All numbers that are evenly divisible by both 6 and 14 are also divisible by which of the following numbers?
Any number \(x\) that is evenly divisible by both \(6\) and \(14\) must be a multiple of their least common multiple (LCM). To find the LCM, we first determine the prime factorizations of both numbers:
\(6 = 2 \times 3\)
\(14 = 2 \times 7\)
The least common multiple is found by taking the highest power of each prime factor present in the factorizations. Here, the prime factors are \(2\), \(3\), and \(7\). Thus, \(\text{lcm}(6, 14) = 2^1 \times 3^1 \times 7^1 = 42\). Therefore, any number \(x\) divisible by both \(6\) and \(14\) is a multiple of \(42\).
Since every such number \(x\) is of the form \(x = 42k\) for some integer \(k\), any number that divides \(42\) will necessarily divide \(x\). We examine the given options to see which one is a divisor of \(42\):
The only number from the options provided that divides \(42\) is \(21\).
Because \(21\) is a divisor of the LCM, it is a factor of \(42\), and since any number \(x\) divisible by both \(6\) and \(14\) is a multiple of \(42\), \(21\) must also divide any such number \(x\). The correct option is C. The final answer is \(\boxed{C}\).
In logistics and scheduling, if two different cargo planes depart from the same airport, one every 6 hours and the other every 14 hours, they will both be on the ground at the same time at intervals corresponding to the least common multiple of their schedules, which is \(42\) hours. If a maintenance crew is scheduled to work every 21 hours, they will be present every time both planes are at the airport simultaneously because 21 is a divisor of 42. This helps in efficiently coordinating airport operations and maintenance windows without conflict.