Find the greatest number that divides each of the four integers 210, 315, 147, and 168 exactly (that is, with zero remainder).

Introduction / Context:
This question asks for the largest common divisor (also called the highest common factor, HCF or greatest common divisor, GCD) of four given integers: 210, 315, 147, and 168. Finding the GCD ensures the number divides each integer without leaving a remainder.

Given Data / Assumptions:

• Numbers: 210, 315, 147, 168.
• We need the largest positive integer that divides all four exactly.
• Standard arithmetic properties of divisibility and GCD apply.

Concept / Approach:
The GCD of multiple integers can be computed stepwise: gcd(a, b, c, d) = gcd(gcd(gcd(a, b), c), d). Prime factorization or the Euclidean algorithm can be used at each step. Euclidean steps are typically quicker for larger numbers.

Step-by-Step Solution:

Verification / Alternative check:
Prime factors: 210 = 2*3*5*7; 315 = 3^2*5*7; 147 = 3*7^2; 168 = 2^3*3*7. Common primes are 3 and 7, but 3 appears in all, 7 appears in all; the minimum powers across all are 3^1 and 7^1, hence 3*7 = 21. Matches the Euclidean result.

Why Other Options Are Wrong:

• 4410: That is a product-like distractor; it does not divide any of the numbers.
• 14 or 7 or 3: These divide the set but are not the greatest possible; 21 is larger and still divides all.

Common Pitfalls:

• Stopping at a common divisor that is not the greatest (e.g., 7 or 3).
• Mixing up LCM and GCD.

Final Answer:
21