congruences A-Z

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

48 Answers

24
eureka123 ·

3)
105000=(102)2500

102=100≡-1mod101
=> (100)2500≡1mod101
=>remainder=1

24
eureka123 ·

1)
22009=22007.22=(23)669.22

23≡1mod7
(23)669=22007≡1mod7

and 22=4≡-3mod7

=>22009≡-3mod7
=>22009≡4mod7

24
eureka123 ·

32009=32008.3=(32)1004.3

32≡-1mod5
=> (32)1004≡1mod5

and 3≡-2mod5

=>32009≡-2mod5
=>32009≡3mod5

24
eureka123 ·

49972

49=72≡-1mod25
=> 49972≡1mod25

1
$ourav @@@ -- WILL Never give ·

thanks bargav..........more plz.....

3
msp ·

eure in two of ur ans u will ended with a negative remainder,i think u didnt complete the soln.

24
eureka123 ·

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]

3
msp ·

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)

3
msp ·

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.

24
eureka123 ·

so it doesnt end there dude... becoz remaindere cant be greater than divisor,,

22≡25mod7
25≡4mod7

=>22≡4mod7

from property no. 4[1]

24
eureka123 ·

@msp..specify Q no. which is wrong.....[16]

3
msp ·

q1) dear.[60]

1
Honey Arora ·

ans to #7
17256=289128
289=711 mod 1000
so 17256=711 mod 1000
Remainder=711

3
msp ·

q2)32009=3.32008

32008=81502=

81=1mod(5)

81502=1 mod(5)

3=-2mod(5)

so 32009=3mod(5)

3
msp ·

q3)100=-1mod(101)

(100)2500=1mod(101)

so remainder is 1.

3
msp ·

[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.

1
Rocky Crazy ·

Great!!

24
eureka123 ·

Sir give more questions...

Your Answer

Close [X]