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### 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
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

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/
```

```Date: 09/02/2014 at 15:51:31
From: Steve
Subject: When the digits of 9x are the digits of x, reversed

I've solved this problem and posted the solution here:

The solution is simple enough that a sharp high school student could

Steve
```
Associated Topics:
High School Discrete Mathematics
High School Number Theory

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