<s>[INST] Let $a_1, a_2, a_3...$ be the sequence of all positive integers relatively prime to 75. Find the value of $a_{2008}$. Let $a_1, a_2, a_3...$ be the sequence of all positive integers relatively prime to 75, where $a_1<a_2<a_3...$ with $a_1=1, a_2=2, a_3=4, a_4=7$. Find the value of $a_{2008}$.
What I have done:
If ${a_n}$ is relatively prime to 75, then it is relatively prime to 3 and 5. Consider the following string of numbers from 1 to 15. $$1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15$$
If multiples of 3 and 5 are removed then, we need to take note that 15 is removed twice, so we need to add it back. In general, the number of integers left in a string of $n$ integers is
$$n-\lfloor \frac{n}{3} \rfloor - \lfloor \frac{n}{5} \rfloor+ \lfloor \frac{n}{15} \rfloor$$
Hence we need to find $n=a_{2008}$ such that
$$n-\lfloor \frac{n}{3} \rfloor - \lfloor \frac{n}{5} \rfloor+ \lfloor \frac{n}{15} \rfloor=2008$$
I cheated a bit here and took away the floor value to solve for $n$ and find that $n=3765$
Can anyone verify for me if this answer is correct? And provide a proper alternative solution? Thank you!
 [/INST] The number of positive integers relatively prime to 75 and less than 75 is 40, as given by Euler's formula.
Thus, in the numbers up to 3750, there are 2000 such numbers. Now we find that the eighth positive integer relatively prime to 75 is 14; hence the answer is 3750+14=3764.
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