Key Concepts and Formulas

• Greatest Common Divisor (GCD): The GCD of two integers $a$ and $b$, denoted as $gcd(a, b)$, is the largest positive integer that divides both $a$ and $b$. If $gcd(a, b) = G$, then $gcd(a/G, b/G) = 1$.
• Prime Factorization and GCD: If $a = p_1^{a_1} p_2^{a_2} \dots$ and $b = p_1^{b_1} p_2^{b_2} \dots$, then $gcd(a, b) = p_1^{\min(a_1, b_1)} p_2^{\min(a_2, b_2)} \dots$.
• Counting Integers in a Range: The number of integers in the range $[A, B]$ (inclusive) is $B - A + 1$.

Step-by-Step Solution

Step 1: Translate the GCD condition.

We are given that $gcd(N, 36) = 2$, where $100 \le N \le 999$. The prime factorization of 36 is $36 = 2^2 \times 3^2$. Since $gcd(N, 36) = 2$, we can write $N = 2k$ for some integer $k$. Then $gcd(2k, 36) = 2$, which means $gcd(k, 18) = 1$. Therefore, $k$ must not be divisible by 2 or 3, since $18 = 2 \times 3^2$. Since $N = 2k$, $N$ is divisible by 2. Since $k$ is not divisible by 2, $N$ is not divisible by 4. Therefore, $N \equiv 2 \pmod{4}$. Also, since $k$ is not divisible by 3, $N = 2k$ is not divisible by 3. So, we want to find the number of 3-digit integers $N$ such that $100 \le N \le 999$, $N \equiv 2 \pmod{4}$, and $N \not\equiv 0 \pmod{3}$.

Step 2: Find the number of 3-digit integers such that $N \equiv 2 \pmod{4}$.

We want to count the number of integers $N$ such that $100 \le N \le 999$ and $N = 4k + 2$ for some integer $k$. Substituting $N = 4k+2$ into the inequality gives $100 \le 4k+2 \le 999$. Subtracting 2 from all parts gives $98 \le 4k \le 997$. Dividing by 4 gives $24.5 \le k \le 249.25$. Since $k$ is an integer, $25 \le k \le 249$. The number of integers $k$ in this range is $249 - 25 + 1 = 225$. So there are 225 such integers $N$. Let this set be $S_1$.

Step 3: Find the number of integers in $S_1$ that are also divisible by 3.

We want to find the number of integers $N$ such that $100 \le N \le 999$, $N \equiv 2 \pmod{4}$, and $N \equiv 0 \pmod{3}$. From $N \equiv 2 \pmod{4}$, we have $N = 4k + 2$. Substituting this into $N \equiv 0 \pmod{3}$ gives $4k + 2 \equiv 0 \pmod{3}$. This simplifies to $k + 2 \equiv 0 \pmod{3}$, so $k \equiv -2 \equiv 1 \pmod{3}$. Thus, $k = 3m + 1$ for some integer $m$. Substituting this back into $N = 4k + 2$ gives $N = 4(3m + 1) + 2 = 12m + 4 + 2 = 12m + 6$. We want to find the number of integers $N$ of the form $12m + 6$ such that $100 \le N \le 999$. Substituting $N = 12m+6$ into the inequality gives $100 \le 12m + 6 \le 999$. Subtracting 6 gives $94 \le 12m \le 993$. Dividing by 12 gives $94/12 \le m \le 993/12$, or $7.833... \le m \le 82.75$. Since $m$ is an integer, $8 \le m \le 82$. The number of integers $m$ in this range is $82 - 8 + 1 = 75$.

Step 4: Calculate the final count.

The number of 3-digit integers $N$ such that $gcd(N, 36) = 2$ is the number of integers $N$ such that $100 \le N \le 999$ and $N \equiv 2 \pmod{4}$, minus the number of those integers that are also divisible by 3. So, the total count is $225 - 75 = 150$.

Common Mistakes & Tips

• Be careful when translating the GCD condition into divisibility rules. Remember that $gcd(a,b) = c$ implies that $a/c$ and $b/c$ are coprime.
• Ensure you are counting integers within the correct range by carefully considering the endpoints.
• When using modular arithmetic, simplify congruences to their simplest form to avoid mistakes.

Summary

We found the number of 3-digit integers $N$ such that $gcd(N, 36) = 2$ by first translating the GCD condition into divisibility rules, then counting the number of integers satisfying these rules, and finally subtracting the number of integers that satisfy both the desired condition and a contradictory condition (divisibility by 3). The final answer is 150.

Final Answer

The final answer is $\boxed{150}$.