Question

still stuck...

Answer 1

on what

Answer 2

* operation is for set of S = {1,2,3,4,5,6} defined according table above. for integer n > 1 and satisfies x^n = x^(n-1) * x^n. what is the value of 5^2015 ?

Answer 3

@dan815

Answer 4

the table u have is for mod 7

Answer 5

honestly i dont know what the question asking ?

Answer 6

what, modulo ?

Answer 7

which part i can see the question is for mod 7 ?

Answer 8

if u look at the table, u see the numbers are the remainders for mod 7

Answer 9

oh, ok... 0,1,2,3,4,5,6 are remainder of mod7. i can see knw

Answer 10

next ?

Answer 11

i dont understand the n>1 and x^n.... part but i with the 5^2015 part they want to know what u get with mod 7

Answer 12

5^2015 mod 7 = ?

Answer 13

oh, i will try to get its remainder... just still process

Answer 14

5^2015 = 5^2014 * 5^2015, it says this equation is satisfied

Answer 15

5^2015 mod7
= 5 * 5^2014 mod7
= 5 * (7-2)^2014 mod7
= 5 * 2^2014 mod7
= 5 * ....
opss.. sorry im forget how to using mod , can you refresh me about it ?

Answer 16

@ganeshie8

Answer 17

best mod for mod

Answer 18

ah, yea i have to setting 2^2014 to 2^(3 x 671 + 1), right ?

Answer 19

since `x^n = x^(n-1) * x^n` property is satisfied, can we also say below will hold too

\[(x^m)^n = x^{mn}\]

Answer 20

from my process above :
= 5 * 2^2014 mod7
= 5 * 2^(3 x 671 + 1) mod7
= 5 * 2 * 1 mod7
= 10 mod7
= 3 mod7
is that right ?

Answer 21

so, the answer is 3 :D

Answer 22

interesting how did all these simplifications come

Answer 23

@ganeshie8 , can you show your argument completly ?

Answer 24

btw, here the options :
a. 3 b. 2 c. 1 d.0

Answer 25

5^x mod 7
here is something im thinking about

i know that for 5^n
5*5*5*5...
5 is lacking 2 in every multiplication

similiarly
(7-2)^n
(7-2)(7-2)(7-2)....=7^n+....+(-2)^n
from binomial expansion u know that
all the other terms have a 7*2 in them except the final -2^n therefore
thats all we need to worry about when dividing by 7 to see the remainder

Answer 26

so u can get that first step atleast

Answer 27

5^2015 mod 7 = 2^2015 mod 7

Answer 28

all the other terms have a 7*

Answer 29

5^2015 mod 7 = \(\color{red}{-}\)2^2015 mod 7

Answer 30

oh right yes

Answer 31

now we can look at powers of 2 that is divisible 7 to break it up again?

Answer 32

or mabe just see where its 1!

Answer 33

2^3 mod 7 is 1 so

Answer 34

is that how u got to the stop
2^(3*671+1)

Answer 35

step*

Answer 36

-2^2015 = -2^(3*671+2) = -4*8^671

Answer 37

is
2^(x+n) mod 7 = 2^x mod 7 * 2^n mod 7?

Answer 38

But we need to first establish the property : (a^m)^n = a^(mn)

Answer 39

we have to establish that wrt to ths mod?

Answer 40

All we are given is that `x^n = x^(n-1) * x^n`

Answer 41

ok right

Answer 42

yes with respect to the defined operation on given set

Answer 43

what are you thinking?

Answer 44

nothing lol it is trivial. .

Answer 45

how do we establish this property then

Answer 46

`x^n = x^(n-1) * x^\(\large \color{Red}{^n}\)

hey i notice a typo just now, that \(\large \color{Red}{^n}\) should not be there rigth ?

Answer 47

to me, from the property of x^n = x^(n-1) * x^n, we can establish the number of 2^2015 becomes :
5^2015 = 5 * (5 * 5 * 5 .... * 5) = 5 * 4 * (4 * 4 * .... * 4) = 6 * (2 * 2 * 2 ... * 2) =6 * 2 * (4 * 4 * ... * 4) = 5 * (2 * 2 * 2 * ... * 2) = 5 * 2 * (2 * 2 ... * 2) = 3 * (2 * 2 * ... ) = 3 * 2 * (2 * 2 * ... 2) = 6 * (2 * 2 * 2 ... 2)
too long, haha ....