let p be a prime >=7. show that p divides at least one of the numbers in the set {1, 11, 111,..}?

Answer 1

Writing the number 11...1 with p-1 digits as 1+10+10^2+...+10^p-2, we see this is a geometric series with sum (10^(p-1)-1)/9. If p is prime and >=7, it is relatively prime to 10, so by Fermat's little theorem, 10^(p-1)-1 is divisible by p. Furthermore, the denominator of 9 in (10^(p-1)-1)/9 will not canel out that factor of p since p is not 3. Thus, the number we started with, 11...1 with p-1 digits, is divisible by p.

Example: 111111 is divisible by 7

Answer 2

111111 = 15873*7
11 = 1*11
111111 = 8547*13
111111111111 = 8547008547*17

The answer will have something to do with the fact that you will get to the multiple before you get to a number with p digits. Try using mods
1 mod 7 is 1
11 mod 7 is 4
111 mod 7 is 6
1111 mod 7 is 5
11111 mod 7 is 2
111111 mod 7 is 0

Starting with 1, 1 mod p is 1
11 mod p is whatever it is
111 mod p is the same thing as (((11 mod p) * 10) + 1) mod p
1111 mod p is the same thing as (((111 mod p) * 10)+1) mod p
This should get you to the answer and your proof. I don't have time to go further right now.

Answer 3

if p is prime number >=7 then it can be 11 which divides element 11 in given set.

Answer 4

I believe you can use induction here

first we make our prime number set as an array P(n) = {p: p is prime, p >=7}
next we make our other set as an array I(n) = {sum(1^m) for m = 1 to n)

let n = 1
P(1) = 7
I(6) / P(1) = 111111/7 = 15873

assume this statment holds for k > n
I(z) / P(k) = a

we need to prove that this holds for k+1

let P(k+1) = t
for every o in N there exsists and r in R such that
r = t * o
also for every i in I, there exsists r1, r2,... rn, sucht that
i = sum(rj) for j = 1 to n
combining these we get
i = sum(t * oj) for j = 1 to n
giving i = t sum(oj) for j = 1 to n
N is closed on addition therefore sum(oj) = q in N
giving i = t * q,
giving i / t = q

implying that for every n in N, there exsits an m, q in N such that
I(m) / P(n) = q