Find the largest number having this property.

Question

The $13$-digit number $1200549600848$ has the property that for any $1 \le n \le 13$, the number formed by the first $n$ digits of $1200549600848$ is divisible by $n$ (e.g. 1|2, 2|12, 3|120, 4|1200, 5|12005, ..., 13|1200549600848 using divisor notation).

Question 1: Find the largest computed number having this property.

Question 2: Is there a theoretical upper bound on the largest possible number with this property?

Edit: Added Question 2 as I believe it is more insightful as compared to brute force computer calculations.

Answer 1

The The On-Line Encyclopedia of Integer Sequences list this series as A109783 and state that 3608528850368400786036725 works for 25 digits, but there is no such 26 digit number.

A thread titled divisor problem at The Math Forum suggests the following argument:

3608528850368400786036725 is the largest number with such a property.

Of course any substring from the left of this has the same property.

Let N be a number with such a property. to be searched a digit d such that 10N+d has the property.

I call a "terminal" number a number N that can't be "expanded" that is for which no digit d exists with 10N+d having the property. The only "number" which can indefinitely be expanded is 0000...

All solutions are then given by the set of these "terminal" numbers. This set is finite and there are only 2492 terminal numbers.

Interesting is a number with all digits differents. The only solution is 3816547290

Edit: to answer question 2 explicitly, the largest is the 25 digit number given above. There is no such 26 digit number and therefore no 27, 28, 29, 30,... digit number.

We can prove this by contradiction. Suppose there was a 30 digit number, then we could chop off the last 4 digits and we'd get a 26 digit number which satisfies the required property, but we know no such number exists. Proof by contradiction.

Answer 2

There are $20456$ such numbers, the biggest is $3608528850368400786036725$. Here's PARI/GP code to print them all:

Q=List([1,2,3,4,5,6,7,8,9])
while(a=Q[1],listpop(Q,1);print(a);a=a*10;d=length(Str(a));for(i=0,9,if(!((a+i)%d),listput(Q,a+i))))

Answer 3

Assuming $n_\max=13$ to be fixed, I got $9848587230963$, which has the property that for any $1 \le n \le 13$, the number formed by the first $n$ digits it is divisible by $n$

$$ \begin{eqnarray} 1&|&9\\ 2&|&98\\ 3&|&984\\ 4&|&9848\\ 5&|&98485\\ 6&|&984858\\ 7&|&9848587\\ 8&|&98485872\\ 9&|&984858723\\ 10&|&9848587230\\ 11&|&98485872309\\ 12&|&984858723096\\ 13&|&9848587230963\\ \end{eqnarray} $$

Answer 4

I wrote a little Mathematica program to find the largest number of this kind. The full tree of numbers with these divisibility properties is very large.