remainder problm
Find the remainder when 10^10+ 10^100+ 101^000 + . . . +10^10000000000 is divided by 7.
hii all
hii...um gettn 6 as ansr...wot u guys r gettn???
HI
i think answer is 2
my approach wz..
10^1/7 remainder 3 10^2/7 remainder 2 10^3/7 remainder 6 10^4/7 remainder 4 10^5/7 remainder 5 10^6/7 remainder 1 n ve know that wen 1 cumes from nxt division onwrds cycle repeats.. SO 10^10/7 leaves remainder of 4 and 10^100/7 leaves remainder of 5 . . . . . so sereies becum 4+5+1+3+2+6+4+5+1+3=34/7 leaves remainder of 6
hi
wud luv to know yr approach..
if you use a calculator
if you use a calculator 10^10 , 10^100 , 10^1000 and so on all leave a remainder of 4 when divided by 7. so the final remainder is the remainder of 4 * 10 by 7 = 5
now my solution involves the
now my solution involves the congruent modulo of power residues. Heres the summary, you guys may/ may not understand. 6 is the minimum power of 10 other than 0 that leaves a reaminder of 1 mod 7 . thus it can be proved that 10^4 is conguent to 10^(4 + 6t) mod 7. clearly 10, 100 , 1000 , so on can be written in the form 6t+4 . .HENCE , all of them leave the same reaminer as 10^4 mod 7 = 4 . YOU CAN CHECK USING A CALCULATOR. we have 10 such terms. so the final remainder is one of 4*10 by 7 = remainder of 40 / 7 = 5
solution
10^10+10^100+..........10^10000000000=(7+3)^10+......(7+3)^10000000000 all d numbers with 7 r multiplied by 7 so remainin sum is 3^10+3^100+..........+3^10000000000 =(7-4)^10+.........+(7-4)^10000000000 we have 2 get remainder wen 4^10+4^100+........+4^10000000000 by 7 d wen 4^1 divided by 7 remainder =4 4^2 divided by 7 rem =2 4^3 divided by 7 rem =1....it repeates......... so wen 4^10+.....+4^10000000000 is divided by 7 remainder is 4+4+4+4+4+4+4+4+4+4=40.........nd 40/7 gives remainder 5 ....answer =5
I know the following is
I know the following is correct
The reminder of ( a x b) = Reminder of a x Reminder of b
That means
The reminder of ( a x b)x = Reminder of ax x Reminder of bx
What I want to know is there any way we can simplify
The reminder of ( a )m x n
try to look up some material
try to look up some material on power residues in congruences on the net, might reach some where. in some case what you are asking can be done but i forgot the cases. sorry !
well ! let a^t = c mod m and
well ! let a^t = c mod m and let t be the minimum power of a (different from 0)such that c=1 mod m, then a^1 will have the same remainder as a^(1+t) , a^(1+2t) and so on. similarly, a^2 will have the same remainder as a^(2+t) , a^(2+2t) and so on. of course here 1,2 etc are < t . just look at my approach while solving the prblem and you will understand what i am saying. and your exact question has an answer if we look into the properties of t which is gonna be though here. thats why i said to look up at the aforementioned material. also ELEMENTARY THEORY OF ALGEBRAIC NUMBERS in google books contains what i have been talking about.its downloadable.