what is the smallest positive integer N such that 13^N \equiv 1 \pmod{2013}
To find the smallest positive integer N such that 13^N ≡ 1 (mod 2013), we can use the concept of order in modular arithmetic.
The order of an integer a modulo n, denoted as ordₙ(a), is the smallest positive integer k such that aᵏ ≡ 1 (mod n). In this case, we need to find the order of 13 modulo 2013.
Step 1: Find the prime factorization of 2013
2013 = 3 * 11 * 61
Step 2: Check the order of 13 modulo each prime factor
- For 13^k ≡ 1 (mod 3), the possible values of k are 1, 2.
- For 13^k ≡ 1 (mod 11), the possible values of k are 5, 10.
- For 13^k ≡ 1 (mod 61), the possible values of k are 30 or some multiple of 30.
Step 3: Find the least common multiple (LCM) of the possible orders
LCM of 2, 10, 30 = 30
Therefore, the smallest positive integer N such that 13^N ≡ 1 (mod 2013) is 30.
To find the smallest positive integer N such that 13^N ≡ 1 (mod 2013), we need to consider the properties of modular arithmetic and use some number theory concepts.
First, let's break down 2013 into its prime factors: 2013 = 3 * 11 * 61.
Now, let's start by finding the smallest positive integer n such that 13^n ≡ 1 (mod 3), then for mod 11, and finally for mod 61.
For mod 3:
To find the smallest n such that 13^n ≡ 1 (mod 3), we can see that 13 ≡ 1 (mod 3). Therefore, any positive integer power of 13 will also be congruent to 1 (mod 3). So, for mod 3, n can be any positive integer.
For mod 11:
To find the smallest n such that 13^n ≡ 1 (mod 11), we can use Euler's theorem. Euler's theorem states that if a and m are coprime (meaning they have no common factors), then a^(ϕ(m)) ≡ 1 (mod m), where ϕ(m) is the Euler's totient function that gives the count of positive integers less than m that are coprime to m.
The Euler's totient function ϕ(11) is equal to 10 since there are 10 positive integers that are coprime to 11 (1, 2, 3, 4, 5, 6, 7, 8, 9, 10). So, we need to find the smallest n such that 13^n ≡ 1 (mod 10).
By trial and error, we can find that 13^6 ≡ 1 (mod 10). Therefore, for mod 11, the smallest n would be 6.
For mod 61:
Again, we can use Euler's theorem to find the smallest n such that 13^n ≡ 1 (mod 61). The Euler's totient function ϕ(61) is equal to 60 since there are 60 positive integers that are coprime to 61 (1, 2, 3, ..., 60). So, we need to find the smallest n such that 13^n ≡ 1 (mod 60).
Using trial and error, we can determine that 13^30 ≡ 1 (mod 60). Therefore, for mod 61, the smallest n would be 30.
Now, we need to find the smallest positive integer N that satisfies all three congruences for mod 3, mod 11, and mod 61. This is where the Chinese Remainder Theorem (CRT) comes into play.
According to CRT, if we have a system of congruences of the form:
x ≡ a (mod m)
x ≡ b (mod n)
x ≡ c (mod p)
Where m, n, and p are pairwise coprime (meaning they have no common factors), then there exists a solution for x, and it is unique modulo the product of the three moduli (m * n * p).
In our case, we have:
N ≡ 0 (mod 3)
N ≡ 6 (mod 11)
N ≡ 30 (mod 61)
Using CRT, we can solve this system of congruences. The solution is:
N ≡ 750 (mod 2013)
So, the smallest positive integer N that satisfies 13^N ≡ 1 (mod 2013) is N = 750.