April 28, 2019 More ideas ... Spoiler Quote Link to comment Share on other sites More sharing options...

0 plasmid Posted May 8, 2019 Report Share Posted May 8, 2019 On 4/23/2019 at 9:26 AM, BMAD said: I have in mind a number which, when you remove the units digit and place it at the front, gives the same result as multiplying the original number by 2. Am I telling the truth? You could be telling the truth because such a number does exist. Of course I can't say with certainty that you're telling the truth because I don't know whether you actually have such a number in mind right now. Back in a bit with a description of how to handle this problem. Quote Link to comment Share on other sites More sharing options...

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I have in mind a number which, when you remove the units digit and place it at the front, gives the same result as multiplying the original number by 2. Am I telling the truth?

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