"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 = ${5 \over 6}$, y z. Then the number of odd divisions of n, including 1, is :"

Question

"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 = ${5 \over 6}$, y > z. Then the number of odd divisions of n, including 1, is :"

Accepted Answer

"y + z = 5 ....... (1)

${1 \over y} + {1 \over z} = {5 \over 6}$

$\Rightarrow {{y + z} \over {yz}} = {5 \over 6}$

$\Rightarrow {5 \over {yz}} = {5 \over 6}$

$\Rightarrow$ yz = 6

Also, (y $-$ z)² = (y + z)² $-$ 4yz

$\Rightarrow$ (y $-$ z)² = (y + z)² $-$ 4yz

$\Rightarrow$ (y $-$ z)² = 25 $-$ 4(6) = 1

$\Rightarrow$ y $-$ z = 1 ..... (2)

from (1) and (2), y = 3 and z = 2

for calculating odd divisor of p = 2^x . 3^y . 5^z

x must be zero

P = 2⁰ . 3³ . 5²

$\Rightarrow$ Total possible cases = (3⁰5⁰ + 3¹5⁰ + 3²5⁰ + 3³5⁰ + .... + 3³5²)

$\therefore$ Total odd divisors must be (3 + 1) ( 2 + 1) = 12"

A natural number has prime factorization given by n = 2^x3^y5^z, where y and z are such
that y + z = 5 and y^$-$1 + z^$-$1 = ${5 \over 6}$, y > z. Then the number of odd divisions of n, including 1, is :

Solution

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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 = ${5 \over 6}$, y > z. Then the number of odd divisions of n, including 1, is :

Solution

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A natural number has prime factorization given by n = 2^x3^y5^z, where y and z are such
that y + z = 5 and y^$-$1 + z^$-$1 = ${5 \over 6}$, y > z. Then the number of odd divisions of n, including 1, is :