11AAre nehi, nishant bhaiya, mera wow matlaab nehi thaa.....If u see my posts, u shall know that even i'm interested more in olympiad stuff than rigorouus JEE routine.....[17]
Keep rocking guys, never mind my coment.......
well this is one f my most faltu googly.........i think nishant bhaiya will beat me up for this....
p is a divisor of ab.......bt p is not a factor of a......prove p divides b...............given p is prime.,....
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30 Answers
49trapped in my own googly[1][1][1][1][2][2][3][3][4][4]...........bt these r useful for olympiads right??????????
1@bhaiyya......
p appears atleast once in the prime factorisation of ab..
so it must appear in the prime factorization of atleast one of a , b.
thats why i think this is obvious.....
or are you looking for a proof using congruencies..?
11
62yes it is prime.
But it is not very obvious.. (Sometimes it may so seem that it is abvious.. because we have been trained from class 6..w)
if p divides ab then p divides a or p divides b
but we are nver told the proof.
1didnt get the question properly...
4 divides 6*2
but 4 divides neither of 6 or 2.
?? or is it mentioned that p is prime ?
62does p is not a factor of a not mean a≡m modp ???
yes it does as long as m is not a multiple of p
Eureka what you have written in 4 is not that obvious..
Try to judge your arguement in 4 had p not been a prime number.
Could you say that easily that if
8X15≡0 (mod 6) implies
8≡0 (mod 6) OR 15≡0 (mod 6) ???
62@Soumya :
so either mn=0 or mn is completely divisible by p.
if m or n is divisible by p then a=0(mod p)or b=0(modp).
if mn=0 then it can be proved also.
I did not \understand what is your arguement where you proved that either m is divisible by p or n is divisible by p.
if mn=0 then it can be proved also.??? How?
1sir another proof
a = m(mod p)
b = n(modp)
so ab = mn(mod p)
so either mn=0 or mn is completely divisible by p.
if m or n is divisible by p then a=0(mod p)or b=0(modp).
if mn=0 then it can be proved also.
1Sir I'm not saying that ab = 0(mod p) implies that a=0(modp) or b=0(modp) , but b = 0(mod p) implies that ab = 0(mod p).
24does p is not a factor of a not mean a≡m modp ???
Can u tell me where i really went wrong in my working ???
i didnt understand ur post#6
62@eureka..
basically you have to prove that if ab≡0 (mod p)
then either a≡0(mod p) or b≡0 (mod p)
Like if p was not a prime then this would obviously not be true.. then why is it so with p being a prime number?
62@Soumya:
or ab ≡ 0 (mod p)
which is true as ab is completely divisible by p .
now the remaining is not very obvious..
as to why this implies a≡0 or b≡0
@Soumik: Then we should remove all the olympiad features from this site!! And make it another site where we do everythign strictly from IIT JEE perspective. And dont you think that congruences is also a deviation from jee syllabus?
62no no this is a very very important fundamental result which is a must know :)
11What do u mean proving it, nishant bhai?
Any linear eqn in integers x and y is solvable if it satisfies gcd(a,b)|k, where ax+by=k.
In the above case gcd(a,b)=1.......
Well, not to hinder these healthy discussions.....don't u guys think we are deviating from JEE track?
24that means i didnt understand ques at all....
can u explain the ques sir ????
1Sir can we do it like this
a = m (mod p)
we assume that b = 0(mod p)
so ab = m x 0 (mod p)
or ab = 0 (mod p)
which is true as ab is completely divisible by p .
so our initial assumption that b = 0 (mod p) is correct . (don't think it is this easy :) )
21continuing from post 7
multiply both sides of eq 1=ax+py by b
b=abx+pby
now p divides abx and p divides pby
hence p divides b
62no subhomoy..
This is not correct...
, integer=integer*(p/b)
Does it mean that p/b = integer?? :O
"Seems that the faaltu question is not so faaltu :D"
49so lets do like this....
ab=kp
so, a=kp/b
so, integer=integer*(p/b)
so p/b =integer....
so, p divides b..
62Now if you are wondering where to start from try this:
Gcd(a,b) is the smallest positive integer of the form as+bt
Hence 1=ax+py for some integers x and y
62But we already know ab≡0 modp
So either m=0 or n=0...
m cant be zero(ATQ)
So obvoiusly n=0
So contradicition...
why is this so?
i mean this is what you have to prove..
if mn≡0 modp then either m≡0 modp or n ≡0 modp
!!! (The proof using congruences will need to to prove that Zp* is a group! which i dont think you can directly at this level!
24Using this all congruency thing fro first time in a geenral ques...so forgive me if there are mistakes
ab≡0 modp
a≡m modp
Lets assume that p is not a factor of b
b≡n modp
ab≡mn modp
But we already know ab≡0 modp
So either m=0 or n=0...
m cant be zero(ATQ)
So obvoiusly n=0
So contradicition...
=> b≡0 modp