Find the 10-digit number divisible by position

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I'm a natural number written with 10 different digits such that for all i∈1,...,10 the number formed by my i first digits (from left to right) is divisible by i. Who am I?

Anyone got any insight? It would be much appreciated.

 

[MarkFL]

We know the ones digit must be 0 because the 10 digit number must be divisible by 10:

$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|}\hline \,\,\, & \,\,\, & \,\,\, & \,\,\, & \,\,\, & \,\,\, & \,\,\, & \,\,\, & \,\,\, & \hline 0\,\end{array}$$

Now, this means the 5th digit must be 5, since numbers divisible by 5 end in either 0 or 5, and 0 is already taken

$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|}\hline \,\,\, & \,\,\, & \,\,\, & \,\,\, & 5 & \,\,\, & \,\,\, & \,\,\, & \,\,\, & \hline 0\,\end{array}$$

Now we know the first digit can be any of the remaining 8 since they are all divisible by 1. The second digit must be even. The sum of the first 3 digits must be divisible by 3. Digits 3 and 4 taken as a 2-digit number must be divisible by 4. The 6th digit must be even and the sum of the first 6 digits must be divisible by 3. Digits 6, 7 and 8 taken as a 3-digit number must be divisible by 8.

Thus, we now know each digit must have the parity of its location, that is, all the odd locations (1st digit, 3rd digit, etc.) must be odd and all the even locations (2nd digit, 4th digit, 6th digit, etc.) must be even.

With a bit of trial and error, you can find the number. :D

 

[MarkFL]

This is the number I found:

Spoiler

$$\begin{array}{|c|c|c|c|c|c|c|c|c|c|}\hline 7 & 8 & 3 & 6 & 5 & 4 & 9 & 2 & 1 & \hline 0\,\end{array}$$

 

[johng1]

Hi there integer 3816547290,

I'm a computer and am very good at integer arithmetic, particularly for "small" integers such as you. I read your post and got to wondering how many cousins do your have? That is how many total 10 digit integers (not necessarily all different digits) have the same divisibility criteria that you enjoy. Here's what I found (it took me less than 1/10 of a second):

All digits different: 3816547290
Total integer cousins found: 2492
First digit Count
1 294
2 268
3 288
4 273
5 268
6 279
7 285
8 269
9 268

The little table shows (surprise to me) that the distribution of your cousins with first digit i is almost uniform. Maybe one of the humans could explain this.

PS. MarkFl's integer 7836549210 is not you; 78365492 is not divisible by 8.

PSS. If you're interested I can show you the little Java program that I executed.

 

[MarkFL]

...MarkFl's integer 7836549210 is not you...

Oops.

 

[johng1]

Hi again,
I got to thinking what n digit (base 10) integers have the divisibility criteria? Namely, if the digits are numbered from 1 to n starting at the most significant digit $d_1d_2\cdots d_n$, for each i the i digit integer $d_1d_2\cdots d_i$ is divisible by i.

I tweaked my program a little and found there is exactly one such 25 digit integer, namely
3608528850368400786036725
Also there are no such 26 digit integers and so there are no such integers for n > 25.

I have no clue about how one would do this by hand. If anyone has any ideas, I'd like to see them.