Question

Prove that 6^18 + 36^20 can be divided by 37. No calculators and dont solve the equation.

Answer 1

@ganeshie8 @perl @texaschic101 @ParthKohli

Answer 2

\(6^{18}\div36^{20}\) is not a integer...

Answer 3

ahh + not /

Answer 4

\[6^{18} + 36^{20} = 36^9 + 36^{20}\equiv (-1)^9 +(-1)^{20} \equiv -1+1\equiv 0 \pmod{37}\]

Answer 5

Well 37 is 6^2+1 I suppose that would be a good starting place, I don't know.

Answer 6

hmmm

Answer 7

if mods dont make sense, start with kainui's step..

Answer 8

(6^18+6^40)/(6^2+1)

Answer 9

ganeshie knows more about these kinds of things than me. In my opinion all numbers are divisible by 37 lol...

Answer 10

ganeshie's answer is optimal here

Answer 11

you may try below if you're familiar with binomial thm \[6^{18}+36^{20} = 36^9 + 36^{20} \\= (37-1)^9 + (37-1)^{20} \\=37M + (-1)^9 + 37N
+ (-1)^{20} \\= 37X -1 + 1\]

Answer 12

what is M?

Answer 13

good question :) that means you have followed upto 3rd line ?

Answer 14

uhhh

Answer 15

hmm interesting! the way you did @ganeshie8

Answer 16

"prove that 6^18 + 36^20 is divisible by 37. "
This is equivalent to finding an integer k such that
6^18 + 36^20 = 37k

Answer 17

so m is instead of k?

Answer 18

tht M is supposed to be all terms that were multiples by 37 when you do binomial theorem or expansion

Answer 19

ok so of course 37M divides by 37

Answer 20

awesome work @ganeshie8 thx

Answer 21

but wait how (37-1)^9=37M-1^9?

Answer 22

soo the final would be 36*37*(37^10+e)?

Answer 23

let me be more rigorous, make this airtight

Answer 24

the mod way works too , that ganeshie suggested

Answer 25

hmmmm how 36^9+36^20=36*(1+36^11)? mybe 36^9*(1+36^11)?

Answer 26

2sd and 3rf line

Answer 27

typo

Answer 28

thx

Answer 29

tthx

Answer 30

6^18 + 36^20
= (6^2)^9 + 36^20
=36^9 + 36^20
=36^9*(1 + 36^11)
= 36^9*( 1 + (37-1) ^11 )
So by binomial theorem
= 36^9( 1 + 37^11 + terms*37 - 1^11 )
= 36^9( 1 + 37^11 + terms*37 -1)
= 36^9( 37^11 + terms*37 )
= 36^9*37*( 37^10 + terms*37)

so thats a multiple of 37

Answer 31

and to be sure about the binomial theorem part.
look at the expansion of a binomial expression
(a + b)^n , show that terms are multiple of the first term 'a', except for the last term