For a positive integer $n≥2$, let $f(n)$ denote the smallest prime divisor of $n$. Find the number of integers $n$, $2≤n≤1000$, such that $f(n)=5$. All the numbers I could think of were $5,25,35,55,...$ This results in a total of $65$ numbers, which I don't know why is incorrect.

  • $\begingroup$ I make it 67? I think you've just miscounted slightly. $\endgroup$ – Christopher Feb 2 '15 at 14:31

Start with the multiples of $5$ - there are $200$ of those in the range specified.

Then we need to remove all those with a factor of $2$ or $3$. There are $100$ of those numbers divisible by $2$ at $10,20, \ldots 1000$. The remaining set - after even numbers are removed - has divisibility by $3$ at $15,45,75,\ldots 975$. There are a total of $33$ such numbers.

This illustrates why it's a good idea to write down some description and notes on your process in finding answers - it allows you to investigate problems with your answer, so you don't end up with "I don't know" as your last word on discrepancies.

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