A natural number has prime factorization given by n=2x3y5z, where y and z are such that y+z=5 and y-1+z-1=56,y>z. Then the number of odd divisors of n, including 1, is

y+z=5,1y+1z=56,y>zN=2x3y52 Solving y=3,z=2 No.of odd divisors of N=(3+1)(2+1)=12

A natural number has prime factorization given by n=2x3y5z, where y and z are such that y+z=5 and y-1+z-1=56,y>z. Then the number of odd divisors of n, including 1, is

y+z=5,1y+1z=56,y>zN=2x3y52 Solving y=3,z=2 No.of odd divisors of N=(3+1)(2+1)=12

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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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$\begin{array}{l} y + z = 5, \frac{1}{y} + \frac{1}{z} = \frac{5}{6}, y > z N = 2^{x} 3^{y} 5^{2} \\ Solving y = 3, z = 2 \end{array}$

$No.of odd divisors of N = (3 + 1) (2 + 1) = 12$

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A natural number has prime factorization given by n=2x3y5z, where y and z are such that y+z=5 and y-1+z-1=56,y>z. Then the number of odd divisors of n, including 1, is

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