Is there a theorem which states the number of integral quotient of $n^2$ divided by $\{1,2,3, ... n^2\}$ is $2n-1$?
Example: If $n=4$,then $16 \div \{1,2,3, ... 16\} = 16,8,5,4,3,2,1$. There are $7$ integral quotients.
Number of integral quotient of $n^2$
4
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number-theory
1 Answers
5
The set $\{\,\lfloor \frac {n^2}d\rfloor\mid 1\le d\le n^2\,\}$ contains the numbers $1,2,\ldots,n$ from taking $d=n,\ldots, n^2$ because $d=n$ leads to $\lfloor\frac{n^2}{n}\rfloor=n$, and because $\lfloor \frac{n^2}{d+1}\rfloor$ is either $\lfloor \frac {n^2}d\rfloor$ or $\lfloor \frac {n^2}d\rfloor-1$ for $d\ge n$ (because $\frac{n^2}d-\frac{n^2}{d+1}=\frac{n^2}{d(d+1)}<1$). So that's $n$ values so far.
On the other hand, the values obtained from $1\le d\le n$ are pairwise distinct because $\frac{n^2}d-\frac{n^2}{d+1}=\frac{n^2}{d(d+1)}>1$ for $d