Log First

First, let's decipher the problem.

• Since we never actually reference $x_1, x_2, \ldots, x_n, x$ by themselves, substitute $a_i = \ln x_i, b = \ln x$ .
• I also assume that we may choose $a_1, a_2, \ldots, a_n, b$ freely. (As the problem is stated now, the answer depends on $a_1, a_2, \ldots, a_n$ ; for example, if it happens that $a_1 = 1, a_2 = a_3 = \ldots = a_n = 0$ , the answer is clearly 0.)

Thus, after removing the fluff:

If $a_1, a_2, \ldots, a_{101}$ are rational numbers and $b$ is an integer, find the maximum value of $b - (\lfloor a_1b \rfloor + \lfloor a_2b \rfloor + \ldots + \lfloor a_{101}b \rfloor)$ .

Now, we know that $\lfloor x \rfloor > x-1$ . Thus,

$\displaystyle\begin{aligned} \lfloor a_1b \rfloor + \lfloor a_2b \rfloor + \ldots + \lfloor a_{101}b \rfloor &> (a_1b - 1) + (a_2b - 1) + \ldots + (a_{101}b - 1) \\ &= (a_1 + a_2 + \ldots + a_{101})b - 101 \\ &= b - 101 \end{aligned}$

Thus, $b - (\lfloor a_1b \rfloor + \lfloor a_2b \rfloor + \ldots + \lfloor a_{101}b \rfloor) < b - (b-101) = 101$ . Since the left hand side of the inequality is an integer, we must then have $b - (\lfloor a_1b \rfloor + \lfloor a_2b \rfloor + \ldots + \lfloor a_{101}b \rfloor) \le 100$ . This is easily achievable with $a_1 = a_2 = \ldots = a_{101} = \frac{1}{101}, b = 100$ , so the answer is $\boxed{100}$ .

• 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=-1. Since {x}+{-x}=1 if x is not an integer and it is given that all are proper fractions. ({.} Represents fractional part)