Problems & Puzzles: Puzzles

Puzzle 57.-  8757193191, another little puzzle from Mike Keith

Mike Keith established that "8757193191 is the provably largest (base 10) number with the property that the first N digits are divisible by the Nth prime, for all appropriate N"

In other words, that for 8757193191:

8 is divisible by 2
87 is divisible by 3
875 is divisible by 5
...
8757193191 is divisible by 29

My questions are:

a)Can you prove (or disapprove) it?

b)Can you find the similar number if we establish the same property but for "the last N digits"?

Solution

Jim Howell wrote at 10/06/99:

"Here is my solution to Puzzle 57, part (b), numbers where the last "n"
digits are divisible by the n-th prime.

Solutions with 9 digits:

632258960
824791960
851750490
904719030
"

***
It was again Jim R. Howell who solved (12/06/99) the part a) of this puzzle with an argument that I would say that it is absolutely simple not without being at the same time complete & clever .

"The following table contains the result of trying all possible
numbers for part (a).  For example, the first line means that
the number
2100 has the property that the first "n" digits are
divisible by the n-th prime, but there is no digit that can be
added on the end to make the result divisible by 11.

So the largest solution is indeed
8757193191 (last line of table).
The only other 10-digit solution is
6300846559.

2   3   5   7  11  13  17  19  23  29  31
---------------------------------------------
2   1   0   0
2   1   0   7   6   9   4   7   0
2   1   5   6   0   5   9   2
2   4   0   1   3   6   9
2   4   0   8
2   4   5   0   8   9   0
2   4   5   7   4
2   7   0   2   7   0
2   7   0   9   3   3
2   7   5   1   1   9   5
2   7   5   8   8   6   2   7

4   2   0   0   9   5   5   1
4   2   0   7   5   8   5
4   2   5   6
4   5   0   1   2   5
4   5   0   8   9   2   7
4   5   5   0   7   8   1   2
4   5   5   7   3
4   8   0   2   6
4   8   0   9   2   2
4   8   5   1   0   8   6
4   8   5   8   7   5   3

6   0   0   6   0   0
6   0   5   5   5   3   6
6   3   0   0   8   4   6   5   5   9
6   3   0   7   4   7   6
6   3   5   6   9
6   6   0   1   1   4
6   6   0   8   8   1   8
6   6   5   0   6   7   2   6   4
6   6   5   7   2
6   9   0   2   5
6   9   0   9   1   1
6   9   5   1
6   9   5   8   6   4   4

8   1   0   6
8   1   5   5   4   2   7   0
8   4   0   0   7   3   7   9
8   4   0   7   3   6   7   0
8   4   5   6   8   9
8   7   0   1   0   3   0
8   7   0   8   7   0   9
8   7   5   0   5   6   3
8   7   5   7   1   9   3   1   9   1"

***
Tiziano Mosconi discovered (17/8/2001) another solution to question b) that was left off by Jim Howell: 224246090

0/2=0
90/3=30
090/5=18
6090/7=870
46090/11=4190
246090/13=18930
4246090/17=249770
24246090/19= 1276110
224246090/23= 9749830

***

 Records   |  Conjectures  |  Problems  |  Puzzles