Question

For each integer $n\ge 1$, define $a_n=\left[\frac{n}{\left[\sqrt{n}\right]}\right]$, where $\left[x\right]$ denotes the largest integer not exceeding $x$, for any real number $x$. Find the number of all $n$ in the set $(1,2,3,......2010)$ for which $a_n>a_{n+1}$.

• A. $42$
• B. $43$
• C. $44$
• D. $45$

Answer 1

The correct option is B $43$

Let us examine the first few natural number: $1,2,3,4,5,6,7,8,9$.
Here we see that $a_n=1,2,3,2,2,3,3,4,3$. We observe that $a_n\le a_{n+1}$ for all $n$ except when $n+1$ is a square in which case $a_n>a_{n+1}$. We prove that this observation is valid in general.
Consider the range
$m^2,m^2+1,m^2+2,,,,..,m^2+m,m^2+m+1,.....m^2+2m$
Let $n$ take values in this range so that $n=m^2+r$, where $0\le r\le 2m$.
Then we see that $\left[\sqrt{n}\right]=m$ and hence
$[\frac{n}{\left[\sqrt{n}\right]}=[\frac{m^2+r}{m}]=m+[\frac{r}{m}]$.
Thus $a_n$ takes the values $\underset{mtimes}{m,m,m,.......m,}\underset{mtimes}{m+1,m+1,m+1,.......m+1,m+2}$, in this range.
But when $n={(m+1)}^2$, we see that $a_n=m+1$.
This shows that $a_{n-1}>a_n$ whenever $n={(m+1)}^2$.
When we take $n$ in the set $(1,2,3,.....2010)$, we see that the only squares are $1^2,2^2,.......{44}^2$(since ${44}^2=1936$ and ${45}^2=2025)$ and $n={(m+1)}^2$ is possible for only $43$ values of $m$.
$\therefore a_n>a_{n+1}$ for $43$ values of $n$. (These are $2^2-1,3^2-1,......{44}^2-1)$