# Remainders

• Mar 28th 2006, 09:21 PM
dooz33
Remainders
Help I'm so confused on how to solve these math problems....
• Find the remainder when 15! is divided by 17
• Find the remainder when 2(26!) is divided by 29
• Verify that 4(29!) + 5! is divisible by 31
• Mar 29th 2006, 03:28 PM
ThePerfectHacker
Quote:

Originally Posted by dooz33
Help I'm so confused on how to solve these math problems....
[list][*]Find the remainder when 15! is divided by 17

Since, you are working modulo 17, you have that the product,
15x14x13x12x11x10x9x8x7x6x5x4x3x2x1
Is the same as,
(-2)x(-3)x(-4)x(-5)x(-6)x(-7)x(-8)x8x7x6x5x4x3x2x1
Thus, you have, (multiply same numbers)
-1x4x9x16x25x36x49x64
Again, since you are working modulo 17 you have,
-1x4x(-8)x(-1)x(-8)x(2)x(-2)x(-4)
Thus, you have, (multiply same numbers)
1x16x64
Thus,
1x(-1)x(-4)
Thus,
4
• Mar 29th 2006, 04:50 PM
topsquark
Now THAT was just too cool! :D

-Dan
• Mar 30th 2006, 10:11 PM
rgep
I suspect that these are exercises on Wilson's Theorem: $(p-1)! \equiv -1 \pmod p$ for prime $p$.

So for example $16.15! \equiv -1 \pmod{17}$ gives $15! \equiv 1$. (ThePerfectHacker dropped some factors from his multiplication.)
• Mar 31st 2006, 09:34 AM
SkyWatcher
Quote:

Originally Posted by rgep
I suspect that these are exercises on Wilson's Theorem: $(p-1)! \equiv -1 \pmod p$ for prime $p$.

So for example $16.15! \equiv -1 \pmod{17}$ gives $15! \equiv 1$. (ThePerfectHacker dropped some factors from his multiplication.)

I suspect there is more then one théorème involved in those three different questions thus they are surely many ways to find the solution
wich is the clever according to the questioner supposed knowledge :)

edit: i am on the third's question from here and now!
----- and i know not much (just a little basis on congruence).

edit: for this one i would take PerfectHacker methodologie