M is the largest three digit number that leaves a remainder of 5 when divided by 6 and a remainder of 3 when divided by 5. What remainder does M leave when it is divided by 11?

Introduction / Context:
This question is a classic example of solving simultaneous congruences and then applying modular arithmetic again with a new divisor. It tests skill with Chinese remainder style reasoning and with searching for a largest three digit number that satisfies two different remainder conditions. After finding this number, we compute one more remainder with respect to a new divisor, which completes the problem.

Given Data / Assumptions:

• M is a three digit integer.
• M leaves remainder 5 when divided by 6, so M ≡ 5 (mod 6).
• M leaves remainder 3 when divided by 5, so M ≡ 3 (mod 5).
• Among all three digit numbers with these properties, M is the largest.
• We need the remainder when M is divided by 11.

Concept / Approach:
We solve the system of congruences M ≡ 5 (mod 6) and M ≡ 3 (mod 5). Because 6 and 5 are coprime, such numbers repeat with period equal to the least common multiple, which is 30. We can find one solution and then generate all other solutions by adding multiples of 30. From those, we pick the largest three digit solution. Finally, we divide this M by 11 to get the required remainder.

Step-by-Step Solution:
We need solutions of M ≡ 5 (mod 6) and M ≡ 3 (mod 5).Let us search for a solution modulo 30, since lcm of 6 and 5 is 30.Check numbers congruent to 5 modulo 6: 5, 11, 17, 23, 29, etc.Among these, test which is also congruent to 3 modulo 5.Take 23: 23 mod 5 = 3, so 23 satisfies both congruences.Thus, all solutions can be written as M = 23 + 30k, where k is a non negative integer.We now want the largest three digit M. So solve 23 + 30k ≤ 999.This gives 30k ≤ 976, so k ≤ 32.53..., hence k can be at most 32.Compute M for k = 32: M = 23 + 30 * 32 = 23 + 960 = 983.Thus, M = 983 is the largest three digit number satisfying the given conditions.Now compute the remainder when 983 is divided by 11.11 * 89 = 979 and 983 - 979 = 4.So the remainder is 4.

Verification / Alternative check:
Check with 6: 983 divided by 6 gives quotient 163 and remainder 5, which matches the first condition.Check with 5: 983 divided by 5 gives remainder 3, which matches the second condition.Check with 11: 983 = 11 * 89 + 4, so the remainder is 4, confirming our answer.

Why Other Options Are Wrong:
Remainders 1, 2, or 3 would correspond to different congruence results when 983 is divided by 11, which contradicts the actual computation.If a different three digit number is mistakenly chosen instead of 983, it will either not be the largest or fail one of the original remainder conditions.

Common Pitfalls:
Some learners incorrectly add 6 and 5 instead of using the least common multiple 30 as the period for repeating solutions.Another mistake is to forget that the number must be three digit, so values above 999 or below 100 must be rejected.It is also easy to make an arithmetic slip when converting from congruences to actual numbers of the form 23 + 30k.

Final Answer:
The remainder when M is divided by 11 is 4.