Series of Num3ers 1111…

The first eight prime numbers:   2, 3, 5, 7, 11, 13, 17, 19
 
Two 1s and 2 at the end, the number formed is even, thus divisible by 2:
112 = 2^4 * 7
 
If we place the digit 2 somewhere else in the number, we get
211   is a prime number.
121 = 11^2   a perfect square.
Each number is not divisible by 2.
 
3 1s and 3 at the end, the number formed is divisible by 3:
1113 = 3 * 7 * 53
 
Placing the digit 3 somewhere else, we get
1131 = 3*13*29
1311 = 3*19*23
3111 = 3*17*61
Each of the number produced is divisible by 3.
 
5 1s and a 5,
111115 = 5 * 71 * 313
The digit 5 placed elsewhere,
511111 is a prime number.   151111 = 137 * 1103,
115111 = 43 * 2677,   111511 = 19 * 5869,   111151 = 41 * 2711
 
7 1s and a 7,
11111117 is a prime number.
71111111 is a prime number.
17111111 = 1063 * 16097,   11711111 = 331 * 35381,
11171111 = 7 * 67 * 23819   divisible by 7.
11117111 = 479 * 23209,   11111711 = 13 * 854747,
11111171 is a prime number.
 
11 1s and 1,
111111111111 = 3 * 7 * 11 * 13 * 37 * 101 * 9901   divisible by 11
11 1s and 11,
1111111111111 = 53 * 79 * 265371653   not divisible by 11.
 
13 1s and 3 at the end, the number formed is divisible by 13:
11111111111113 = 13 * 95003 * 8996567
 
But 13 1s and 13 at the end does not produce a number divisible by 13:
111111111111113 = 1163 * 95538358651
 
17 1s and 7,
111111111111111117 = 3 * 17 * 29 * 67 * 262331 * 4274299
the number formed is divisible by 17.
 
But the number formed with 17 1s and 17 is not divisible by 17:
1111111111111111117 = 7 * 54011 * 2938848729521
 
19 1s and 9 at the end, the number formed is divisible by 19:
11111111111111111119 = 19 * 463 * 50683 * 24920719969
 
The number formed with 19 1s and 19 is not divisible by 19:
111111111111111111119 = 7 * 211 * 75227563379222147
 
 
 
 
In summary:
 
2: 112 divisible by 2, 211 and 121 are not.
3: 1113, 1131, 1311, 3111 are divisible by 3
5: If the last number is either 0 or 5, the entire number is divisible by 5.
7:
11111117, 71111111 and 11111171 are prime numbers.
11171111 divisible by 7 and 11111711 divisible by 13
11: 111111111111 (12 decimal digits) divisible by 3, 7, 11 and 13
13: 11111111111113 (14 decimal digits) divisible by 13
17:
111111111111111117 (18 decimal digits)divisible by 3 and 17
1111111111111111117 (19 decimal digits)divisible by 7
19:
11111111111111111119 (20 decimal digits) divisible by 19
111111111111111111119 (21 decimal digits) divisible by 7
 
Questions :
 
Let N be the number 1111..X…111
The number X is inserted somewhere in the number 111…111.
The number length is (n+1) decimal digits. The number is formed with n 1s and the number X = 2, 3, 5, 7, 11 (or 1), 13, 17, 19 (or 9)
 
I ask, if, for example, X = 3, what are the smallest N that can be divisible by the other primes, 7, 11, 13, 17, 19?
 
Similarly, if X = 2, 5, 7, 11, 13, 17 and 19 (or 9)
 
 

About benvitalis

math grad - Interest: Number theory
This entry was posted in Math Beauty and tagged . Bookmark the permalink.

1 Response to Series of Num3ers 1111…

  1. Pingback: Curious properties of 11, 111, 1111, … | Fun With Num3ers

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