Question

Find the remainder when 5^{2009}+13^{2009} is divided by 18.

Answer 1

@amistre64

Answer 2

@phi and @kelliegirl33

Answer 3

I would look for a pattern of remainders when you divide 18 into 5, 5^1, 5^2, ...
eventually it will start repeating. then reduce 5^2009 using modulo

Answer 4

i don't understand

Answer 5

can you list the remainders you get when you divide 18 into
5^0, 5^1, 5^2, 5^3, 5^4, 5^5, 5^6, 5^7 ?

Answer 6

can you give me shortest method to solve it? it seems to be longer

Answer 7

if you know a faster way, what is it ?

Answer 8

i don't know

Answer 9

mod 18 would have been my suggestions as well.

Answer 10

5^0, 5^1, 5^2, 5^3, 5^4, 5^5, 5^6, 5^7 ?
rem: 1 5 7 17 13 11 1 5

it repeats modulo 6
for example 5^7/18 is the same as 5^(7 mod 6)/18 = 5/18 = 5

that means 5^2009 / 18 will have the same remainder as
5^n /18 where n = 2009 mod 6 = 5

looking at the table, we see 5^5/18 has a remainder of 11

now do the same thing for 13
add the two numbers, and find the remainder after dividing by 18

Answer 11

another quick way is to notice that, 'leaving 13 as remainder' is same as 'leaving -5 as remainder -5',
so both terms wud simply cancel out giving u 0 :
5^{2009}+13^{2009}
5^{2009} - 5^{2009}

Answer 12

ooh, that is quicker. @ganeshie8

Answer 13

alternative :
a^n + b^n = (a+b)(bla bla bla bla)
for a,b are odd numbers

so,
5^{2009}+13^{2009 = (5+13)(.........) = 18(.......)

obvious, if it divided by 18, then its remainder is zero

Answer 14

yes, that is a good way too. @RadEn

Answer 15

@ganeshie8 can you explain your answer please ?

Answer 16

@RadEn if a,b are even no.then

Answer 17

@phi i don't understand your answer

Answer 18

@amistre64 , @UnkleRhaukus @mukushla

Answer 19

and i don't understand "mod" also ?What is the significance of this ?

Answer 20

yeah, it just for a,b are odd's

Answer 21

if a,b are even, is there any shortcut method?

Answer 22

i never see for a,b are even

Answer 23

ohkk

Answer 24

can you tell me what is "mod"?

Answer 25

\(\large 5^{2009}+13^{2009}\)

first observation : if \(a\) leaves a remainder 'r' when divided by some number, then \(a^n\) leaves a remainder of \(r^n\)

next : since \(13\) leaves a remainder of \(-5\) when divided by 18, \(13^{2009}\) leaves a remainder of \((-5)^{2009}\)

plugging them cancels out the left hand side completely giving u 0

Answer 26

mod is just a notation referring to remainders :

\(a \equiv b \mod n\)

means

\(a-b\) is divisible by \(n\)

Answer 27

-5? how?

Answer 28

not +5

Answer 29

13/18 = (18*1 - 5) / 18 = 1 R (-5) = 0 R (13)

Answer 30

leaving 13 remainder is same as saying leaving 13-18 = -5 remainder

Answer 31

okk

Answer 32

basically, we can have both negative and positive remainders

Answer 33

how?

Answer 34

2/3 = 0 R (2) = 1 R (-1)

Answer 35

when u divide 3 in 2, u can ask two questions :
1) how much im overflowing
2) how much im lacking

Answer 36

answer to first question gives u positive remainder
answer to second question gives u negative remainder

Answer 37

both are correct in remainder arithmetic

Answer 38

so then, what you want to say ?

Answer 39

wat do u mean ?

Answer 40

i mean that the remainder is not +ve in this case,right?

Answer 41

u get to choose, if u wanto use positive / negative remainder


when u talk about dividing wid 18 :
13 = -5 = 31 = 13 + 18k

Answer 42

more precicely :
\(13 \equiv -5 \equiv 31 \mod 18\)

Answer 43

forget it, if that doesnt make sense... just see this :-

2/3 = 0 R (2) = 1 R (-1)