Question: How many integers between $1$ and $200$, inclusive, are congruent to $3 \pmod
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- \text{lcm}(36, 121) = \text{lcm}(2^2 \cdot 3^2, 11^2) = 2^2 \cdot 3^2 \cdot 11^2 = 4 \cdot 9 \cdot 121 = 36 \cdot 121 = 4356
- Since $4356$ is a four-digit number and divisible by both $36$ and $121$, it is the smallest such number.
- \boxed{4356}
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