Recall that the analytic rank $r^{\rm an}(E)$ of a (modular) elliptic curve $E$ is defined to be the order of vanishing of its Hasse-Weil $L$-function $L(E,s)$ at $s=1$. A conjecture due to Ralph Greenberg in [MR1260957 (95a:11059)] implies in particular the following:

Basic fact: Let $\Sigma$ be a finite set
of prime integers. Then, as $E$ ranges over all rational elliptic
curves of conductor divisible *only* by primes in $\Sigma$, we
have that $r^{\rm an}(E)$ is $0$ or $1$ except for finitely many
$E$'s.

Immediate as it is, this fact is a special case of the (rather deep, as formulated in the paper above) Greenberg's conjecture, namely that of normalized newforms of weight $2$ with rational Fourier coefficients, and it shows that (rational) elliptic curves with larger and larger Mordell-Weil rank will tend to have larger and larger conductor; more precisely, such elliptic curves will tend to have conductor divisible by a larger and larger number of primes.

So I am wondering if we can give an easy answer to the following:

-Can we show that there exist rational elliptic curves with conductor divisible by an arbitrarily large number of distinct primes?

-Can we do better and given any finite set $\Sigma$ of
distinct primes, show the existence of a rational elliptic curve
with conductor divisible by *each* of the primes in $\Sigma$?

-Can we show that in fact there exist a number of non-isogenous such elliptic curves for each of the previous questions that grows (e.g. linearly!) with the cardinality of $\Sigma$?

Qwithexactlybad primes in $\Sigma$, then the answer is no. Take $\Sigma$ empty for example. :) Or $\Sigma=\{5\}$, or $\Sigma=\{13\}$, and such. $$ $$ If the question is as reads, then the answer is yes, via twisting. To get an elliptic curve divisible by all primes in $\Sigma$, consider the $2^\#\Sigma$ curves obtained by quadratic twists of subsets of $\Sigma$, and (at least) one will have the conductor divisible by all $p\in\Sigma$. There are $\infty$ many such curves, seen by twisting $p\not\in\Sigma$. – Junkie Apr 15 '11 at 4:34