Find the missing value

[BMAD]

 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

[plasmid]

  Reveal hidden contents

Call the original number N and the new number M. Define N₁ as the units digit of N, N₂ as the tens digit of N, etc. If you use the same numbering system for M then the fact that M = 2N means that M₁ = 2N₁ mod 10, M₂ = 2N₂ mod 10 + int(N₁/5), M₃ = 2N₃ mod 10 + int(N₂/5), etc. And the fact that you can move the units digit of N to the front to also generate M means that M₁ = N₂, M₂ = N₃, … M_max = N₁.

Combining the facts that M₁ = 2N₁ mod 10 and M₁ = N₂ means that N₂ = 2N₁ mod 10

M₂ = 2N₂ mod 10 + int(N₁/5) and M₂ = N₃ means N₃ = 2N₂ mod 10 + int(N₁/5)

M₃ = 2N₃ mod 10 + int(N₂/5)  …

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 N₁ = 0 or N₁ = 1 because in those cases it would be impossible for M (where N₁ is now the lead digit) to be 2N. I wrote a spreadsheet to do those calculations for the various potential starting values N₁ and will paste the results.

Now we’re looking for a number where the original N₁ 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 N₁ is even we’re looking for a number with an N_max of N₁/2 and N_max-1 less than 5, and if N₁ is odd then we’re looking for an N_max of N₁/2 rounded down and N_max-1 of 5 or greater. So taking the case of N₁ = 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.

 

[rocdocmac]

  Reveal hidden contents

Answer withdrawn

 

Edited April 24, 2019 by rocdocmac

[rocdocmac]

  Reveal hidden 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..

 

[BMAD]

  On 4/24/2019 at 10:59 AM, rocdocmac said:

  Reveal hidden 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..

 

Expand  

you have not proven nor disproven this

[rocdocmac]

I know, but somebody will perhaps!

  Reveal hidden contents

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

[janakiraman]

the answer will be '00'...lol

 

[rocdocmac]

The way I see it.

  Reveal hidden contents

[rocdocmac]

... this ... (not thus)!

[BMAD]

I think you are on the right track rocdocmac

[Wilson]

Off base probably......

  Reveal hidden contents

Binary  01.......10  ?

 

[rocdocmac]

More ideas ...

  Reveal hidden contents

 

[plasmid]

  On 4/23/2019 at 4:26 PM, 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?

Expand  

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.