A computer is programmed to choose an integer between $1$ and $99,$ inclusive, so that the probability that it selects the integer $x$ is equal to $\displaystyle\log_{100}\left(1+\frac{1}{x}\right).$ Suppose that the probability that $81 ≤ x ≤ 99$ is equal to $2$ times the probability that $x = n$ for some integer $n$. What is the value of $n?$
According to question$,$ $$\log_{100}\left(1+\frac{1}{x}\right)=2\log_{100}\left(1+\frac{1}{n}\right)$$$$\implies\left(1+\frac{1}{x}\right)=\left(1+\frac{1}{n}\right)^2$$ $$\implies 81\le x=\frac{n^3}{2n^2+n}\le99$$ Since $n$ is an integer $\implies n=\{163,164,\cdots,198\}$
What is wrong with my approach $?$ Why I am not getting a unique answer$?$
Any help is greatly appreciated.
Hint: Let $X$ be the random variable with distribution $P(X=x)=\log_{100}(1+\frac1x)$. Then $P(81\le X\le99)=\sum_{x=81}^{99}P(X=x)$, and the question states $2P(X=n)=P(81\le X\le99)$, so the formula $$\log_{100}\left(1+\frac{1}{x}\right)=2\log_{100}\left(1+\frac{1}{n}\right)$$ is not a correct interpretation of the question. Do you know what it should be?