What is the least positive number that, when divided by both 63 and 105, leaves a remainder of 4 in each case?

Introduction / Context:
This problem is a standard application of greatest common divisors and least common multiples in modular arithmetic. It asks for the smallest number that leaves the same remainder when divided by two different divisors. Such questions are common in number system topics of aptitude exams and help learners understand congruences and how to use least common multiples to synchronise remainder conditions.

Given Data / Assumptions:

• We are looking for a number N.
• When N is divided by 63, it leaves a remainder of 4.
• When N is divided by 105, it leaves the same remainder of 4.
• N must be the least positive integer satisfying both conditions.

Concept / Approach:
If a number N leaves the same remainder r when divided by two numbers, say m and n, then N - r is divisible by both m and n. In this problem, N leaves remainder 4 when divided by 63 and 105, so N - 4 must be divisible by both 63 and 105. Therefore, N - 4 is a common multiple of 63 and 105. The smallest positive N will correspond to N - 4 being the least common multiple of 63 and 105, and then we add the remainder back.

Step-by-Step Solution:
Step 1: Let the required number be N.Step 2: According to the problem, N ≡ 4 (mod 63) and N ≡ 4 (mod 105).Step 3: This means that N - 4 is divisible by both 63 and 105.Step 4: Therefore, N - 4 must be a common multiple of 63 and 105.Step 5: To get the smallest such N, find the least common multiple (LCM) of 63 and 105.Step 6: Factorise: 63 = 3^2 * 7 and 105 = 3 * 5 * 7.Step 7: The LCM takes the highest powers of all primes present: LCM = 3^2 * 5 * 7 = 9 * 5 * 7 = 315.Step 8: So N - 4 = 315 for the smallest positive solution.Step 9: Therefore, N = 315 + 4 = 319.

Verification / Alternative check:
Check divisibility conditions for N = 319. When 319 is divided by 63, we compute 63 * 5 = 315 and the remainder is 319 - 315 = 4, which matches the requirement. When 319 is divided by 105, 105 * 3 = 315 and again the remainder is 319 - 315 = 4. Since both conditions are satisfied and 319 is based on the least common multiple, there is no smaller positive N that works for both divisors simultaneously.

Why Other Options Are Wrong:
Option 634: 634 - 4 = 630, which is divisible by 63 and 105, but 630 is actually 2 * 315, so 634 is a larger valid solution, not the least one.
Option 256: 256 - 4 = 252, which is divisible by 63 but not by 105.
Option 214: 214 - 4 = 210, which is divisible by 105 but not by 63.
Option 172: 172 - 4 = 168, which is divisible by 63 but not by 105.

Common Pitfalls:
A common mistake is to set up separate equations for N and try to solve them directly rather than shifting by the remainder. Some may also confuse greatest common divisor (GCD) with least common multiple (LCM) and attempt to use the GCD instead, which does not synchronise the remainder conditions properly. Remember that when the remainders are the same, subtracting the remainder reduces the problem to finding a common multiple.

Final Answer:
The least number that leaves a remainder of 4 when divided by both 63 and 105 is 319.