Find the least non residue

Question

Find the least non residue of the following

$41 × 42 × · · · 54 modulo 19$

$41 × 42 × · · · 54=54!/40!$

$41 ≡ 3 mod 19$

$54 ≡ 16 mod 19$

That is as far I can get.

Any help of how to continue would be appreciated.

Answer 1

As you have seen $41 \equiv 3 \pmod{19}$ and $54 \equiv 16 \pmod{19}$. Can you see then that $41\cdot 42 \cdot\dots \cdot 54 \equiv 3\cdot 4\cdot\dots 16 \pmod{19}$? Also notice that

$4\cdot5 \equiv 1 \pmod{19}$

$6 \cdot 3 \equiv -1 \pmod{19}$

$7\cdot8 \equiv -1 \pmod{19}$

$13\cdot 16 \equiv -1 \pmod{19}$

$14\cdot 15 \equiv 1 \pmod{19}$

$11\cdot12 \equiv -1 \pmod{19}$

$9\cdot10 \equiv 14 \pmod{19}$.

Answer 2

As you say, $$41\times 42\times\cdots\times 54\equiv3\times4\times\cdots\times16\pmod{19}.$$ But $$3\times4\times\cdots\times16=\frac{18!}{2\times17\times 18}.$$ If there was a nice way to calculate $18!$ modulo $19$, that would surely help.