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 :

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=(5/6), y > z . Then the number of odd divisors of n, including 1 , is : (A) 11 (B) 6 (C) 6 x (D) 12

Tardigrade Admin · Accepted Answer

y+z=5 (1/y)+(1/z)=(5/6) y>z ⇒ y =3, z =2 ⇒ n =2 x ⋅ 33 ⋅ 52=(2.2 .2 ldots)(3.3 .3)(5.5) Number of odd divisors =4 × 3=12

Q. 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

6

6 x

12

$y + z = 5$
$\frac{1}{y} + \frac{1}{z} = \frac{5}{6} y > z$
$\Rightarrow y = 3, z = 2$
$\Rightarrow n = 2^{x} \cdot 3^{3} \cdot 5^{2} = (2.2.2 \dots) (3.3.3) (5.5)$
Number of odd divisors $= 4 \times 3 = 12$

Q. 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=65​,y>z. Then the number of odd divisors of n, including 1 , is :

11

6

6 x

12

y+z=5 y1​+z1​=65​y>z ⇒y=3,z=2 ⇒n=2x⋅33⋅52=(2.2.2…)(3.3.3)(5.5) Number of odd divisors =4×3=12

Q. 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 :

$y + z = 5$
$\frac{1}{y} + \frac{1}{z} = \frac{5}{6} y > z$
$\Rightarrow y = 3, z = 2$
$\Rightarrow n = 2^{x} \cdot 3^{3} \cdot 5^{2} = (2.2.2 \dots) (3.3.3) (5.5)$
Number of odd divisors $= 4 \times 3 = 12$