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### A Special Ten-digit Number

```
Date: 02/17/99 at 14:20:36
From: Donna Allen
Subject: A special Ten-digit Number

Someone was offered a million dollars if he could create a ten-digit
number that met these conditions:

It contains the digits 0-9;
its first 2 digits are divisible by 2;
its first 3 digits are divisible by 3;
its first 4 digits are divisible by 4;

this pattern continues all the way through the number.

How close can you get to this special number?

I figured that the first number has to be odd and the second number
even and then continue with odd, even odd even and the fifth number
could either be a 5 or a 0. I got as far as 1,2,3,6,5,4. I know
there must be a mathematical pattern but I do not know how to figure
it.
```

```
Date: 02/18/99 at 03:08:53
From: Doctor Schwa
Subject: Re: A special Ten-digit Number

You are on the right track, and there is no simple method from here.
You need to look at lots of possibilities, but narrow them down as
you have been doing.

In other words, you know it is oeoe5eoeo0 (o stands for odd digit, e
even). Now keep going. The first four digits have to be divisible by 4.
So to do that with "oe" it has to be 12, 16, 32, 36; that is, the
fourth digit has to be 2 or 6. The same logic shows that the 8th digit
has to be 2 or 6. So now we have two possibilities, oeo25eo6o0 or
oeo65eo2o0.

In order to be divisible by 3, 6, 9, the first 3, 6, 9 digits have to
total to a multiple of 3, so that means each group of 3 digits is a
multiple of 3. That means the middle 3 digits are either 258 or 654,
and so we have eithero4o258o6o0 or o8o654o2o0.

Now you have to fill in the remaining odd digits so that the initial
three are divisible by 3. There are a few ways to do that. You also
have to check whether the first seven digits are divisible by 7, which
will eventually lead you to eliminating all but one possible solution.

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