A natural number, when divided by each of 9, 10, 12, and 15, leaves a remainder of 3 in every case. Find the smallest such natural number that satisfies all these remainder conditions simultaneously.

Introduction / Context:
This problem involves congruences and least common multiples. The condition that a number leaves the same remainder when divided by several divisors can be rephrased in terms of a number being a fixed amount greater than a common multiple of those divisors. This is a typical pattern in modular arithmetic questions that appear in competitive exams.

Given Data / Assumptions:

• Let the required number be N.
• N leaves a remainder of 3 when divided by 9, 10, 12, and 15.
• We want the smallest natural number N satisfying all these conditions.

Concept / Approach:
If N leaves remainder 3 when divided by each of the given divisors, then N minus 3 is exactly divisible by 9, 10, 12, and 15. In other words, N - 3 is a common multiple of these numbers. The smallest such N corresponds to N - 3 being the least common multiple (LCM) of these divisors, and then we add 3 back to find N.

Step-by-Step Solution:
Compute LCM of 9, 10, 12, and 15. 9 = 3^2 10 = 2 * 5 12 = 2^2 * 3 15 = 3 * 5 Take highest powers: 2^2, 3^2, 5. LCM = 2^2 * 3^2 * 5 = 4 * 9 * 5 = 180. So N - 3 must be a multiple of 180. For the smallest N, take N - 3 = 180, so N = 180 + 3 = 183.

Verification / Alternative check:
Check N = 183: 183 / 9 gives remainder 3. 183 / 10 gives remainder 3. 183 / 12 gives remainder 3. 183 / 15 gives remainder 3. Thus 183 satisfies all conditions. Any smaller number would correspond to N - 3 being a smaller common multiple than 180, which is impossible since 180 is the least common multiple.

Why Other Options Are Wrong:
153, 63, 123, and 303 either fail to give remainder 3 with one or more of the divisors or are not the smallest such number. For example, 153 divided by 10 does not leave remainder 3, and 303 is larger than the minimal solution 183.

Common Pitfalls:
A frequent mistake is to misinterpret the condition and look for a number divisible by the divisors instead of one leaving a fixed remainder. Another common issue is computing the LCM incorrectly by not taking the highest powers of primes. Always convert remainder conditions to a common multiple condition by subtracting the remainder, then use the LCM to find the smallest solution.

Final Answer:
183