LOG second

Algebra Level 5

$x_{1} \times x_{2}... \times x_{n}=e$ and each $\ln(x_{i})$ is a proper fraction. Find the greatest integral value of $\ln(x)-\lfloor(\ln x_{1}) \times \ln(x)\rfloor -\lfloor(\ln x_{2}) \times \ln(x)\rfloor -\cdots-\lfloor (\ln x_{n}) \times \ln(x)\rfloor,$

where $n$ is the least 2-digit prime number , and $\ln x$ is a natural number

Notation : $\lfloor \cdot \rfloor$ denotes the floor function .

The answer is 10.

1 solution

Prince Loomba
May 1, 2016

let $ln x_{i}$ to be some $X_{i}$ , and then put $X=X \times (X_{1}+X_{2}+...)$ . The equation becomes sum of n fractional parts whose max value is n-1. It can be shown that this function attains maximum value n-1 at $lnx=u-1$ , where u is LCM of the denominators of $X_{i}$ . Since {x}+{-x}=1 if x is not an integer and it is given that all are proper fractions. ({.} Represents fractional part)

LOG second

The answer is 10.

1 solution

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