Question

A natural number has a prime factorization given by \[n = {2^x}{3^y}{5^z}\], where \[y\] and \[z\] are such that \[y + z = 5\] and \[{y^{ - 1}} + {z^{ - 1}} = \dfrac{5}{6}\], \[y > z\]. Then find the number of odd divisors of \[n\], including 1.
A. \[11\]
B. \[6x\]
C. \[12\]
D. 6

Answer 1

Hint: Solved the given equations and find the values of \[y\] and \[z\]. Then apply the formula of the total number of divisors of a number and find the number of odd divisors of \[n\].

Formula used:
If \[a,b,c\] are the prime factors of a number \[n\] and \[{n_1},{n_2},{n_3}\] are the integral powers of them, then the total numbers of divisors of a number \[n\] are: \[d = \left( {{n_1} + 1} \right)\left( {{n_2} + 1} \right)\left( {{n_3} + 1} \right)\]

Complete step by step solution:
Given:
The prime factorization of a number is: \[n = {2^x}{3^y}{5^z}\]
\[y + z = 5\] and \[{y^{ - 1}} + {z^{ - 1}} = \dfrac{5}{6}\], \[y > z\]
Now simplify the equation.
\[{y^{ - 1}} + {z^{ - 1}} = \dfrac{5}{6}\]
\[ \Rightarrow \]\[\dfrac{1}{y} + \dfrac{1}{z} = \dfrac{5}{6}\]
\[ \Rightarrow \]\[\dfrac{{y + z}}{{yz}} = \dfrac{5}{6}\]
Substitute \[y + z = 5\] in above equation.
\[\dfrac{5}{{yz}} = \dfrac{5}{6}\]
Cross multiplies the above terms.
\[yz = 6\]
By using the identities \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\] and \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\], we get
\[{\left( {y - z} \right)^2} = {\left( {y + z} \right)^2} - 4yz\]
Substitute \[yz = 6\] and \[y + z = 5\] in the above equation.
\[{\left( {y - z} \right)^2} = {\left( 5 \right)^2} - 4\left( 6 \right)\]
\[ \Rightarrow \]\[{\left( {y - z} \right)^2} = 25 - 24\]
\[ \Rightarrow \]\[{\left( {y - z} \right)^2} = 1\]
Take square root on both sides.
\[y - z = \pm 1\]
Now solve the equations \[y + z = 5\] and \[y - z = \pm 1\].
Add equations \[y + z = 5\] and \[y - z = 1\].
\[2y = 6\]
\[ \Rightarrow \]\[y = 3\] and \[z = 2\]
Add equations \[y + z = 5\] and \[y - z = - 1\].
\[2y = 4\]
\[ \Rightarrow \]\[y = 2\] and \[z = 3\]
It is given that \[y > z\].
So, possible solution is \[y = 3\] and \[z = 2\].
Now substitute the above values in \[n = {2^x}{3^y}{5^z}\].
\[n = {2^x}{3^3}{5^2}\]
Since 2 is an even number.
So \[x\] must be 0.
Apply the formula for the total number of divisors of a number.
Therefore, the odd divisors of a number \[n\] are:
\[d = \left( {3 + 1} \right)\left( {2 + 1} \right)\]
\[ \Rightarrow \]\[d = \left( 4 \right)\left( 3 \right)\]
\[ \Rightarrow \]\[d = 12\]
Hence the correct option is C.

Note: To find the total number of divisors of a number, first find the prime factorization of the number and then find the integral power of each distinct prime number. Then add one to all the exponents and then find their product. The product is the total number of divisors of a number.