Reversing the DigitsDate: 07/28/99 at 08:15:15 From: Kelly Wray Subject: Reversing the digits We have investigated this question, and we know ho to do it, but we do not know why it is so, which is a major part of our maths investigation. We would greatly appreciate it if you could explain why when you: 1. Take any two-digit number, 2. Reverse the digits of this number, 3. Subtract the reversed number from the original one, 4. Divide the result by the difference of the two digits of the original number the answer is always nine. Example: 73 - 37 = 36/(7-3) = 9 Thank-you very much. Kelly Date: 07/28/99 at 10:29:19 From: Doctor Rick Subject: Re: Reversing the digits Hi, Kelly. I don't know what level of math you can work with. I hope you know some algebra, because that's the best way to understand this phenomenon. Here is what to do. Call the tens digit of the bigger number, a. Call the ones digit of the bigger number, b. Then the bigger number is 10a + b. What do you get when you reverse the digits? Follow the steps you listed, using these algebraic expressions instead of particular numbers. You will get an expression that you can simplify. See what happens then. By the way, will this really work with ANY two-digit number? Your step 3 should be changed to "subtract the smaller from the greater of the numbers from steps 1 and 2." Even then, it won't always work. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 07/29/99 at 04:38:04 From: Kelly Wray Subject: Re: Reversing the digits Thank you very much for your reply. Unfortunately, I still do not understand why the answer always equals nine. I understand the concept that the bigger number is 10a + b, and when you reverse that it is 10b + a. So 10a-b - 10b-a = x/(a-b) = 9 But I still do not understand why this is the case. Also I am aware that there are many exceptions to this rule, such as the multiples of eleven. Any further help would be greatly appreciated. Thank you. Kelly Wray Date: 07/29/99 at 11:40:15 From: Doctor Rick Subject: Re: Reversing the digits Hi, Kelly. Yes, there are exactly 9 exceptions, because 0/0 does not equal 9; it is "indeterminate." You start out well in writing out the process as an expression, but you need to follow through and be careful. Watch as I follow the steps: 1. Take any two-digit number: 10a + b 2. Reverse the digits of this number: 10b + a 3. Subtract the reversed number from the original one: (10a + b) - (10b + a) [Note the parentheses you omitted] 4. Divide the result by the difference in the two digits of the original number: (10a + b) - (10b + a) --------------------- (a - b) Now simplify this expression and see what you get. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/