Question

(20202)^4

find the sum of its last 4 digits?

Answer 1

http://www.wolframalpha.com/input/?i=20202%5E4

your sum is 6+4+1+6=17

Answer 2

how this come?

Answer 3

(2^n)+1 your n=4

Answer 4

it occurs to me that (20000+202)^4 expanded into the binomial terms might give an answer

Answer 5

mr hero go to http://www.wolframalpha.com/input/?i=20202%5E4 & type (20202)^5 & then see what answer it give

Answer 6

I WANT TRICK FOR SOLVE THIS

Answer 7

OK TRY THIS LATER

Answer 8

there might be an easy trick, but doing as I say gives an answer (20000+202)^4 expanded to the binomial terms

Answer 9

terms are 20000⁴+4*20000³(202)... the 20000 are always irrelevant

Answer 10

THIS PROCESS TAKING SO MUCH TIME DEAR

Answer 11

BUT IF U HAVE ANY EXPLANATION ABOUT THIS THEN GIVE

Answer 12

2⁴*10101⁴ ?

Answer 13

THEN

Answer 14

im sure ther was a trick to potentiating those things, just cant remember it,

Answer 15

again, since the 10000 is irrelevant you only need to calculate 101⁴

Answer 16

OK WHEN U FIND THEN PLZ REPLY TO ME

Answer 17

okay, first do 2⁴*10101⁴, then, calculate 101^4, by doing (100+1)^4, multiply last 4 digits, sum, you are done

Answer 18

U R RIGHT DEAR BUT U MISSING ONE THING THE MULTIPLY RESULT OF (100+1)^4 IS AGAIN MULTIPLY BY 16

Answer 19

BUT THIS QUESTION ALSO DO BY DIRECT TRICK I WANT THAT

Answer 20

the only trick that occurs to me right now is to use (a+b)²=a²+2ab+b² and only calculate relevant terms

Answer 21

HOW CAN BE THIS APPLY?

Answer 22

we start with ((200+2)²)²= (200²+800+4))², 200² is irrelevant, so we calculate (800+4)², 800² is irrelevant, 2*4*800 and 4² are

Answer 23

20202=2x3x7x13x37

Answer 24

WHAT IS THIS?

Answer 25

prime factorization may hold the trick if one exists

Answer 26

how we calculate?

Answer 27

((200+2)²)²= (200²+800+4))², =(200²+804)²=200⁴+2*804+(800+4)², last term expands to 800²+2*4*800+16, 200⁴ is not important, 800² is also not important since they both end up in 4 zeroes, so 1608+64000+16

Answer 28

might have miscalculated some of the terms, in the end, one gets 6400+16 to be the relevant ones