| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 28 Mar 2008 22:53:47 IST
|
|
|
find last 3 digits of (43)^211
sorry 4 the prev sum...tht was really stupid..
hope this is a lil tougher
and the same conditions apply
|
Nitwit Blubber Odment Tweak
|
|
|
|
28 Mar 2008 23:05:43 IST
|
|
|
last 2 r 2,7 rite or wrong?
|
B.Tech Mechanical
NIT Trichy |
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
28 Mar 2008 23:14:12 IST
|
|
|
707?
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
28 Mar 2008 23:15:21 IST
|
|
|
yes 707 :) wats ur method elasti?
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
28 Mar 2008 23:27:43 IST
|
|
|
hey elasti watsur method of doing this problem.plz reply soon......
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
Disclaimer:
I donno abc of mods.. All I've seen is just 1 or 2 sums bein solvd by mods so heres my atempt. Der must surely be an easy way out
Basically u gotta find the remainder wen 43 ^211 is divided by 1000
43^3 ends wid 507 . 43^6=49 mod 1000
now write 43^211 =43^210*43
(43^6)^35*43
so the abv thing mod 1000 will be
49^35 mod 1000 *43 mod 1000
(50-1)^35 mod 1000 *43 mod 1000
= 249 *43 mod 1000
wich is 707
|
this reply: 15 points
(with 3
in 3 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
29 Mar 2008 01:19:53 IST
|
|
|
(43)^211
(43)^4=1mod8
so (43)^208=1mod8
(43)^3=3mod 8
so (43)^211=3mod8
(43)^100=1mod125
so (43)^200=1mod125
(43)^11=mod125
(43)^211=-43mod125
so 8k+3=82mod125
so 8k=79mod125
so k=125t+88
so 8k+3=1000t+707
so last 3 dig are 707
|
Nitwit Blubber Odment Tweak
|
this reply: 10 points
(with 2
in 2 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
