Question

Show that if \(n\) is composite then the below number with \(n\) ones is composte :

\[\large 111111\ldots \text{n times} \]

Answer 1

if n is even, then you'll have an even number of 1s

so let's say n = 4

we'd have 1111 and that's divisible by 11 because the first and third '1' are added while the other '1's are subtracted off

so for some general 4 digit number abcd it is divisible by 1 if a+c-(b+d) = 0 or is divisible by 11

http://www.mathsisfun.com/divisibility-rules.html

in this case, 1+1-(1+1) = 0 so 1111 is proven to be divisible by 11. In general, any even number of 1s will work the same way.

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If n is a multiple of 3, then you'll have the digits add up to that number which proves the number is divisible by 3

example: 111,111 is divisible by 3 because 1+1+1+1+1+1 = 6 which is a multiple of 3

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That's all I can think of at the moment. There are probably a few holes somewhere.

Answer 2

That looks interesting XD but it only proves for the cases when n is divisible by 2, 3

Answer 3

yeah true

Answer 4

```
example: 111,111 is divisible by 3 because 1+1+1+1+1+1 = 6 which is a multiple of 3
```

I am still working on this..
from that example I am guessing we can factor out `111` from `111,111`
so 3 and 111 are factors of 111,111

Answer 5

If \(n = 15\)
\(5\) divides \(15\) and the number with \(15\) ones will be of form
\[\color{blue}{11111},\color{blue}{11111},\color{blue}{11111}\]

it is possible to factor out \(\color{blue}{11111}\)

Answer 6

http://www.wolframalpha.com/input/?i=factor(111111111111111)&t=crmtb01

that says 11111 is not a factor

Answer 7

oh nvm, I was looking in the wrong spot lol

Answer 8

haha it is not listed in prime factorization yeah
11111 might be a composite number but it has 5 ones hmm weird so the converse of original statemetn isn't true

Answer 9

I noticed every number is of the form 3k+1 for some integer k.

For instance, with 1111, we have

\[\Large {\color{red}{1}}{\color{blue}{1}}{\color{green}{1}}1 = {\color{red}{1000}}+{\color{blue}{100}}+{\color{green}{10}}+1\]

\[\Large {\color{red}{1}}{\color{blue}{1}}{\color{green}{1}}1 = {\color{red}{999+1}}+{\color{blue}{99+1}}+{\color{green}{9+1}}+1\]

\[\Large {\color{red}{1}}{\color{blue}{1}}{\color{green}{1}}1 = {\color{red}{999+1}}+{\color{blue}{99+1}}+{\color{green}{9+1}}+1\]

\[\Large {\color{red}{1}}{\color{blue}{1}}{\color{green}{1}}1 = ({\color{red}{999}}+{\color{blue}{99}}+{\color{green}{9}})+({\color{red}{1}}+{\color{blue}{1}}+{\color{green}{1}})+1\]

\[\Large {\color{red}{1}}{\color{blue}{1}}{\color{green}{1}}1 = 3({\color{red}{333}}+{\color{blue}{33}}+{\color{green}{3}})+3+1\]

\[\Large {\color{red}{1}}{\color{blue}{1}}{\color{green}{1}}1 = 3[({\color{red}{333}}+{\color{blue}{33}}+{\color{green}{3}})+1]+1\]

\[\Large 1111 = 3(370)+1\]

This trick of expanding the number out and rewriting it in the form 3k+1 should work for any string of 1s. But I get stuck here as well.

Answer 10

so we gotta show that (10^n-1)/9 is composite when n is composite