Difficulty: Medium
Correct Answer: 183
Explanation:
Introduction:
This question examines your understanding of modular arithmetic and the use of least common multiple (LCM) to handle conditions involving the same remainder for several divisors. It is a typical number system problem in aptitude tests.
Given Data / Assumptions:
Concept / Approach:
If a number N leaves the same remainder r when divided by several divisors, then N − r is exactly divisible by each of those divisors. Therefore, N − 3 must be a common multiple of 9, 10, 12, and 15. The smallest possible N occurs when N − 3 is the LCM of these divisors.
Step-by-Step Solution:
Let N be the required number. Given: N ≡ 3 (mod 9), N ≡ 3 (mod 10), N ≡ 3 (mod 12), N ≡ 3 (mod 15) So N − 3 is divisible by 9, 10, 12, and 15. Compute LCM of 9, 10, 12, and 15. 9 = 3^2 10 = 2 × 5 12 = 2^2 × 3 15 = 3 × 5 LCM needs highest powers: 2^2, 3^2, 5 LCM = 2^2 × 3^2 × 5 = 4 × 9 × 5 = 180 Thus, N − 3 = 180 for the smallest N. So N = 180 + 3 = 183
Verification / Alternative check:
Check divisibility: 183 ÷ 9 leaves remainder 3. 183 ÷ 10 leaves remainder 3. 183 ÷ 12 leaves remainder 3. 183 ÷ 15 leaves remainder 3. These verifications confirm that 183 satisfies all conditions. Any smaller number would correspond to a smaller common multiple than 180, which is impossible because 180 is the LCM.
Why Other Options Are Wrong:
63, 123, and 153 do not leave remainder 3 for all four divisors simultaneously. At least one of the divisors gives a different remainder for each of these numbers, so they cannot satisfy the given conditions. Only 183 works for every divisor.
Common Pitfalls:
Students sometimes attempt random trial and error or only check some of the divisors. Another common mistake is to confuse HCF with LCM and use the HCF instead of the LCM, which leads to much smaller and incorrect values of N.
Final Answer:
The smallest natural number satisfying all the conditions is 183.
Discussion & Comments