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### 10-Digit Number Puzzle

```
Date: 02/10/2001 at 08:18:42
From: ihsan
Subject: Turkish math olympiads - May

"abcdefghij" is a ten-digit-number. All of the digits (a,b,c,d...)
are different from each other. If 11111 divides it evenly, how many
possibilities are there for abcdefghij?
```

```
Date: 02/12/2001 at 13:19:13
From: Doctor Rob
Subject: Re: Turkish math olympiads - May

Thanks for writing to Ask Dr. Math, Ihsan.

If all digits are different, and there are ten of them, then all the
numbers from 1 to 10 must appear. That implies that the digits of
abcdefghij add up to 45 = 0 + 1 + 2 + ... + 9. That implies that
abcdefghij is divisible by 9.

Since 9 and 11111 are relatively prime, their least common multiple
is 9*11111 = 99999, and so that must divide evenly into abcdefghij.
Now consider what number 99999 must be multiplied by. It must be
larger than abcde, because abcde*100000 = abcde00000 < abcdefghij. It
can't be abcde+2 (or larger), because:

99999*(abcde+2) = (100000-1)*(abcde+2)
= abcde00000 + 199998 - abcde

Looking at the 10000's digit of that, 0 + 9 - e would not require a
borrow, regardless of the value of e. Therefore the 100000's digit
would be e + 1, which doesn't equal the 100000's digit of abcdefghij.
That implies that:

abcdefghij = 99999*(abcde+1)
= (100000-1)*(abcde+1)

abcde00000 + fghij = abcde00000 + (99999 - abcde)

fghij = 99999 - abcde

Then the following equations must all hold:

f = 9 - a
g = 9 - b
h = 9 - c
i = 9 - d
j = 9 - e

Now there are nine choices for a (it cannot be 0), and then f is
known. That leaves eight choices for b, and then g is known. That
leaves six choices for c, and then h is known. That leaves four
choices for d, and then i is known. That leaves two choices for e, and
then j is known. Thus the total number of such numbers abcdefghij is
9*8*6*4*2 = 3456.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
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