QJEEM21Mac443: JEE Main 2021 Mathematics - Class 11 JEEM21Mac443 Solution: A natural number has prime factorization given by n = 2 x 3 y 5 …

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