Question

find the remainder of

\(\Large\tt \color{black}{30^{72^{87}}}\) divided by 11

Answer 1

work 72^87 mod 10 first

Answer 2

its 8

Answer 3

can i apply fermat number here

Answer 4

yes thats right! may i know how you got 8 ?

Answer 5

its 2^87/10

Answer 6

to be honest, idk the method you're using but it seems very interesting
can you teach me quick if possible ? xD

Answer 7

its the 2,4,8,6, rule :P

Answer 8

Oh cycle ?

Answer 9

lol its very easy (7*10+2)10 leaves remainder 2^87/10

Answer 10

i mean (7*10+2)&87/10 leaves remainder 2^87/10

Answer 11

this one (7*10+2)^87/10 leaves remainder 2^87/10

Answer 12

\[\large \dfrac{72^{87}}{10} = \dfrac{(7*10 + 2)^{87}}{10} = \dfrac{2^{87}}{10}\]

?

Answer 13

nice :) how do you reduce 2^87 ?

Answer 14

its the old remainder theorm i think (ax+b)^n mod a leaves b

Answer 15

by binomial theorem : (ax+b)^n mod a leaves b^n

Answer 16

leaves b^n i mean

Answer 17

(ax+b)^n = aM + b^n = b^n mod a

Answer 18

okay whats the trick for reducing 2^87 ?

Answer 19

2*2^(86) mod 10= 2^3*16^(21) mod 10 ..and so on

u need even power to reduce

Answer 20

if u follow the cycle its easiest

Answer 21

2^87 = 2^3*2^84 = 2^3*(16)^21 = 8*6 = 8

like that ?

Answer 22

yes

Answer 23

Very clever indeed, i use congruences to reduce which is something similar to this but i like your method as it looks more natural xD

Answer 24

yes but 1 month earlier u told congruencies are universal

Answer 25

Ok, so what do we have so far

30^72^87 = 30^(10M+8)

right ?

Answer 26

yes they are still universal, but it requires some theory to be known to appreciate them

Answer 27

we use fermat number

Answer 28

you mean we use Fermat little theorem

Answer 29

yes this one

Answer 30

30^(10M+8) = 30^8*30^(10M)
= 30^8*1 mod 11
= ...

Answer 31

u mean we can use fermat here ?

Answer 32

yes since 11 is a prime :

\[\large a^{10} \equiv 1 \pmod {11}\]

Answer 33

see you can't fully avoid congruences :P

Answer 34

i always avoid it

Answer 35

i use remainder theorm ,but this had a^b^c so i cant use it here

Answer 36

hmm what do you get after applying fermat little theorem ?

Answer 37

we got rid of exponent right

Answer 38

30^72^87 = 30^(10M+8)

Answer 39

its 6!!

Answer 40

show work

Answer 41

30^8/11
=8^8/11
=64^4/11
=(-2)^4/11
=16/11

its 5 sry

Answer 42

am i right

Answer 43

Excellent !!

Answer 44

so for fermat do i have to always reduce b^c/10 in a^b^c

Answer 45

thats not fermet, but it depends on what number you're dividing by

Answer 46

if you're dividing by 13, you reduce b^c/12

Answer 47

if you're dividing by 19, you reduce b^c/18

Answer 48

because fermat theorem says this :
\[\large a^{p-1}\equiv 1 \pmod p\]

Answer 49

ok thanks very much for help

Answer 50

@ganeshie8 please help me!!