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⋯(1) 1y+1z=56 ⇒ yz=6 Also (y z)^2=(y+z)^2 4yz ⇒ y z=±1 nbsp; nbsp;...(2) from (1) and (2), y=3 or 2 and z=2 or 3 for calculating odd divisor of p=2^x. ...

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Standard XII

Mathematics

Direct Method of Validation

A natural num...

Question

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:

11

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Solution

The correct option is C $12$
$y + z = 5 \dots (1)$
$\frac{1}{y} + \frac{1}{z} = \frac{5}{6}$
$\Rightarrow y z = 6$
Also $(y - z)^{2} = (y + z)^{2} - 4 y z$
$\Rightarrow y - z = \pm 1 . . . (2)$
from $(1)$ and $(2)$ , $y = 3$ or $2$ and $z = 2$ or $3$
for calculating odd divisor of $p = 2^{x} {.3}^{y} {.5}^{z}$
$x$ must be zero
$P = 2^{0} {.3}^{3} {.5}^{2}$
$∴$ total odd divisors must be $(3 + 1) (2 + 1) = 12$

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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:

The correct option is C 12y+z=5⋯(1) 1y+1z=56 ⇒yz=6 Also (y−z)2=(y+z)2−4yz ⇒y−z=±1...(2) from (1) and (2), y=3 or 2 and z=2 or 3 for calculating odd divisor of p=2x.3y.5z x must be zero P=20.33.52 ∴ total odd divisors must be (3+1)(2+1)=12

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: