3)
105000=(102)2500
102=100≡-1mod101
=> (100)2500≡1mod101
=>remainder=1
this thread is mainly intended to give some insight into congruences and how to solve various jee related problems using that concept..
definition of congruent:
A number `a` is said to be congruent to `b` modulo `m` if `m` divides (a-b)
its written as a ≡ b mod m
so we can say that when we write a≡b mod m ,, `b` is the remainder that is obtained when` a` is divided by `m`.
now let us take a few examples:
9≡1mod2
13≡1mod4
but now u get this doubt, when 13 is divided by 4 , remainder is why only 1 ? why cant we say the remainder is -3 ????
infact this doubt is correct. we can even write
13≡-3mod4
so we can write in various ways as we wish to..
SOME PROPERTIES OF CONGRUENCES:
1) we have say
a≡ bmod p
c≡ k mod p
then we can write,
(a+c)≡ (b+k) mod p
but here u need to observe that the number with which we are dividing a and c is the same number `p`. so that has to be kept in mind..
similarly we can aslo write
(a-c)≡ (b-k) mod p
and ac≡ bk mod p
here also we see that the number with which we are dividing a and c by the same number `p`. so alwyas rememebr that .
now its obvious that division cannot be defined here ( as cogruence itself is a way of decribing the process of division)
so the foloowing properties are trivial to observe:
1) a≡ bmod 0 implies a=b
2) a≡ a mod m
3) a≡ b mod m implies b≡ a mod m
4) a≡ b mod m and b≡ c mod m implies a≡ c mod m
5) a≡ b mod m implies ka≡ kb mod m
6)a≡ b mod m implies an≡ bn mod m
i think these are enough keeping jee syllabus in mind. now i want to know if u guys have any doubts whatsoever i have posted till now here?
after that we can start with questions
3)
105000=(102)2500
102=100≡-1mod101
=> (100)2500≡1mod101
=>remainder=1
1)
22009=22007.22=(23)669.22
23≡1mod7
(23)669=22007≡1mod7
and 22=4≡-3mod7
=>22009≡-3mod7
=>22009≡4mod7
32009=32008.3=(32)1004.3
32≡-1mod5
=> (32)1004≡1mod5
and 3≡-2mod5
=>32009≡-2mod5
=>32009≡3mod5
eure in two of ur ans u will ended with a negative remainder,i think u didnt complete the soln.
Wow.....atleast someone did that....I wanted to do the smae..but didnt ahve guts to start an olympiad topic here....
thanx b555[1][1]
1)22009=2222007
22007≡1mod(7)
2≡-5mod(7)
22=25mod(7) (i am rong in this step,i dunno y i got such a weird remiander ,can ne1 explain me this step.)
if this true the
final ans is 22009≡25mod(7)
dude wat i told is dat u din finish up ur soln,
since u have got a≡-bmod(m) which doesnt means dat b is a remainder.And y u have edited.see ur edited one seems rong.
so it doesnt end there dude... becoz remaindere cant be greater than divisor,,
22≡25mod7
25≡4mod7
=>22≡4mod7
from property no. 4[1]
ans to #7
17256=289128
289=711 mod 1000
so 17256=711 mod 1000
Remainder=711
q2)32009=3.32008
32008=81502=
81=1mod(5)
81502=1 mod(5)
3=-2mod(5)
so 32009=3mod(5)
[50][51][52][53][54][56][55][57][58][61]
thanq sir and bargave,atlast i have learnt something in this chapter,which i have never seen.