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In mathematics, two integers are considered coprime if their greatest common divisor (GCD) is 1. This means they have no common factors other than 1. For example, in the original exercise, we examined whether 16 and 103 are coprime. Since 103 is a prime number, it only has two divisors: 1 and 103 itself. Prime numbers are very handy when checking for coprimality. If a prime number does not divide another number, these two numbers are automatically coprime. To confirm that 16 and 103 are coprime, we find that 103 does not divide 16. Therefore, their GCD is indeed 1, which lets us move forward in finding the modular inverse.

The Extended Euclidean Algorithm is a method to find integers, usually called coefficients, that satisfy Bézout's identity for two numbers. It goes beyond just finding the GCD. It actually helps to express the GCD as a linear combination of the two numbers.For our exercise, we use this algorithm to derive the multiplicative inverse of 16 modulo 103. The inverse is a number such that, when multiplied by 16, yields 1 when considering modulo 103.Here's how it works:

• We start with the division: 103 divided by 16, which results in the remainder 7. This is the first step.
• We iterate this process, continually using remainders, until we isolate a remainder of 1.

Through back substitution, the calculations reveal the inverse is 6 because it satisfies the equation: \( 16 \times 6 \equiv 1 \pmod{103} \).

The multiplicative inverse of a number is another number which, when multiplied with the original number, results in an identity element under multiplication, which, in the modular system, is 1.In the context of modular arithmetic, finding the multiplicative inverse of a number is crucial for solving congruences like the one in the exercise: \(16 \cdot x \equiv 21 \pmod{103}\).Using the Extended Euclidean Algorithm, we found that 6 is the multiplicative inverse of 16 modulo 103. This means

• If you multiply 16 by 6, it equals 1 in the realm of modulo 103.
• This allows us to solve the equation step by step, by isolating \(x\).

Multiplying both sides of our equation by 6 facilitates the cancellation of 16 on the left-hand side, leading to: \( x \equiv 126 \pmod{103} \), which simplifies eventually to \( x \equiv 23 \pmod{103} \). Thus, 23 is the integer solution to the original equation, showing the powerful utility of the multiplicative inverse in modular arithmetic.