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 \(y^{-1} + z^{-1} = \frac{5}{6}\), \(y \gt z\). Then the number of odd divisions of \(n\), including 1, is :

• A. \(11\)
• B. \(6\)
• C. \(12\)
• D. \(6x\)

Hint:

Find \(y, z\) from \(y+z=5\) and \(yz=6\); odd divisors come only from the factors \(3^y 5^z\), giving \((y+1)(z+1)\).

Answer 1

The Correct Option is C

\(\frac1y + \frac1z = \frac{y+z}{yz} = \frac56\) and \(y+z=5\) give \(yz = 6\).
So \(y, z\) are roots of \(t^2 - 5t + 6 = 0\), i.e. 2 and 3. Since \(y \gt z\): \(y = 3\), \(z = 2\).
An odd divisor of \(n\) cannot contain the factor 2, so it has the form \(3^p 5^q\) with \(0 \le p \le 3\), \(0 \le q \le 2\).
Number \(= (3+1)(2+1) = 12\).
Answer: (3) \(12\)