Question

Find the last two digits of 6 raised to the power of 13.

Answer 1

alll wrong it is

Answer 2

what class is this for?

Answer 3

are you doing stuff with modulo 10?

Answer 4

A prereq math class

Answer 5

No module 10

Answer 6

so it is not 36?

Answer 7

no!

Answer 8

6 is the last number

Answer 9

2nd last number trying to figure out for odd powers

Answer 10

96???

Answer 11

Thank you

Answer 12

ya i think so too i did something messed up in there lol

Answer 13

lemme tell u what i did, tell me if i messsed up

Answer 14

I am with a study group right now who also thinks it may be 16 now that it has been mentioned.

Answer 15

6^0 = 1 (mod 100)

6^1 = 6 (mod 100)

6^2 = 36 (mod 100)

6^3 = 6^2*6^1 = 36*6 = 216 = 16 (mod 100)

6^4 = (6^3)*6^1 = 16*6 = 96 (mod 100)

6^5 = (6^4)*6^1 = 96*6 = 576 = 76 (mod 100)

6^6 = (6^5)*6^1 = 76*6 = 456 = 56 (mod 100)

6^7 = (6^6)*6^1 = 56*6 = 336 = 36 (mod 100)


So to sum things up

6^0 = 1 (mod 100)
6^1 = 6 (mod 100)
6^2 = 36 (mod 100)
6^3 = 16 (mod 100)
6^4 = 96 (mod 100)
6^5 = 76 (mod 100)
6^6 = 56 (mod 100)
6^7 = 36 (mod 100)

the pattern starts off with 1, 6, but then gets into a loop of 36, 16, 96, 76, 56 and this loop repeats forever

Answer 16

Aww comment got deleted somehow...aww lol...I'm trying to figure it out too!

Answer 17

Now that @jim_thompson5910 has shown that...it seems easy! lol

Answer 18

ohh its 16 i multiplied 36 in the end

Answer 19

i did it for 36^7 oops