A natural number has prime factorization given by n=2x3y5z, where y an – InterviewSolution

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1.	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=56,y>z. Then the number of odd divisors of n, including 1, is:
Answer» 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:

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