How many consecutive 1's are necessary for divisibility by a number?

In summary, the conversation discusses trying to figure out the necessary number of successive 1's for a number composed solely of 1's to be divisible by another number x. Examples are given for various values of x, and a general rule is hypothesized to be (p-1)n for prime numbers p, but exceptions are found for x = 11 and x = 13. It is also noted that for composite numbers, the necessary number of 1's may depend on the divisibility properties of their prime factors. Further insight is requested to determine a pattern for this problem.
  • #1
Vespero
28
0

Homework Statement


I'm trying to figure out how many successive 1's are necessary for a number composed solely of 1's to be divisible by another number x. For example, how many 1's are necessary for 1...1 to be divisible by 7? Simply performing the calculation shows that the first such number is 111111 (six 1's) which gives 15873 when divided by 7. The next such number is 111111111111 (twelve 1's) which gives 15873015873 when divided by 7. However, I am looking for a general rule or property here.



Homework Equations





The Attempt at a Solution



Clearly, x cannot be even, as 1...1 is never even. Similarly, x cannot be divisible by 5. x divisible by 3 also seldom works.

So, I have checked the number of 1's necessary for divisibility by several numbers x:

Choice for x: Number of 1's
x = 7: [itex]\hspace{1 in}[/itex] 6, 12, 18, ..., 6n for integer n
x = 11: [itex]\hspace{1 in}[/itex] 2, 4, 6, 8, ..., 2n for integer n
x = 13: [itex]\hspace{1 in}[/itex] 6, 12, 18, ..., 6n for integer n
x = 17: [itex]\hspace{1 in}[/itex] 16, 32, ..., 16n
x = 19: [itex]\hspace{1 in}[/itex] 18, 36, ..., 18n
x = 23: [itex]\hspace{1 in}[/itex] 22, 44, ..., 22n
x = 29: [itex]\hspace{1 in}[/itex] 28, 56, ..., 28n
x = 49: [itex]\hspace{1 in}[/itex] 42, 84, ..., 42n
x = 77: [itex]\hspace{1 in}[/itex] 6, 12, 18, ..., 6n

So, for most prime numbers p, it seems like the necessary number of successive 1's is (p-1)n for positive integers n. However, for x = 11, only multiples of 2 are necessary, and for x = 13, we have multiples of 6. For the last two values of x, I tried composite numbers that were multiples of values of x I had already tried. For x = 49 = 7*7, notice that our numbers of 1's are 7 times the corresponding values for x = 7. For x = 77, they are 3 times the corresponding values for x = 11. So, in general it seems like if x = p a prime, we have (p-1)n, but x = 11 and x = 13 break this rule. Similarly, composites seem to behave differently depending on the divisibility properties of their prime factors. Could anyone offer any insight and help me figure out a pattern?
 
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  • #2
Vespero said:
x divisible by 3 also seldom works.

It actually works 1/3 of the time, only when the number of 1's in the number equals 3n, with n being any non-negative integer.
 

1. How many consecutive 1's do I need to make a number divisible by 3?

The number of consecutive 1's needed to make a number divisible by 3 is 2. This is because a number is divisible by 3 if the sum of its digits is also divisible by 3.

2. Can I use any number of consecutive 1's to make a number divisible by 7?

No, the number of consecutive 1's needed to make a number divisible by 7 must follow a specific pattern. The pattern is 1, 3, 2, 6, 4, 5, repeating. For example, 111 is divisible by 7, but 1111 is not. However, 11111 is divisible by 7 because it follows the pattern of 1, 3, 2, 6, 4, 5.

3. Is there a limit to how many consecutive 1's I can use to make a number divisible by a certain number?

Yes, there is a limit. The maximum number of consecutive 1's needed to make a number divisible by a certain number is one less than the number itself. For example, to make a number divisible by 8, you would need a maximum of 7 consecutive 1's.

4. Can I use any other number besides 1 to make a number divisible by another number?

Yes, you can use other numbers besides 1 to make a number divisible by another number. For example, to make a number divisible by 4, you can use the pattern of 2, 4, 8, 6, repeating. So, a number like 222 is divisible by 4 because it follows the pattern of 2, 4, 8, 6.

5. Is the concept of using consecutive 1's to make a number divisible by another number applicable to all numbers?

No, this concept is not applicable to all numbers. It only works for certain numbers that have a specific pattern or rule for divisibility. For example, this concept does not work for prime numbers, as they have no other factors besides 1 and themselves.

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