24Q2
23n+1=23n.2
23n≡1mod7
2≡2mod7
=>23n+1=2mod7
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
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48 Answers
24@iota..
http://get-set-go-jee-math.blogspot.com/2009/07/chinese-remainder-theorem.html
11There was a particular reason i gave the qsn 17256....It was not simple mod-bashing honey.....Point is it requires a nice theorem called "Carmichael's Theorem" (If i remember properly)....I remember prophet sir remarkably solved this using that theorem in goiit once....It's really quite useful!
39q5) What is the least positive number that you can divide by 7 and get a remainder 4, divide by 8 and get a remainder 5, and divide by 9 and get a remainder 6.
49many doubts.......i am seeing this link for first time.......what if....
a≡ b mod(m)
so, an≡ bnmod(m)
but what if bn≥m??????????
help..........help............
62Prove that
Question 1) http://targetiit.com/iit-jee-forum/posts/proof-of-fermat-s-little-theorem-using-congruences-11744.html
Question 2) 23n+1 ≡ 2 (mod 7)
11Deepak, number 5 can also be done by the general soln of the linear Diophantine Eqns, well it's out of JEE range!
21okie soumik[1].....well is der any method for q5 which is well within jee syllab......i guess der is none
39eragon ; is it so for sure?
anwyays this congruencies itself no t in jee syllabus
21@ b555 i din get wat u said......
btw crt is one of the way of doing it...der may be many ....i dont know of them...and i meant to say tat can tat q be solved without using congruecies...
1q5 just a try
let the smallest number be x
let x=9m+6
9m+6≡ 5 mod 8
let 9m≡k mod 8
6≡ -2mod 8
so k-2=5 k=7
so 9m≡7 mod 8
or 9m≡63 mod 8
0r m≡ 7 mod 8
let m =8r+7
also 9m+6≡4 mod 7
let 9m≡v mod 7
6≡ -1 mod 7
so v-1=4 v=5
so 9m≡ 5 mod 7
or 9m≡ 54 mod 7
m≡6mod 7
8r+7≡ 6 mod 7
so 8r≡6 mod 7
or r≡ 6 mod 7
minimum value of r is 6
so x= 501
39u havent givena try ?
atleast post here till where r u getting struck
may be we can help u further
24sorry sanky..i am also doing this topic first tiem..so not much idea as to where soln ends...I ahve corrected now
19yes bhargav,,,,you have done a great help to us ! thanks for this !
62Using congruences and not binomial theorem..
Try and find the remainder when
1) 22009 is divided by 7
2) 32009 is divided by 5
3) 105000 divided by 101
4) 49972 divided by 25
111 more Nishant sir,
Use congruencies to find rem when 17256 is divided by 1000....