Question

Find the remainder when \( 4^{2015} + 5^{2015}\) is divided by \(9\)

Answer 1

In other words, work/reduce : \[( 4^{2015} + 5^{2015}) \mod 9 \]

Answer 2

using binomial 9=4+5.
\[4^{2014} +5^{2014}= (4+5)[4^{2013} +5 *4^{2012}+.....+5^{2013}]\]

Answer 3

Clever ! if i understand correctly it seems you have used the formula \(x^n - y^n\) and factored 4+5 which shows the number is a multiple of 9. Interesting xD

Answer 4

yea

Answer 5

\(x^n+y^n\)
and there will be alternate negatives right??
it won't affect the final result though :)

Answer 6

any other methods ?

Answer 7

Oh right! thats a good catch, the terms in sum should alternate signs..

Answer 8

\(x^n~ mod ~q\)
there's any theorem ? when q is prime ?

Answer 9

im looking forward to see many more clever methods xD
i didn't think of the catch.me's trick.. it was pretty neat :)

Answer 10

yes but 9 is not prime here so we need to try something else i think

Answer 11

\(x^n~ mod ~ab = x^n~ mod~a \times x^n ~mod~b\)
is something like that true ?

Answer 12

the last digit of \(5^n\) is always \(5\)
so u need to work only \(4^{2015}\)

Answer 13

their is clever method lol

Answer 14

lets see..

consider \(2^3\pmod{9}\)

\(2^3 \mod{3}\) gives 2
so \(2^3 \mod{3} \times 2^3 \mod{3} = 2 \times 2 = 4\)

however we know that \(2^3\pmod{9}\) equals 8 so this relation will not hold in general @hartnn

Answer 15

math, if 5^n last digit is 5, then we can ignore it in calculation mod 9 ?
i didn't get that point

Answer 16

\(\large\tt \begin{align} \color{black}{
\dfrac{4^{2015} + 5^{2015}}{9}\\~\\
\implies \dfrac{(9-5)^{2015} + 5^{2015}}{9}\\~\\
\implies \dfrac{(-5)^{2015} + 5^{2015}}{9}\\~\\

\Large \equiv 0}\end{align}\)

Answer 17

Brilliant !!!!!!

Answer 18

oh i see yes the last digit only works on mod 10 sry

Answer 19

that really cute :D

Answer 20

if shorter methods are done, post some longer traditional methods too :P

Answer 21

we can generalize this i think
\(a^{2k+1} + b^{2k+1}\) is always divisible by \(a+b\)

Answer 22

@hartnn

tell me remainder

\(\dfrac{6^{159}}{(10)}\)

Answer 23

i think you need the final equation no proof!

Answer 24

"traditional" reminds me of binomial theorem, it seems bit lengthy...

Answer 25

6 is correct? math

Answer 26

yes how u got that

Answer 27

last digit
6^1 =6
6^2= 6
6^3 =6
so on,

Answer 28

thats a fun problem haha! guess 6 is the only number that stays 6 after taking any number of powers

Answer 29

1 is boring :

1^1 = 1
1^2 = 1
1^3 = 1
...

Answer 30

yes thast what i was saying if its mod 10 ,then the last digit trick is very powerful

Answer 31

2 has a cycle of
(2,4,8,6)

Answer 32

5^1
5^2..........has unit digit 5

Answer 33

Oh yes ! so 1, 5 and 6

Answer 34

9 has 9,1,9,1,91,9,1,9,1,9,1,9,1,9,1,9......

Answer 35

1 -- (1)
2 -- (2,4,8,6)
3 -- (3, 9,7, 1)
4 -- (4, 6)
5 -- (5)
6 -- (6)
7 --(7,9,3,1)
8 -- (8,4,2,6)
9 -- (9,1)

all correct ?

Answer 36

i observe, whenever 1 comes in the cycle, thats the end...obviously :P

Answer 37

u missed 0 , lol

Answer 38

0 -- (0)
:P

Answer 39

whenever 6 comes in the cycle, that is also the end...

6*2 =1 `2`
6*4 =2 `4`
6*6 = 3 `6`
6*8 =4 `8`

Answer 40

Messy one
\[4 \times4^{2014} +5\times 4^{2014}-5\times 4^{2014} +5\times5^{2014}+4\times5^{2014}-4\times 5^{2014}\]
\[9\times4^{2014}-9\times4^{2013}+9\times5^{2014}-9\times5^{2013}.\]
thus remainder is Zero

Answer 41

this is too long

Answer 42

why long you just plus and minus what makes it a multiple of 9

Answer 43

i mean the length is long to write such things in exam

Answer 44

thats kinda cute :D
i see what you're doing but the second line is not matching with the first line

Answer 45

revise it again

Answer 46

\[\color{Red}{4 \times4^{2014} +5\times 4^{2014}}-5\times 4^{2014} +\color{blue }{5\times5^{2014}+4\times5^{2014}}-4\times 5^{2014}\]

Answer 47

\[\color{Red}{9 \times4^{2014} }-5\times 4^{2014} +\color{blue }{9\times5^{2014}}-4\times 5^{2014}\]

Answer 48

how do you get a 9 out of remaining two terms ?

Answer 49

yea I made 20 =9 sorry :(

Answer 50

can i use phi :P
phi(9)=phi(3^2)=3^2-3=6

5^6=1 mod 9
5^2010=1 mod 9
5^5= 2 mod 9
5^2015=2 mod 9---1

4^6=1 mod 9
4^2010=1 mod 9
4^5=7 mod 9
4^2015=7 mod 9----2

from 1 +2
4^2015+5^2015=0 mod 9

Answer 51

Neat :)

Answer 52

but the winner is this method for keeping it short and simple :D
http://gyazo.com/566f88e1d0c62946d936d758dd62aae2

Answer 53

:P

Answer 54

uhh... sorry for nitpicking, guys, but uhmm
Catch.me's method is awesome, but his example was kind of faulty, even WITH the alternating signs.
\[4^{2014} +5^{2014}= (4+5)[4^{2013} -5 *4^{2012}+.....-5^{2013}]\]
is not true.

This factoring method for \(\large x^n + y^n \)
Only works if n is odd.

For instance, say n is 4.
\[\Large 4^4 + 5^4 = 256+625 = 881\]

881 is not even divisible by 9
which is 4+5

Answer 55

thats right! we can factor like that only when the exponent is odd :

\[4^{2015} +5^{2015}= 4^{2015} -(-5)^{2015} = (4+5)[stuff]\]

Answer 56

Good thing 2015 is odd, yeah? haha

Answer 57

speaking of 2015, happy new year, Ganeshie :D

Answer 58

Hahah xD happy new year to you and hardik :)

Answer 59

that was the whole point of this question!
to wish everyone Happy New Year 2015 ;)

Answer 60

oh i see the \(2015\)

Answer 61

did'nt notice at first the power was the upcoming year :) ,lol

Answer 62

I cooked up this problem to be moderately challenging and I'm convinced that I was successful xD Happy new year mathmath :D

Answer 63

happy new year to u @ganeshie8 and other friends :)

Answer 64

Sweet question and answers! :)
Very flavorful!