Key Concepts and Formulas

• Greatest Common Divisor (GCD): The largest positive integer that divides two or more integers without leaving a remainder. If $\text{gcd}(a, N) = K$, then $a$ is a multiple of $K$, and $\text{gcd}(a/K, N/K) = 1$.
• Prime Factorization: Expressing a number as a product of its prime factors.
• Principle of Inclusion-Exclusion (Subtraction Principle): $N(\text{A and not B}) = N(\text{A}) - N(\text{A and B})$

Step-by-Step Solution

Step 1: Understanding the Problem

We need to find the number of 4-digit numbers $a$ such that $\text{gcd}(a, 54) = 2$. This means $a$ must be a multiple of 2, and after dividing both $a$ and 54 by 2, the resulting numbers must be coprime.

Step 2: Prime Factorization of 54

We find the prime factorization of 54: $54 = 2 \times 27 = 2 \times 3^3$ This tells us the prime factors of 54 are 2 and 3.

Step 3: Applying GCD Properties

Given $\text{gcd}(a, 54) = 2$, we have two conditions:

1. $a$ must be a multiple of 2, meaning $a$ is even.
2. $\text{gcd}(a/2, 54/2) = \text{gcd}(a/2, 27) = 1$. This means $a/2$ is not divisible by 3. Consequently, $a$ is not divisible by 3. If $a$ were divisible by 3, then $a$ would be divisible by 6, and $\text{gcd}(a, 54)$ would be a multiple of 6, not 2.

Therefore, we need to count 4-digit numbers $a$ that are divisible by 2 but not by 3.

Step 4: Defining the Range

The range of 4-digit numbers is $1000 \le a \le 9999$.

Step 5: Formulating the Counting Strategy

We use the Principle of Inclusion-Exclusion: $N(\text{divisible by 2 and not divisible by 3}) = N(\text{divisible by 2}) - N(\text{divisible by 2 and divisible by 3})$ Since numbers divisible by both 2 and 3 are divisible by 6, the strategy is: $N(\text{desired numbers}) = N(\text{4-digit numbers divisible by 2}) - N(\text{4-digit numbers divisible by 6})$

Step 6: Counting 4-Digit Numbers Divisible by 2

The first 4-digit even number is 1000, and the last is 9998. The arithmetic progression is $1000, 1002, \dots, 9998$. The number of terms is: $N(\text{divisible by 2}) = \frac{9998 - 1000}{2} + 1 = \frac{8998}{2} + 1 = 4499 + 1 = 4500$

Step 7: Counting 4-Digit Numbers Divisible by 6

The first 4-digit multiple of 6 is 1002 ($6 \times 167$), and the last is 9996 ($6 \times 1666$). The arithmetic progression is $1002, 1008, \dots, 9996$. The number of terms is: $N(\text{divisible by 6}) = \frac{9996 - 1002}{6} + 1 = \frac{8994}{6} + 1 = 1499 + 1 = 1500$

Step 8: Final Calculation

Applying the subtraction principle: $N(\text{desired numbers}) = N(\text{divisible by 2}) - N(\text{divisible by 6}) = 4500 - 1500 = 3000$

Therefore, there are 3000 four-digit numbers whose greatest common divisor with 54 is 2.

Common Mistakes & Tips

• Incorrect application of GCD properties: Ensure you correctly interpret the implications of $\text{gcd}(a, N) = K$ on the prime factors of $a$.
• Miscounting terms in arithmetic progressions: Double-check the first and last terms, and the common difference. Remember to add 1 in the formula.
• Forgetting the Subtraction Principle: Recognize when to use inclusion-exclusion to count elements satisfying certain conditions.

Summary

We determined that a 4-digit number $a$ has a GCD of 2 with 54 if and only if $a$ is divisible by 2 but not by 3. We used the Principle of Inclusion-Exclusion to count the number of such numbers by subtracting the number of 4-digit multiples of 6 from the number of 4-digit multiples of 2. This resulted in 3000 such numbers.

The final answer is \boxed{3000}.