Establish divisibility conditions

A number is divisible by 12 if it is divisible by both 3 and 4. The rule for 3 is that the sum of the digits must be divisible by 3, and for 4 is that the last two digits form a number divisible by 4.

$n \equiv 0 \pmod{12} \iff n \equiv 0 \pmod{3} \text{ and } n \equiv 0 \pmod{4}$

Testing Options A, B, and D

For 144 (sum 9, ends 44), 288 (sum 18, ends 88), and 432 (sum 9, ends 32), all are divisible by both 3 and 4, confirming they are divisible by 12.

$144/12 = 12, \quad 288/12 = 24, \quad 432/12 = 36$

Testing Option C

For 316, the sum of digits is $3+1+6 = 10$. Since 10 is not divisible by 3, 316 is not divisible by 3, and consequently, not divisible by 12.

$316 \div 3 = 105.33, \quad 316 \div 12 = 26.33$