There is an infinite set of numbers with nothing in the middle.

By that I mean a positive whole number whose first digit and last digit are not zero, but any and all digits in between are zero.

The 81 numbers between 11 and 99 that are not multiples of ten are numbers with nothing in the middle. So are the 81 numbers of form x0y between 101 and 909. So are the 81 numbers of form x00y between 1001 and 9009. And so on.

I started wondering, which of these numbers is divisible by each prime number? What is the lowest prime number whose multiples include a complete set of numbers with nothing in the middle?

12 | 22 | 32 | 42 | 52 | 62 | 72 | 82 | 92 |

14 | 24 | 34 | 44 | 54 | 64 | 74 | 84 | 94 |

16 | 26 | 36 | 46 | 56 | 66 | 76 | 86 | 96 |

18 | 28 | 38 | 48 | 58 | 68 | 78 | 88 | 98 |

and likewise for 102, 104, etc; but obviously 2 is not a divisor of odd numbers with nothing in the middle.

21 | 51 | 81 | ||||||

12 | 42 | 72 | ||||||

33 | 63 | 93 | ||||||

24 | 54 | 84 | ||||||

15 | 45 | 75 | ||||||

36 | 66 | 96 | ||||||

27 | 57 | 87 | ||||||

18 | 48 | 78 | ||||||

39 | 69 | 99 |

and likewise for 201, etc; but obviously 3 is not a divisor of numbers with nothing in the middle whose digits to not add up to a multiple of 3.

15 25 35 45 55 65 75 85 95

and so on for 105, 205, etc.

14 | 21 | 35 | 42 | 56 | 63 | 77 | 84 | 91 |

28 | 49 | 98 | ||||||

105 | 203 | 301 | 406 | 504 | 602 | 805 | 903 | |

308 | 609 | |||||||

1,001 | 2,002 | 3,003 | 4,004 | 5,005 | 6,006 | 8,001 | 9,002 | |

1,008 | 2,009 | 8,008 | 9,009 | |||||

10,003 | 20,006 | 30,002 | 40,005 | 50,001 | 60,004 | 80,003 | 90,006 | |

30,009 | 50,008 | |||||||

100,002 | 200,004 | 300,006 | 400,001 | 500,003 | 600,005 | 800,002 | 900,004 | |

100,009 | 400,008 | 800,009 | ||||||

1,000,006 | 2,000,005 | 3,000,004 | 4,000,003 | 5,000,002 | 6,000,001 | 8,000,006 | 9,000,005 | |

5,000,009 | 6,000,008 |

And so on with 10,000,004, etc. So that's 65 of the possible 81 combinations covered. Can we do better?

11 | 22 | 33 | 44 | 55 | 66 | 77 | 88 | 99 |

209 | 308 | 407 | 506 | 605 | 704 | 803 | 902 |

and then 1001, 2002… and 20009, 30008… etc

A mere 17 of the 81 configurations. Well, let's try another.

13 | 26 | 39 | 52 | 65 | 78 | 91 | ||

104 | 208 | 403 | 507 | 702 | 806 | |||

1,001 | 2,002 | 3,003 | 4,004 | 5,005 | 6,006 | 7,007 | 8,008 | 9,009 |

20,007 | 30,004 | 40,001 | 60,008 | 70,005 | 80,002 | |||

100,009 | 200,005 | 300,001 | 500,006 | 600,002 | 800,007 | 900,003 | ||

4,000,009 | 5,000,008 | 6,000,007 | 7,000,006 | 8,000,005 | 9,000,004 |

After that, we are back to 10,000,003, 20,000,006, etc because 13 is also a divisor of 999,999.

Well, does this mean that there is no prime number which is a divisor of a full set of possible configurations of numbers with nothing in the middle?

17 | 34 | 51 | 68 | 85 | ||||

102 | 204 | 306 | 408 | 901 | ||||

1,003 | 2,006 | 3,009 | 6,001 | 7,004 | 8,007 | |||

20,009 | 30,005 | 40,001 | 70,006 | 80,002 | ||||

200,005 | 500,004 | 700,009 | 800,003 | |||||

1,000,008 | 3,000,007 | 5,000,006 | 7,000,005 | 9,000,004 | ||||

2*10^{7}+7 |
3*10^{7}+2 |
5*10^{7}+9 |
6*10^{7}+4 |
9*10^{7}+6 |
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10^{8}+1 |
2*10^{8}+2 |
3*10^{8}+3 |
4*10^{8}+4 |
5*10^{8}+5 |
6*10^{8}+6 |
7*10^{8}+7 |
8*10^{8}+8 |
9*10^{8}+9 |

2*10^{9}+3 |
4*10^{9}+6 |
6*10^{9}+9 |
7*10^{9}+2 |
9*10^{9}+5 |
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4*10^{10}+9 |
5*10^{10}+7 |
6*10^{10}+5 |
7*10^{10}+3 |
8*10^{10}+1 |
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3*10^{11}+8 |
4*10^{11}+5 |
5*10^{11}+2 |
9*10^{10}+7 |
|||||

10^{12}+4 |
2*10^{12}+8 |
5*10^{12}+3 |
6*10^{12}+7 |
9*10^{11}+2 |
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10^{13}+6 |
3*10^{13}+1 |
4*10^{13}+7 |
6*10^{13}+2 |
7*10^{14}+8 |
9*10^{12}+3 |
|||

10^{14}+9 |
2*10^{14}+1 |
4*10^{14}+2 |
6*10^{14}+3 |
8*10^{14}+4 |
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10^{15}+5 |
4*10^{15}+3 |
5*10^{15}+8 |
7*10^{15}+01 |
8*10^{15}+6 |
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8*10^{16}+9 |
9*10^{16}+8 |

And then we are back to 10^{17}+7, and the cycle starts again.

So, indeed, 17 is the first prime number which is a divisor of the full set of possible configurations of numbers with holes in the middle.

There, I thought you needed to know that.