Arun has some books, not less than 15 and not more than 30 in number. If he gives these books equally to 2 friends, 1 book remains. If he gives them equally to 3 friends, 2 books remain. If he gives them equally to 4 friends, 3 books remain. How many books does he have?

Introduction / Context:
This question is a nice application of remainders and modular arithmetic. It describes a real world situation involving distributing books among friends and uses congruence conditions to determine the unknown total number of books.

Given Data / Assumptions:

• Let the number of books be N.
• 15 ≤ N ≤ 30.
• When divided among 2 friends, remainder is 1 ⇒ N ≡ 1 (mod 2).
• When divided among 3 friends, remainder is 2 ⇒ N ≡ 2 (mod 3).
• When divided among 4 friends, remainder is 3 ⇒ N ≡ 3 (mod 4).

Concept / Approach:
We interpret the conditions using congruences. The pattern N ≡ 1 (mod 2), N ≡ 2 (mod 3) and N ≡ 3 (mod 4) suggests that N is one less than a multiple of 2, 3 and 4, or equivalently, N + 1 is divisible by all three numbers. We then search within the allowed range for the value that satisfies these conditions.

Step-by-Step Solution:
Step 1: From N ≡ 1 (mod 2), N is odd.Step 2: From N ≡ 2 (mod 3), N + 1 is divisible by 3.Step 3: From N ≡ 3 (mod 4), N + 1 is divisible by 4.Step 4: Therefore, N + 1 must be a common multiple of 2, 3 and 4. The least common multiple of 2, 3 and 4 is 12, so N + 1 is a multiple of 12.Step 5: Within the range 15 ≤ N ≤ 30, N + 1 must be a multiple of 12. Possible values for N + 1 in this range are 24 and 36, corresponding to N = 23 and N = 35, but 35 is outside the allowed range.Step 6: Thus N = 23 is the only candidate.Step 7: Check: 23 ÷ 2 gives remainder 1, 23 ÷ 3 gives remainder 2, and 23 ÷ 4 gives remainder 3. All conditions are satisfied.

Verification / Alternative check:
You can test the answer choice values directly: 19, 23, 17 and 15. Only 23 satisfies all three remainder conditions simultaneously.

Why Other Options Are Wrong:
19 fails because 19 ÷ 3 leaves remainder 1, not 2.17 fails because 17 ÷ 4 leaves remainder 1, not 3.15 fails because 15 ÷ 2 leaves remainder 1 but 15 ÷ 3 leaves remainder 0, not 2.

Common Pitfalls:
Forgetting to restrict N to the given range can lead to extraneous solutions.Some students misinterpret the remainder conditions and work with N - 1 instead of N + 1.Recognising that N + 1 must be a multiple of the least common multiple of 2, 3 and 4 makes the solution quick and robust.

Final Answer:
Arun has 23 books.