A natural number has prime factorization given by n = 2 x 3 y 5 z , where y and z are such that y + z = 5 and y - 1 + z - 1 = 5 6 , y > z . Then the number of odd divisors of n , including 1 , is:

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MathematicsPermutation & CombinationnPr & nCrHard2 minPYQ_2021

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A natural number has prime factorization given by $n = 2^{x} 3^{y} 5^{z}$ , where $y$ and $z$ are such that $y + z = 5$ and $y^{- 1} + z^{- 1} = \frac{5}{6}, y > z$ . Then the number of odd divisors of $n$ , including $1$ , is:

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Answer:

Solution:

$y + z = 5$

$\frac{1}{y} + \frac{1}{z} = \frac{5}{6} y > z$

$\Rightarrow y = 3, z = 2$

$\Rightarrow n = 2^{x} . 3^{3} . 5^{2} = (2.2.2 \dots) (3.3.3) (5.5)$

For calculating the odd divisor of $n = 2^{x} 3^{y} 5^{z},$ $x$ must be $0$ .

Hence, Number of odd divisors $= (3 + 1) \times (2 + 1) = 4 \times 3 = 12$ .

Stream:JEESubject:MathematicsTopic:Permutation & CombinationSubtopic:nPr & nCr

⏱ 2mℹ️ Source: PYQ_2021

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