Consider the product of any four consecutive natural numbers, n(n + 1)(n + 2)(n + 3). What is the largest natural number that always divides this product?

Introduction / Context:
This question asks for a universal divisor of every product of four consecutive integers. We use parity and divisibility arguments to identify guaranteed factors (powers of 2 and 3) present in any block of four consecutive numbers.

Given Data / Assumptions:

• General product: n(n + 1)(n + 2)(n + 3) for integer n
• We seek a constant divisor valid for all n

Concept / Approach:
Among any four consecutive integers: (i) there are two even numbers, one of which is a multiple of 4, ensuring a factor of 8; (ii) there is always a multiple of 3, ensuring a factor of 3. Combine these guaranteed factors to find the fixed divisor.

Step-by-Step Solution:
Among n, n+1, n+2, n+3: exactly two are even; one of the even numbers is divisible by 4 ⇒ contributes 4.The other even number contributes an extra factor 2 ⇒ total factor from parity = 8.At least one of the four is divisible by 3 ⇒ factor 3 present.Fixed divisor = 8 * 3 = 24.

Verification / Alternative check:
Try n = 1: product = 1*2*3*4 = 24, divisible by 24. Try n = 2: 2*3*4*5 = 120, also divisible by 24.

Why Other Options Are Wrong:
6 and 12 are too small (not maximal). 48 and 120 do not always divide the product (48 would require a guaranteed factor 16, which is not always present).

Common Pitfalls:
Overestimating the fixed power of 2 (thinking 16 is always present) or forgetting to include a guaranteed multiple of 3.

Final Answer:
24