Remainder!!!!!!!!!!!!!!!!!!!!!!!

Well, if you want, you can use the fact that $(n+1)!=(n+1)\cdot n!$ in conjunction with modular arithmetic identities so as to ease the calculation of the remainder for the factorials without calculating the factorial itself. Let me give a demonstration.

Take $a=4$ and $b=5$ . Since, $a!$ can be computed orally, we find that $a!\equiv (-4) \pmod{14}$ . Now, you can write $b!=b\cdot a!$ and then use modular arithmetic identities to get $5!\equiv 5\cdot (-4) \equiv (-20) \equiv (-6) \equiv 8 \pmod{14}$ . Checking on a calculator shows that the remainder when $5!$ is divided by $14$ is $8$ indeed. Note that we calculated this without evaluating the factorial. Likewise, we can find the remainder for higher factorials without calculating the factorial itself.

Hope this helps. :)

Wow! Sudoku Subbu.Your method of division is great!!.Is 873/14 is 218 with remainder 1 ?

Did you know 14 x 218=3052.Then how can you say that 873/14=218.