A natural number $a$ is selected from the first $100$ natural numbers. The probability that $a=[\frac a2]+[\frac a3]+[\frac a5]$, where [.] represents greatest integer function, is $\frac mn$ where $m,n$ are coprime then $(m+n)$ is equal to
My Attempt:
Let $a=30n+\gamma$, where $0\le\gamma\lt30$
Putting this in the given equation, I get,
$n=\gamma-[\frac{\gamma}{2}]-[\frac{\gamma}{3}]-[\frac{\gamma}{5}]$
$\gamma=0$ doesn't satisfy but $\gamma=1, 2, …, 29$ satisfy.
So, the probability is $\frac{29}{100}$.
Is this correct?
Best Answer
The answer looks correct. But you probably should demonstrate that $a\le 100$ when $\gamma=1,2,…,29$. It is sufficient to show that $n\le 2$. It is true since $$n=\gamma-[\gamma/2]-[\gamma/3]-[\gamma/5]\le $$
$$\le \gamma-(\gamma/2-1/2)-(\gamma/3-2/3)-(\gamma/5-4/5)=$$
$$=59/30-\gamma/30\le 2.$$
It is also worth showing that $n\ge 0$:
$$n=\gamma-[\gamma/2]-[\gamma/3]-[\gamma/5]\ge$$
$$\ge\gamma-\gamma/2-\gamma/3-\gamma/5=-\gamma/30.$$
Since $\gamma <30$, we get $n\ge-\gamma/30>-1\ge 0$.
Finally, it would be good to show that different values of $\gamma$ give different values of $a$. If different values of $\gamma$ lead to different values of $n$ then values of $a$ will be different since $\gamma<30$. Otherwise, if values of $n$ are the same, then values of $a=30n+\gamma$ obviously differ.