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How many numbers such $n$ are there that $n<100,\lfloor{\sqrt{n}} \rfloor \mid n$.

I just take the perfect squares and then wrote their multiply to the next perfect square and got the answer $27$.Am I right?

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Yes. To write a solution formally, note that we have to find numbers such that $\lfloor{\sqrt{n}} \rfloor \mid n$.

Also note that if $\lfloor{\sqrt{n}} \rfloor=k$, then $k^2 \le n < k^2+2k+1$.

Thus if $\lfloor{\sqrt{n}} \rfloor=k$, we have from the conditions $n=k^2, k^2+k, k^2+2k$. There are $9$ numbers of the form $k^2$, $9$ numbers of the form $k^2+k$, and $9$ numbers of the form $k^2+2k$, so there are indeed $27$ of them.

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