This exercise seemed straightforward but I have not managed to do the following proof. Let $p\#$ be the product of primes not exceeding p. Let $c(n)$ be the nth coprime to $p\#$ (mod 2,3,...,p). Let $X =\prod_{p_i\leq p}\phi(p_i).$ Of course there are X coprimes for given p, beginning in numerical order with 1.
To show: In the limit as p gets large, the sum $$\sum_{n\leq X}\frac{c(n)}{n}\sim p\#. $$
Some evidence--
\begin{array} {|r|r|} \hline p & \sum c(n)/n &p\#~~~~& \sum c(n)n^{-1}/p\# \\ \hline 5 &24.89 &30& 0.829\\ \hline 7 &201.05 &210&0.957 \\ \hline 11&2295.08&2310& 0.9935\\ \hline 13&30008.69 &30030&0.99929 \\ \hline 17& 510480.82&510510&0.999943\\ \hline \end{array}
Putting $C(x)=\sum_{n\leq x}\frac{c(n)}{n}$ we can use Abel summation to put
$$\sum_{n\leq x}\frac{c(n)}{n}= \frac{1}{x}C(x)+\int_1^x\frac{C(t)}{t^2}dt$$
and if $x=X,$ I think it's easy to show the first term on the right is $\frac{p\#}{2}.$ Assuming that's true, is there enough information to argue that the integral has the same value? If not, is there a better way?
Let $\# k = \prod_{p \le k} p = e^{\sum_{p \le k} \log p} = e^{k+o(k)}$ (by the PNT) then
$$\sum_{n \le N, gcd(n,\# k) = 1} 1 = \sum_{d | \# k} \mu(d) \lfloor \frac{N}d \rfloor = \sum_{d | \# k} (\mu(d)\frac{N}d + O(1)) = N \prod_{p \le k} (1-p^{-1}) +O(2^{\pi(k)})$$
With $c_k(m)$ the $m$-th integer coprime to $\# k$ then $m = \sum_{n \le N, gcd(n,\# k) = 1} 1$ and $gcd(N,\# k) = 1$ implies $c_k(m) = N$ so $$c_k(N \prod_{p \le k} (1-p^{-1}) +O(2^{\pi(k)})) = N+O(2^{\pi(k)}) , \qquad c_k(m) = \frac{m}{\prod_{p \le k} (1-p^{-1})}+O(2^{1+\pi(k)})$$
Whence $$\sum_{m \le \varphi(\# k)} \frac{c_k(m)}{m}=\sum_{m \le \varphi(\# k)}\frac{1}{\prod_{p \le k} (1-p^{-1})}+\frac{O(2^{1+\pi(k)})}{m} \\=\frac{\varphi(\# k)}{\varphi(\#k)/\#k}+O(2^{1+\pi(k)} \log \varphi(\# k))= \# k+O(2^{1+\pi(k)} k) $$
where all the $O$ constants are $1$