Let a, b, c in mathbb{N} and a+b+c=5. Let L a | vedclass

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(A) Given $a, b, c \in \mathbb{N}$ (positive integers) such that $a+b+c=5$. The possible triplets $(a, b, c)$ are $(1, 1, 3), (1, 3, 1), (3, 1, 1), (2

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$2 \cdot 3^2 \cdot 5 \cdot 7$

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(A) Given $a, b, c \in \mathbb{N}$ (positive integers) such that $a+b+c=5$. The possible triplets $(a, b, c)$ are $(1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1), (1, 2, 2), (2, 1, 2)$.Calculating the values of $2^a 3^b 5^c$ for each triplet:$(1, 1, 3) \implies 2^1 \cdot 3^1 \cdot 5^3 = 6 \cdot 125 = 750$$(1, 3, 1) \implies 2^1 \cdot 3^3 \cdot 5^1 = 2 \cdot 27 \cdot 5 = 270$$(3, 1, 1) \implies 2^3 \cdot 3^1 \cdot 5^1 = 8 \cdot 3 \cdot 5 = 120$$(2, 2, 1) \implies 2^2 \cdot 3^2 \cdot 5^1 = 4 \cdot 9 \cdot 5 = 180$$(1, 2, 2) \implies 2^1 \cdot 3^2 \cdot 5^2 = 2 \cdot 9 \cdot 25 = 450$$(2, 1, 2) \implies 2^2 \cdot 3^1 \cdot 5^2 = 4 \cdot 3 \cdot 25 = 300$The greatest value $M = 750$ and the least value $L = 120$.Then $M - L = 750 - 120 = 630$.Prime factorization of $630 = 2 \cdot 3^2 \cdot 5 \cdot 7$.Thus,option $A$ is correct.

Let $a, b, c \in \mathbb{N}$ and $a+b+c=5$. Let $L$ and $M$ be the least and greatest values of $2^a 3^b 5^c$ respectively. Then $M-L=$

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