Reversing a Number by Multiplying by 9
Date: 08/23/99 at 06:32:01 From: Jacky Chan Subject: Inverse the digits of a number by multiplying by 9 When some numbers are multiplied by 9, why is the result the reverse of the original number? For example, 1089*9 = 9801. Are there any rules or theories about this? What is this called? What must be the thousands digit, hundreds digit, tens digit or units digit be? What about n-digit numbers? Is it possible to write out all these types of numbers? Can you tell me more about it and give me an explanation in detail as well as a conclusion?
Date: 08/23/99 at 12:42:53 From: Doctor Rob Subject: Re: Inverse the digits of a number by multiplying by 9 Thanks for writing to Ask Dr. Math. As far as I know, there is no name for this kind of situation or this kind of number. Such numbers exist because solutions exist to the kind of equation I'll give below. You can write out all such numbers for any given number n of digits. To write a rule giving all of them would be much more difficult, perhaps not possible. For 1-digit numbers, the equation would be 9*a = a whose only solution is a = 0. For 2-digit numbers, let a and b be the digits, 1 <= a <= 9, 1 <= b <= 9. Then the equation is 9*(10*a + b) = 10*b + a which simplifies to 89*a = b This is impossible for values of a and b between 1 and 9, so there is no solution. For 3-digit numbers, let a, b, and c be the digits, with a and c between 1 and 9, and b between 0 and 9. Then the equation is 9*(100*a + 10*b + c) = 100*c + 10*b + a, which simplifies to 899*a + 80*b = 91*c. Even if b = 0 and a = 1, their minimum values, and c = 9, its maximum value, the left side is larger than the right side, so this equation also has no solution. For 4-digit numbers, let a, b, c, and d be the digits, so 9*(1000*a+100*b+10*c+d) = 1000*d + 100*c + 10*b + a 8999*a + 890*b = 10*c + 991*d This implies that d = 10 - a, by looking at last digits, which we substitute, and get 9990*a + 890*b = 10*c + 9910 When we divide by 10, we get 999*a + 89*b = c + 991 This implies that a = 1, so 89*b = c - 8 This implies that b = 0 and c = 8, and there is only one solution 9*1089 = 9801. For 5-digit numbers, the equation would be 9*(10000*a+1000*b+100*c+10*d+e) = 10000*e+1000*d+100*c+10*b+a 89999*a + 8990*b + 800*c = 910*d + 9991*e. This implies that a = 1 and e = 9, by looking at last digits, as before, and then 8990*b + 800*c = 910*d + 80 899*b + 80*c = 91*d + 8 This implies by looking at sizes that either b = 1 (so d = 1 and 899 + 80*c = 99, which has no solutions) or b = 0 (so 91*d - 80*c = 8, which has the unique solution d = 8, c = 9). Thus there is just one 5-digit solution 10989. This process can continue indefinitely. The equations become harder and harder to solve, even knowing that the first digit has to be 1 and the last 9. But there are more and more variables, just the one equation, and all variables are restricted to being digits, that is, integers between 0 and 9 inclusive. It stands to reason that with more digits to play with, there will be more solutions possible. I don't know that for a fact, however. From these first two solutions, it isn't hard to write down a solution which always works for at least 4 digits: 10...89, where the dots represent any number of 9's: 1089, 10989, 109989, 1099989, 10999989, 109999989, and so on. These can also be written in the form 11*(10^n-1) for n >= 2. Perhaps with this assistance, you will be able to finish the problem. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
Date: 09/11/99 at 04:30:07 From: Chen Xu Chu Subject: Re: Inverse the digits of a number by multiplying by 9 I have a few more questions about this problem. Is it possible to say exactly how many 9-flips there are with precisely n digits? Is it possible to describe all 9-flips, or all 9-flips with n digits, in some simple way? A simple check reveals that this number 109...989 (n digits) is certainly a 9-flip. How can we prove that this is the only one with n digits? If you use the same technique with n = 8 and n = 9, I can find that there are other 9-flips, notably 10891089 (8 digits) and 108901089 (9 digits). So you know there are more than just one! Of course, the new numbers have some pleasant pattern, can you tell me what the pattern is for these new numbers? Thank you.
Date: 10/07/1999 at 15:03:37 From: Doctor Rob Subject: Re: Inverse the digits of a number by multiplying by 9 Thanks for writing back, Jacky. You are asking questions to which I don't know the answers, for the most part. The new examples you have found have the form 11*990...099, where the dots represent any number of zeroes. Your examples have 0 and 1 zeroes where the dots are. It isn't hard to see that a number of the form 11 times any number, which is formed from alternating blocks of zeros and nines, each of length at least 2, beginning and ending with a block of nines, and its own digit-reverse, will satisfy the conditions you want, such as 11*9999999900000990000099999999 = 109999998900010890001099999989 The number of these of a given length can be counted, at least using a recursive formula, if not in closed form. It is not clear to me that this would be all of the possibilities, however. The recursion is as follows. If N(n) is the number of such numbers containing n digits, then N(1) = 0 N(2) = 0 N(3) = 0 N(4) = 1 N(5) = 1 [(n-11)/2] N(n) = SUM (j+1)*N(n-2*j-8) + [n/2] - 2, n >= 6, j=0 where [x] = integer part of x. This implies that N(2*k+1) = N(2*k), for all k > 0. If we let M(k) = N(2*k), then we have M(1) = 0 M(2) = 1 k-6 M(k) = SUM (j+1)*M(k-j-4) + k - 2, k >= 3. j=0 and so we can compute M(3) = 1 M(4) = 2 M(5) = 3 M(6) = 5 M(7) = 8 M(8) = 13 and so on. These numbers are easily recognized: they are the Fibonacci Numbers, M(k) = F(k-1), so N(n) = M([n/2]) = F([n/2]-1), for all n >= 2. This is a surprising and pleasing result. To prove it, you would have to show that the Fibonacci numbers satisfy the recursion given above. That I leave to you. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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