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 $\Sigma$?



Elliptic curves with arbitrarily large conductor
The conductor is divisible by the same primes as the discriminant. The discriminant of $y^2=x^3+b$ is $-432b^2$. So you can certainly make the conductor divisible by an arbitrarily large number of distinct primes, and by whichever primes you want (as long as you want 2 and 3). 


