Floating 2Date: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/