The number of odd proper positive divisors of $3^{a} 6^{b} 21^{c}$ is

(a + 1) (b + 1) (c + 1) - 2

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(a + b + c + 1)(c + 1) - 1

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(a + 1)(b + 1)(c + 1) - 1

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(a + 1)(b + 1)(c + 1)

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Solution

The correct option is B (a + b + c + 1)(c + 1) - 1
We have to break $3^{a} 6^{b} 21^{c}$ in prime numbers for finding proper divisors.
$3^{a} 6^{b} 21^{c}$ $=$ $2^{b} 3^{a + b + c} 7^{c}$
Number of odd proper divisors are $=$ $(a + b + c + 1) (c + 1) - 1$ (Have to exclude case of 2)

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