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/