Question

prove that

\(\large\tt \begin{align} \color{black}{n^{(n-1)}+5~\mod (n+1)\equiv 4,6\\~\\\normalsize\text{ for }\\n>5\hspace{.33em}\\~\\}\end{align}\)

Answer 1

\(\)

Answer 2

n=-1 mod n+1

Answer 3

ok got it

Answer 4

nice

Answer 5

\(
n^{(n-1)}+5 \mod (n+1) \equiv (-1)^{(n-1)}+5\mod (n+1) \\ \text {when n is odd then}\\
n^{(n-1)}+5 \mod (n+1) \equiv 1+5\mod (n+1) \equiv 6\mod (n+1) \\ \text {when n is even then}\\
n^{(n-1)}+5 \mod (n+1) \equiv -1+5\mod (n+1) \equiv 4\mod (n+1) \)

Answer 6

oh hehe

Answer 7

again mod......... ^.^ >,<

Answer 8

you're familiar with mods @Nnesha

Answer 9

lol how u got that \(\huge (-1)^{n+1}\) @Marki

Answer 10

\[n \equiv -1 \pmod {n+1}\]

Answer 11

n+1=0 mod n+1
n=-1 mod n+1

Answer 12

take (n-1) power both sides

Answer 13

\[n \equiv -1 \pmod {n+1}\]

\[n^{n-1} \equiv (-1)^{n-1} \pmod {n+1}\]

Answer 14

oh i see O-0

Answer 15

wanna see ur method then @mathmath333

Answer 16

we have this :

\[a\equiv b \pmod{n} \implies a^k \equiv b^k \pmod{n}\]

Answer 17

lol it was unsolved by me

Answer 18

oh :O

Answer 19

yes i know few mods
here is a list
ganeshie8
iambatman
abhisar
hero
zarkon
tk
......................................
but i never used this mod word <-- in math

Answer 20

this mod is much friendlier than the ones you listed above ^

Answer 21

for starting, think of "mod" as another fancy name for "remainder"

Answer 22

3 mod 10 evaluates to 1

because

10/3 leaves a remainder 1

Answer 23

so ur cool with it now ? @mathmath333 ?
and this condition n>5
to show
5=5 mod k for any k >5 and n+1>5

Answer 24

see if you can work below :

12 mod 5 = ?

Answer 25

okay wait :)

Answer 26

yea i m kind of cool guy :)
it thought this question was harder lol

Answer 27

lol

Answer 28

i thught its tough for first sight

Answer 29

ok , have to go
Cya !

Answer 30

nope!!!