MHB Common Multiple, what is the fourth term

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To determine the fourth term that, along with 12a^2bc, 8ab, and 4a^2cd, results in a least common multiple of 24a^3bc^2d, the necessary components are identified. The existing terms provide a least common multiple of 24a^2bcd, requiring an additional "a" and "c" in the fourth term. The coefficient of this new term must be an odd prime number that divides 24. The discussion highlights that the only odd prime factor of 24 is 3. Therefore, the fourth term should be 3a^1c^1, leading to the desired least common multiple.
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Can you show what you've tried or your thoughts on how to begin?
 
Three of the terms are 12a^2bc, 8ab, and 4a^2cd. To find the least common multiple of those 3, note that 12= 2^2(3), 8= 2^3, and 4= 2^2. The least common multiple of those three is 2^3(3)= 24. The highest power of a is a^2 and the highest power of b, c, and d is 1 for all three. So the least common multiple of those three terms is 24a^2bcd. We want another term such that the least common multiple of all four terms is 24a^3bc^2d. We already have the "24", the "b" and "d", two of the three "a"s, and one of the two "c"s. It looks like we need just one more "a" and one more "c". Any coefficient must be already included in the "24". What is an odd prime number that divides 24?
 
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