If n is a natural number then 10^n ≡ 1 mod 9???

Question

I'm struggling with these problems, as stated in the title...

If n is a natural number then 10^n ≡ 1 mod 9, I'm not sure how to go about proving or disproving this.

Answer 1

First of all, let's see if you agree with the statement. Congruence mod 9 means here that 9 divides 10^n-1, right? Well, 10 to any power will be a 1 with some number of 0's behind it. Subtract one and you end up with a string of 9's, which is certainly divisible by 9.

EDIT: There's also another, much faster proof that I hadn't thought of. One of the properties of modular congruences is that you can raise each side to a positive integer power. Since clearly 10 is congruent to 1, 10^n is congruent to 1^n which is congruent to 1.
//EDIT

Okay, so now we're out to prove it. The best way I can think of is by induction. It works with n=0 (or n=1, if you prefer). So assume that it works for n=k, where k is some integer >=0. Now we want to know if 9 divides 10^(k+1)-1.
10^(k+1) - 1
=10*10^k - 1
=9*10^k + 10^k - 1.
But we know that 9 divides 9*10^k, and 9 also divides 10^k-1 (by our induction hypothesis). So 9 must divide the whole thing, which happens to be 10^(k+1)-1, so by induction, the statement holds for all n>=0.

Answer 2

We can try and prove it by induction. What we want to prove:

If n is a natural number, then 10^n ≡ 1 (mod 9).
Proof by induction:

Base case: Let n = 1.
Then 10^1 (mod 9) ≡ 10 (mod 9) ≡ 1 (mod 9), so the formula holds true for n = 1.

Induction hypothesis:

Assume the formula holds true up to some value k. That is, assume

10^k ≡ 1 mod 9

Then

10^(k + 1) ≡ 10 * 10^k ≡ (1) (1) {by our induction hypothesis}
≡ 1 (mod 9)

Implying the formula holds true for n = k + 1.

Therefore, by the principle of mathematical induction, the formula holds true for all natural numbers n.

Answer 3

look at a few simple examples:
10^1 = 9 + 1
10^2 = 99 + 1 = 11•9 + 1
10^3 = 999 + 1 = 111•9 + 1

so we need to show 10^n = 9k + 1 for some integer k. look at the 1st example: 10 - 1 = 9. in the 2nd, 10² - 1 = 9 • 11, and that from alg 1 is a difference of squares: 10² - 1 = (10 - 1)(10 + 1).

what about 10^3 - 1? in alg 2 we learn to factor it as (10 - 1)(10² + 10 + 1). it's not a large step to factoring

10^n - 1 = (10 - 1)[10^(n-1) + 10^(n-2) + ... + 10 + 1]

and since the (10 - 1) factor is 9, the other factor is the necessary value of k. notice for n = 3, k = 111. for n = 4, k = 1111. n gives you the number of 1's in the k value.

Answer 4

I'll do an inductive proof because I know it will work (there are probably easier ways to do this, starting from the fact that 10 ≡ 1 mod 9).

Base case 1
10^0 ≡ 1 mod 9 -> 1 ≡ 1 mod 9


Inductive case:
Assume 10^k ≡ 1 mod 9
Need 10^(k+1) ≡ 1 mod 9

10^k ≡ 1 mod 9 -> 9 | (10^k - 1)
-> there exists an integer d such that 9d = 10^k - 1
-> 90d = 10^(k+1) - 10
-> 90d + 9 = 10^(k+1) - 1
-> 9(10d+1) = 10^(k+1) - 1
-> 10^(k+1) ≡ 1 mod 9

Answer 5

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