Show that the integers 1111, 111111, 11111111, ... (numbers formed by an even number of numbers 1) are all composed.

Asked Nov 20 '20 at 20:42

Active Nov 21 '20 at 18:39

Viewed 86 times

I am studying congruence and I am trying to solve this problem, but I cannot think of a way to do this.

Would anyone be able to help me?

Show that the integers 1111, 111111, 11111111, ... (numbers formed by an even number of numbers 1) are all composed.

edited Nov 21 '20 at 18:39

hardmath

asked Nov 20 '20 at 20:42

Marina

3

The one such number that you didn't include is a big hint. – Arthur Nov 20 '20 at 20:44
3

Hint:Divisibility by 11 – Lion Heart Nov 20 '20 at 20:47
Your numbers can be written as $$11\cdots1 = \frac{99\cdots 9}{9} = \frac{10^{2n}-1}9.$$
Now can you think of a number $m$ relatively prime to $9$ such that $10^{2n}-1 \equiv 0 \pmod m$?
– mechanodroid Nov 20 '20 at 20:51
1

I think by "composed" in the title you meant composite. – hardmath Nov 20 '20 at 21:07
and "composed" in the body – J. W. Tanner Nov 20 '20 at 21:10
11 is composite? – Mars Nov 20 '20 at 21:24
@hardmath You're right, I don't goofed. Sorry, Tanner. – Arthur Nov 21 '20 at 07:29
@Arthur: I rolled back Lieutenant Zipp's edit because it introduced the error noted by yourself and fellow Commenter Mars... – hardmath Nov 21 '20 at 18:41
Why roll back my edit? Why not just re-edit it to remove the offending 11 from the problem statement? Now it's relatively worse-formed and worse-formatted. – Lieutenant Zipp Nov 22 '20 at 17:35

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