Question

Hi, how do we prove that 3^(2m)*m+3m, is divisible by 4 ?

Answer 1

you wanto use congruences / binomial ?

Answer 2

\(3^{2m}*m + 3m \equiv (-1)^{2m}*m - m \equiv 0 \mod 4\)

Answer 3

@ganeshie8 Could you explain ?

Answer 4

Got it

Answer 5

good :)
just need to see that \(3 \equiv-1 \mod 4\)

Answer 6

but how does \((−1)^{2m}∗m−m = 0 mod 4\) ?

Answer 7

whats the value of \((-1)^{2m}\) after simplifying ?

Answer 8

OOOOh

Answer 9

:)

Answer 10

Exactly

Answer 11

What about induction ?

Answer 12

definitely worth trying ! induction proof looke even more beautiful

Answer 13

Yes let's get started.

Answer 14

with \(0\) it's ok.

Answer 15

yes go straight to induction hypothesis

Answer 16

\(m(3^{2m}+3)=0mod4\) is true. :)

Answer 17

how about, \(m+1(3^{2m}*3+3)\)

Answer 18

sorry \((m+1)\)

Answer 19

assume : \(k(3^{2k}+3) = 4n \)

and prove : \((k+1)(3^{2(k+1)}+3) = 4t\)

Answer 20

Yes

Answer 21

@ganeshie8 from where can i find the 4 ?

Answer 22

its getting nasty lol... @Abhishek619 wana try :)

Answer 23

It's difficult-

Answer 24

@ganeshie8 \(-2m=0 mod4\) ?

Answer 25

induction theorem

Answer 26

?

Answer 27

It is divisible by 4.