Senseless promotions? Or, the wisdom of the tea people

[bonanova]

Things are not going well at the Acme Company.

Executive talent is hard to come by, and it is not cheap. Folks at the water cooler have no ideas, and the coffee-breakers can't imagine how to improve things either. But those who party around the teapot, they came up with something. They suggested to the Board that Acme promote the newest hire in the mail room and make him the CEO! We need to shake things up, but good. Qualifications, job experience, brains, judgment, integrity, these are all things of the past.

Some were not so sure. Doesn't make sense at all, the old timers said. Almost like appointing some guy with orange complexion to be President. That's exactly the idea, said the tea-people. Turn things on end, let the bull loose in the china shop, and see what happens. Hey -- how could it be worse than what we have now? Not surprisingly, the debate was long and heated.

Such a risk merited proof of possible gain, so the old guard posed a challenge: produce a concrete example of where the idea had been tried with incontrovertible benefit. In fact, make it mathematical. You know, something that might make a good BrainDen puzzle.

We'll promote the mail room guy, they said, if you can show us an integer that doubles in value when its least significant digit is promoted to its most-significant position.

That is, give us a number { some digits } q that has half the value of q { same digits }.    

Spoiler

102 doesn't quite work, since 210 does not quite equal 204

That all happened last week, and now we're looking for the mail room guy. Was he promoted? Did the tea people find such a number? Is there one?

We need a number or a proof that one does not exist.

T.L.D.R.

What number doubles in value by by moving its last digit to the first position (if there is one)?

[Molly Mae]

Ouch

 

I wasn't expecting a number this long and I almost gave up thinking it wouldn't work.  After hunting for a few random 2- and 3-digit numbers, I decided to try it mathematically.  I started with a number beginning with the number 1, so I know the last number must be 2, which means the last number of the re-arranged number must be 4.  Then, since the number would have to be doubled, I multiplied the 4 by 2, so the rearranged number must end in ...84.  I kept this up until I arrived at 105,263,157,894,736,842.  Shifting the 2 to the front returns 210,526,315,789,473,684, which is double the original number. 

I can't guarantee it's the smallest number with this property, but it's the first one that begins with a 1.

EDIT: Another note, I use Google's calculator function to verify that the second number is indeed double the first and discovered that these are called parasitic numbers.  I think that's a fun name.

  

Edited March 8, 2018 by bonanova
Gold Star

[Molly Mae]

1 hour ago, Quantum.Mechanic said:

I must be thick today, I can't follow the given solution. Can someone paste a little bit more hand holding? Like the first 5 values in the sequence?

EDIT: For clarity, I use "number" to reference an individual digit.  I use "sequence" to reference the string of numbers.
EDIT2: Added spoiler tag.

Sorry, I'm pretty bad at explaining things.  You're basically building two sequences: n and 2n.  I arbitrarily chose 1 as the first number of the first sequence and we don't know the length of the sequence, so it looks like this:

First: 1????...???
Second: ?????...???

Since 1 is the first number of the first sequence and the second sequence is double the first sequence, we can update our second sequence.

First: 1????...???
Second: 2????...???

From the OP, we are promoting the least significant number of the first sequence to the most prominent position.  So we now know that the first sequence ends with a 2.  Let's update that.

First: 1????...??2
Second: 2????...???

Again, since our second sequence is double the first, we know that the second sequence ends in a 4.

First: 1????...??2
Second: 2????...??4

Remember that the second sequence is the first sequence except that the last number is moved to the front.  So the second to last number of the first sequence is a 4.

First: 1????...?42
Second: 2????...??4

Four doubled is 8.

First: 1????...?42
Second: 2????...?84

The 8 moves up to the first sequence:

First: 1????...842
Second: 2????...?84

Eight doubled is 16.

First: 1????...842
Second: 2????...684

You can keep doing this until your second sequence can begin with a 2 without having to carry over a number.  It's a neat puzzle where you build two sequences simultaneously using new information from one sequence to build the next part of the other sequence.

I hope that clears it up.

Edited March 9, 2018 by Molly Mae

[CaptainEd]

Nice approach, Molly Mae!

[bonanova]

1 hour ago, Molly Mae said:

Ouch

  Reveal hidden contents

I wasn't expecting a number this long and I almost gave up thinking it wouldn't work.  After hunting for a few random 2- and 3-digit numbers, I decided to try it mathematically.  I started with a number beginning with the number 1, so I know the last number must be 2, which means the last number of the re-arranged number must be 4.  Then, since the number would have to be doubled, I multiplied the 4 by 2, so the rearranged number must end in ...84.  I kept this up until I arrived at 105,263,157,894,736,842.  Shifting the 2 to the front returns 210,526,315,789,473,684, which is double the original number. 

I can't guarantee it's the smallest number with this property, but it's the first one that begins with a 1.

EDIT: Another note, I use Google's calculator function to verify that the second number is indeed double the first and discovered that these are called parasitic numbers.  I think that's a fun name.

  

@Molly Mae gets the coveted bonanova gold star for finding an algorithm and getting the answer in less than one day.

So now that the cat (algorithm) is out of the bag, so to speak, it should be trivial to find whether an integer can triple, (or quadruple, or quintuple) under the same manipulation.

It's also interesting that his solution forms a ring that shows the doubling process also occurs for four other ending digits (not 1, so his is the smallest.)  But which ones? Incidentally, this also shows that all such integers have 18 digits.

[Molly Mae]

10 minutes ago, bonanova said:

@Molly Mae gets the coveted bonanova gold star for finding an algorithm and getting the answer in less than one day.

Signature updated!

[bonanova]

Just now, Molly Mae said:

Signature updated!

I'm humbled.

btw I sharpened (with humor) the flavor text and posted this on another favorite site. It will be interesting to see how quickly it's solved there.

[Quantum.Mechanic]

I must be thick today, I can't follow the given solution. Can someone paste a little bit more hand holding? Like the first 5 values in the sequence?

[Quantum.Mechanic]

On 09/03/2018 at 5:54 PM, Molly Mae said:

EDIT: For clarity, I use "number" to reference an individual digit.  I use "sequence" to reference the string of numbers.
EDIT2: Added spoiler tag.

 

  Hide contents

 

Sorry, I'm pretty bad at explaining things.  You're basically building two sequences: n and 2n.  I arbitrarily chose 1 as the first number of the first sequence and we don't know the length of the sequence, so it looks like this:

First: 1????...???
Second: ?????...???

Since 1 is the first number of the first sequence and the second sequence is double the first sequence, we can update our second sequence.

First: 1????...???
Second: 2????...???

From the OP, we are promoting the least significant number of the first sequence to the most prominent position.  So we now know that the first sequence ends with a 2.  Let's update that.

First: 1????...??2
Second: 2????...???

Again, since our second sequence is double the first, we know that the second sequence ends in a 4.

First: 1????...??2
Second: 2????...??4

Remember that the second sequence is the first sequence except that the last number is moved to the front.  So the second to last number of the first sequence is a 4.

First: 1????...?42
Second: 2????...??4

Four doubled is 8.

First: 1????...?42
Second: 2????...?84

The 8 moves up to the first sequence:

First: 1????...842
Second: 2????...?84

Eight doubled is 16.

First: 1????...842
Second: 2????...684

You can keep doing this until your second sequence can begin with a 2 without having to carry over a number.  It's a neat puzzle where you build two sequences simultaneously using new information from one sequence to build the next part of the other sequence.

I hope that clears it up.

 

Yes, thanks very much. I got lost trying to do this myself, but "it's obvious in hindsight", as they say.