1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Oct 26, 2011 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations

    3. 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?
    Last edited: Oct 26, 2011
  2. jcsd
  3. Oct 26, 2011 #2
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook