Proving Something is Always SoDate: 04/01/98 at 02:48:12 From: Emma Subject: Prove result is always 1089 I have a question about proving something is always so. John said, "Write down any 3 digit integer and write down the number obtained by reversing the digits. Find the difference between these two numbers. Repeat this process with your answer but this time add the two numbers. The result is always 1089." EXAMPLE: 734 - 437 = 297 and 297 + 792 = 1089. My first question is how do I go about writing a proof? What is necessary for me to prove that the statement is always correct? My next question is a problem that I have come upon while experimenting with different numbers. I have found that my result is a negative number. Is this still correct? EXAMPLE: 123 - 321 = -198 and -198 + -891 = -1089. How do I explain my findings? Help would be much appreciated. Thanks, Emma Date: 04/01/98 at 05:04:45 From: Doctor Pete Subject: Re: Prove result is always 1089 Hi, Usually the hard part about proving these kinds of problems is figuring out what to start with. The idea we need is "how do we express 3-digit numbers?" Well, a 3-digit number, say, 426, is actually a representation of: 426 = 400 + 20 + 6 = 4(100) + 2(10) + 6(1) So we represent any 3-digit number N in terms of its hundreds, tens, and units digits: N = 100h + 10t + u where h may be an integer between 1 and 9 (if h is 0, then N is only 2 digits), and t and u may be integers from 0 to 9. That said, the process of reversing the digits gives you a number: M = 100u + 10t + h Now, suppose that in reversing the digits of N to get M, we find that M < N. Then we want the difference N-M, which is N - M = 100(h-u) + 10(t-t) + (u-h) = 100(h-u) + (u-h) > 0 since M < N implies N-M > 0. It follows that h-u > 0, or u-h < 0. (What happens when M = N?) Well, you can't have a negative units digit, which is how we have so far represented N-M, so we "borrow" 100 from the hundreds digit (we have no tens to borrow but we carry this through): N - M = 100(h-u-1) + 100 + (u-h) = 100(h-u-1) + 90 + 10 + (u-h) = 100(h-u-1) + 10(9) + (10+(u-h)) = 100(h-u-1) + 10(9) + (10-(h-u)) Note that the units digit is now written 10-(h-u), which is clearly between 0 and 9 since h-u > 0. So now we take this number, call it P. Reverse the digits to get Q: Q = 100(10-(h-u)) + 10(9) + (h-u-1) Add the two: P + Q = 100(10-(h-u) + h-u-1) + 10(9+9) + (10-(h-u) + h-u-1) = 100(9) + 10(18) + 9 = 1089 So we're done. The only thing left is to consider the case when N < M, which gives you negative numbers if you take the difference N-M. (Note that it would be the same if you took the absolute value difference, |N-M| = |M-N|), but you would use similar reasoning.) The only case in which the process fails, which I hinted at earlier, is when N = M. For example, 464, or 121, reversing these numbers and subtracting gives 0, and 0 + 0 = 0. Can you see where the proof "breaks down" in this case? Also, observe that a crucial step in the proof is to obtain N-M = P in its hundreds-tens-units form; i.e., its digits. If we did not do this, we could not reverse them to find Q. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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