Step 1: Understanding the Concept:
We need to find numbers $N$ between 100 and 999 such that the greatest common divisor of $N$ and 36 is exactly 2.
Prime factorization of $36$ is $2^2 \times 3^2$.
If $\gcd(N, 36) = 2$, this imposes two conditions on $N$:
1. $N$ must be divisible by 2, but NOT divisible by 4 (otherwise the gcd would have a factor of 4).
2. $N$ must NOT be divisible by 3 (otherwise the gcd would have a factor of 3).
Step 2: Key Formula or Approach:
Condition 1 means $N$ must be of the form $4k + 2$.
Condition 2 means $N$ must not be a multiple of 3.
We will count how many numbers in the 3-digit range satisfy the first condition, and then subtract the count of numbers that violate the second condition.
Step 3: Detailed Explanation:
The range for 3-digit numbers is $100 \le N \le 999$.
Step A: Count numbers of the form $N = 4k + 2$.
\[ 100 \le 4k + 2 \le 999 \]
\[ 98 \le 4k \le 997 \]
\[ 24.5 \le k \le 249.25 \]
Since $k$ must be an integer, $k \in \{25, 26, 27, \dots, 249\}$.
Let $n_1$ be the number of such integers:
\[ n_1 = 249 - 25 + 1 = 225 \]
These 225 numbers satisfy Condition 1.
Step B: Remove numbers that are multiples of 3.
We need to find how many of the numbers generated by $4k+2$ are divisible by 3.
\[ 4k + 2 \equiv 0 \pmod 3 \]
\[ (3k + k) + 2 \equiv 0 \pmod 3 \]
\[ k + 2 \equiv 0 \pmod 3 \]
\[ k \equiv 1 \pmod 3 \]
So, for $N$ to be a multiple of $3$, $k$ must be of the form $3m + 1$.
Let's find the valid values for $k$ in our range $[25, 249]$:
The first $k$ in range satisfying $k \equiv 1 \pmod 3$ is 25 (since $25 = 3 \times 8 + 1$).
The sequence of bad $k$ values is an arithmetic progression: $25, 28, 31, \dots$
We need to find the largest term in this AP that is $\le 249$.
$25 + (m - 1)3 \le 249$
$3(m - 1) \le 224$
$m - 1 \le 74.66 \dots$
Since $m$ is an integer, the maximum value is $m - 1 = 74 \implies m = 75$.
So there are 75 values of $k$ that produce a number divisible by 3.
Let $n_2$ be the count of these bad numbers, $n_2 = 75$.
Step C: Final Calculation.
The total valid numbers is the total $N$'s minus the bad $N$'s:
\[ \text{Total} = n_1 - n_2 = 225 - 75 = 150 \]
Step 4: Final Answer:
The total number of such 3-digit numbers is 150.