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,


$\gamma=0$ doesn't satisfy but $\gamma=1, 2, ..., 29$ satisfy.

So, the probability is $\frac{29}{100}$.

Is this correct?


3 Answers 3


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 $n\ge 0$:



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.


Yes, the answer is $\frac{29}{100}$. There are a few details that need to be taken care of.
You are setting $\gamma$ to be one of $\{1,2,\dots,29\}$ and each value of $\gamma$ generates a value for $n$, which then determines $a=30n+\gamma$.

Why is $\gamma\ge \left\lfloor \frac{\gamma}{2} \right\rfloor + \left\lfloor \frac{\gamma}{3} \right\rfloor + \left\lfloor \frac{\gamma}{5} \right\rfloor$ for all $\gamma \in \{1,2,\dots,29\}$ ?

What about two different $\gamma$'s giving the same value for $a$ ?

Is the condition $a\le 100$ satisfied in each case ?


Let $a=6q+r$. Then we need $r =\left[\dfrac{r}{2} \right]+\left[\dfrac{r}{3} \right]+\left[\dfrac{q+r}{5} \right]$

When $r=0$, $q$ can be $1,2,3,4$

For $r=1,2,3,4,5$ it is easy to see that $q$ has $5$ solutions each (none exceeding $100$)

Thus there are $29$ solutions


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