Question

(2222^5555+5555^2222) what is the remainder when divided by 7 ?

Answer 1

Did you learn module before?

Answer 2

no ,what is that ?
can u guide me?

Answer 3

@hartnn PLZ SEE THIS

Answer 4

2222 = 2*1111 = 2* 11 * 101
5555= 5*11*101
101 can be factored further ?

Answer 5

what will be the answer plz show the further steps ?

Answer 6

It is more easy to do with module. I can't see any simple solution to this question.

Answer 7

there is a simple solution actually...

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = last digit 6
2^5 last digit 2

so you see... 2 repeats the loop for every exponent of 4

you have 2222^5555 no matter how big the answer is...since the last digit is 2 and the exponent is divisible by 5... the last digit of 2222^5555 will be 4

then...
5^1 = 5
5^2 = 5
5^3 = 5

as you can see...all powers of 5 have 5 as the last digit..

so...5555^2222..no matter how big this is...the last digit is just 5


do these make sense?

Answer 8

Form the pattern as Lg told...

Answer 9

well..

2222 = 3 mod 7
=>2222^5555 = 3^5555 mod7
and 5555 = 4 mod 7
=>5555^2222 = 4^2222 mod7

now,, 3^3 = -1 mod7

(3^3)^(1851) = (-1)^(1951) mod 7
3^5553 = (-1) mod 7
=>3^(5555) = -9 mod(7) = 2 mod(7)

same way,,we can see,,

4^3 = 1 mod 7
(4^3)^(740) = 1^740 mod7
4^2220 = 1 mod7
=>4^2222 = 16mod 7 = 2 mod 7

so final remainder will be 2+2 = 4..

Answer 10

\[\frac{ 10+5 }{ 7 } = \frac{ 10 }{ 7 } + \frac{ 5 }{ 7 }\]

Combine this concept and @lgbasallote ' s method...)

Answer 11

@igba
last digit of 2222^5555 should be 8 and not 4

anyways,,what use will you make of that last digit thing anyways ?

Answer 12

am not arguing,just curious..

Answer 13

i mean the pattern
(remainder)

2^1 /7 = 2
2^2 /7 = 4

2^3 /7 = 1

2^4 /7 = 2

etc....this pattern continues.......adn Find Such pattern for 5^1/7 ......

Answer 14

let's let wolfram judge @shubhamsrg

Answer 15

http://www.wolframalpha.com/input/?i=+2222%5E5555

seems 4 is the last digit

Answer 16

I do not think premkishan could understand what you guys said.

Answer 17

i think theres' a ... there after "4" + a option of "more digits"

anyways, i dont need wolfram to crosscheck..its easy to see.. 2^(4n+1) has last digit 2
2^(4n+2) has last digit 4
and 4n+3 -->8

5555 = 4n+3

Answer 18

@NewbieCarrot how sure are you he cares?

Answer 19

atleast i didnt tell anything wrong ! o.O

Answer 20

he has to learn modulus for such kinda questions,,as simple as that..

Answer 21

I think there are some reason why he have this question without learning module.

Answer 22

Or he was not telling the truth.

Answer 23

if you say you don't need wolfram to solve these....then can you tell me why you did not get the same result for the remainder as wolfram did? http://www.wolframalpha.com/input/?i=%282222%5E5555%2B5555%5E2222%29+mod+7

Answer 24

ohh..lol..leme recheck..

Answer 25

-9 mod(7) = 5 mod(7) not 3 (mod 7)

Answer 26

yes,,right !! silly of me !! hmm..

Answer 27

so with that correction,,ans indeed becomes 0 ..
long live moduli ! :P