Floating 2

Date: 06/12/2001 at 10:34:59
From: Britney Turner
Subject: Algebra

The floating 2:

Start with the number 2 on the far left side, then float the number 2 
to the far right side. The new number must be three times more than 
the old number.
 
For example: 268  682.

The only problem is that they're not three times apart.

Date: 06/12/2001 at 12:55:22
From: Doctor Greenie
Subject: Re: Algebra

Hi, Britney -

I'm hoping that you haven't written your question the way you really 
meant to. If the new number is "three times more than" the old number, 
then it is FOUR times AS LARGE AS the old number. For that problem, 
your question has no solution.

However, if you meant to say that the new number is three times AS 
LARGE AS the old number, then there are solutions.

So let's look at this problem: You want to find a string of digits 
"dd...dd" (with an unknown number of digits) such that the number 
"dd...dd"2 is three times as large as the number 2"dd...dd".

(This is awkward notation.  Remember that "dd...dd" is a string of 
digits; there is no multiplication implied by writing the "d"s next to 
each other.)

We can get a series of equations to try to solve by trying different 
numbers of digits in the string "dd...dd".

If we try a string of a single digit, then we have

    "d"2 = 3 * 2"d"

"d" is now a 1-digit number; call it x.  Then algebraically we have

    10x+2 = 3(20+x)
    10x+2 = 60+3x
       7x = 58

The solution to this equation is not an integer.

So next let's try a string of 2 digits. We now have

    "dd"2 = 3 * 2"dd"

"dd" is now a 2-digit number; call it x. Then algebraically we have

    10x+2 = 3(200+x)
    10x+2 = 600+3x
       7x = 598

The solution to this equation also is not an integer.

Let's go one step further and try a string of 3 digits. We now have

    "ddd"2 = 3 * 2"ddd"

"ddd" is now a 3-digit number; call it x. Then algebraically we have

    10x+2 = 3(2000+x)
    10x+2 = 6000+3x
       7x = 5998

The solution to this equation also is not an integer.

If we pause now and compare the calculations we have made for these 
first three cases, we can see that what we are going to need is an 
equation of the form

    7x = 599...998

which has an integer solution.

To find equations of this form, we can be a bit clever. We are 
currently looking for numbers of the form 599...998 which, when 
divided by 7, leave no remainder. Let's look instead for numbers of 
the form 600...000 which, when divided by 7, leave a remainder of 2.

The repeating decimal equivalent of the fraction 6/7 is

   6/7 = 0.857142857142857142...

From this we can see that

         60/7 =      8.57142857142857142... =        8 remainder 4
        600/7 =     85.7142857142857142...  =       85 remainder 6
       6000/7 =    857.142857142857142...   =      857 remainder 1
      60000/7 =   8571.42857142857142...    =     8571 remainder 3
     600000/7 =  85714.2857142857142...     =    85714 remainder 2

We have now that 600000/7 leaves remainder 2, so 599998/7 is an 
integer (=85714).  So our first solution to the problem is

    x = 285714;  3x = 857142

And if we think about continuing looking for numbers of the form 
600...000 which leave remainder 2 when divided by 7, we can see that 
we will have an infinite number of similar solutions to the problem:

    x = 285714285714;  3x = 857142857142

    x = 285714285714285714; 3x = 857142857142857142

and so on....

And not only do we have an infinite number of solutions to the 
problem; this infinite set of solutions of this form provides the only 
solutions to the problem.

Let's go back and see why the question as you asked it has no 
solutions (that is, there are no solutions if dd...dd2 is four times 
as large as 2dd...dd instead of three times as large).

In the above analysis of the problem where "dd...dd2" is three times 
as large as 2dd...dd, we encountered the following series of equations 
to try to 
solve:

    7x = 58    where x is a 1-digit integer
    7x = 598   where x is a 2-digit integer
    7x = 5998  where x is a 3-digit integer
   ...

If we do the same type of analysis for the problem where dd...dd2 is 
supposed to be four times as large as 2dd...dd, then the corresponding 
series of equations we get is the following:

    6x = 78    where x is a 1-digit integer
    6x = 798   where x is a 2-digit integer
    6x = 7998  where x is a 3-digit integer
   ...

In every case, these equations have integer solutions (every number of 
the form 799...998 is divisible by both 2 and 3, and therefore by 6) -
but they are always the wrong number of digits.

Thanks for the interesting problem. It turned out to be a lot more 
interesting than it sounded when I first read it!

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/