JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Permutations and Combinations

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 :

Answer: (D)

Solution

From

y^{- 1} + z^{- 1} = \frac{y + z}{y z} = \frac{5}{6}

together with

y + z = 5

comes

y z = 6

, so

y

and

z

are the roots of

t^{2} - 5 t + 6 = 0 \Rightarrow t = 3, 2

and

y > z

forces

y = 3

z = 2

, giving

n = 2^{x} 3^{3} 5^{2}

.
A divisor of

n

is odd exactly when it carries no factor

2

, that is when it has the form

3^{p} 5^{q}

with

0 \leq p \leq 3

and

0 \leq q \leq 2

, so their number is

(3 + 1) (2 + 1) = 12

The number of odd divisors of

n

, including

1

, is

12

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Single correct option