Question

Given a number : 111111...11 (19 times)
how to prove that is a prime number ?

Answer 1

use didvisblitytests http://www.jimloy.com/number/divis.htm

Answer 2

thogh i think it is not divisible by any other number...i.e prime

Answer 3

@hartnn , @UnkleRhaukus , @Hero , @.Sam. , @campbell_st , @mukushla or anyone, help me...
@akash809 , can u explain to me how do this ?

Answer 4

to check whether a number is prime or not you have to see whether it is divisible by anyother no. than 1 and itself....do u know this

Answer 5

trial and error method ? how long do it ? :)

Answer 6

do it till 11 alll other divisiblity tests are repetitive

Answer 7

well its not even, ends in 5 or 0 and the digits don't add to a multiply of 3

so that eliminates

2, 3, 4, 5, 6, 8,9, 10

Answer 8

i am still confusing ...:(

Answer 9

use tests of divisiblity
if it isnt divisible, then it is a prime number ;)

Here are divisibility tests for the first few integers (2 through 11):

2.Even (last digit is 0, 2, 4, 6, or 8).
3.Sum of digits is divisible by 3.
4.Last two digits divisible by 4 (even0, even4, even8, odd2, odd6).
5.Last digit is 0 or 5.
6.Divisible by 2 and 3 (even, and sum of digits is divisible by 3).
8.Last three digits divisible by 8.
9.Sum of digits is divisible by 9.
10.Last digit is 0.
11.The difference between the sum of the odd numbered digits (1st, 3rd, 5th...) and the sum of the even numbered digits (2nd, 4th...) is divisible by 11.

if its not divisible by any of the numbers above, then the number is prime ;)

Answer 10

already mentioned it...but it won't be sufficient ....but if looking for more fun on prime numbers

use this algorithm to create a program in c++
Input is n
output is prime or not

Start
{
for k<-2 to root(n)
{
if n mod k=0 then {n is not prime stop}
else {continue}
}
print(n is prime)
}

input the number and see whether it is prme or not

Answer 11

http://www.wolframalpha.com/input/?i=1111111111111111111

Answer 12

Well this may work:
11111111111111111111=10^18 +10^17+10^16+...+10^2+10^1+10^0
Now, let x=10 then
11111111111111111111=x^18+x^17+x^16+...x^2+x+1

Answer 13

Let f(x)=x^18+x^17+x^16+...x^2+x+1

Answer 14

Just need to prove that f(x) cannot be factored

Answer 15

@ganeshie8 Can u help?

Answer 16

Numbers of form 10^n-1/9 are known as repunits.....

Answer 17

Oh.. there is a problem with my approach

Answer 18

it's is binomial series right

Answer 19

http://en.wikipedia.org/wiki/Repunit