The question is , if f(n)= 1+10+10^2 + + 10^n , where n is

[sigh1342]

the question is , if f(n)= 1+10+10^2 +... + 10^n , where n is integer.
find the least n s.t. f(n) is divisible by 17 , I have no idea about it.

 

[mfb]

What is the remainder if you divide...
1 by 17?
10 by 17?
100 by 17?
...
a+b by 17, if you know it for a and b?

That should help.

 

[ramsey2879]

the question is , if f(n)= 1+10+10^2 +... + 10^n , where n is integer.
find the least n s.t. f(n) is divisible by 17 , I have no idea about it.

It should also help to recognize that 9 times the sum plus 1 is a power of 10. Thus Fermats Little theorem re primes P dividing A^(P-1) - 1 may apply.

 

[Edgardo]

You can also find n by "brute force", i.e. check if f(1), f(2), f(3), ... is divisible by 17. Use http://www.wolframalpha.com/ or write a small program in your programming language of choice.

 

[blue_raver22]

I would approach this problem by first analyzing the pattern of the given function. We can see that each term in the function is a power of 10, starting from 10^0 to 10^n. This suggests that the function is a geometric series with a common ratio of 10.

Next, we can use the formula for the sum of a geometric series to express f(n) as: f(n) = (10^(n+1) - 1) / (10 - 1).

To find the least value of n that makes f(n) divisible by 17, we can set up the following equation:

(10^(n+1) - 1) / (10 - 1) = 17k, where k is an integer.

Simplifying this equation, we get:

10^(n+1) - 1 = 17k(9)

We can see that this equation has a solution when n+1 is a multiple of 3, since 10^(n+1) - 1 will be divisible by 9. Therefore, the least value of n that makes f(n) divisible by 17 is n = 2, which corresponds to f(n) = 111.