NOTATION:   $P(x)$   stands for the product of all primes which do not exceed real   $x$;   e.g.   $P(10)=210$.

QUESTION:   Given any real   $x\ge 3$,   compute or estimate the smallest natural number   $n:=n(x)\ge 3$   such the remainder of the division of   $n$   by any odd prime   $p\le x$   is   $1$ or $2$   (it may be any combination of   $1$s   and   $2$s).

In particular, improve upon my simple theorem below, and still better upon its consecutive improvements provided by the MO participants.

THEOREM $$ n(x)\ > \ \left\lceil\sqrt{P(x)}\right\rceil + 1 $$

EXAMPLEs (small calculations):

• $n(3) = 4$
• $n(5) = 7$
• $n(7) = 16$

REMARK (warning): First I talk about all primes (see NOTATION), then about odd primes (see QUESTION).

PROOF of the THEOREM

Let integer   $n\ge 3$   be such that all mentioned remainders are   $1$   or   $2$.   Let   $A$   be the product of all odd primes   $p\le x$   for which   $n=1\mod p$, and   $B$   be the same for remainder   $2$.   Thus   $A|n-1$   and   $B|n-2$,   hence

$$A\cdot B\ =\ \frac{P(x)}2 $$

Furthermore,   $2|n-1$   or   $2|n-2$,   hence   $A'|n-1$   and   $B'|n-2$   for certain natural numbers such that

$$ A'\cdot B'\ =\ P(x) $$

It follows that

$$ n\ \ \ge\ \ \max(A'\ B') + 1$$

while

$$ \max(A'\ B')\ \ \ge\ \ \left\lceil\sqrt{A'\cdot B'}\right\rceil\ \ =\ \ \left\lceil\sqrt{P(x)}\right\rceil $$

This proves the required inequality:

$$ n\ \ \ge\ \ \left\lceil\sqrt{P(x)}\right\rceil + 1 $$