Well This congruency is a good thing to solve. Well it is very easy not a difficult one but still I m posting it !
p^2 is congruent to -11 modulo 36. Well you have to find all integral solution for p and you have to prove that no more solutions exist.
is p=5?
ReplyDeleteor p=-5
ReplyDeleteconfirm the answer, i'll give you my solution
didnt get the question at all
ReplyDeletecan smone please explain
well iske infinite solutions hai....
ReplyDeletep=36k+-5........
There are no infinite solutions.. agar hote toh questions hi naa post maarta..
ReplyDeleteThere are only 4 distinct real positive integer solution yaar tabhi toh..
@ sambhav bhaiya.. yaar 5 aur 13 toh mera bhi aaya tha.. wo next two nahi aaye.. and you have to also prove that only 4 exists.. not more than that.. :D
abe koi question toh samjha do-aakash
ReplyDeletedividend=(divisor)(quotient) + remainder
ReplyDeleteD=dq+r
D-dq=r
according to the definition, r is the smallest positive integer that remains when anything is divided
so, for that, dq shud be the greatest integer less than D
hence, dq=-36
so, r=25
so p^2=+5, -5
i'd be happy to know how 13 is a solution
ok, the questioner doesn't want to stick to the definition of remainder
ReplyDeleteso, simply
D-dq = r, put r=p^2
(-11) - (36)(q)=p^2
now, we have to find integral values of q so that p is an integer
i couldn't do this...:-(
p^2 = 36k+25,
ReplyDeleteSambhav, it does satisfy, 169=180-11
so we do know it is a solution
p^2-25=36k
(p-5)(p+5)=36k
RHS has at least 36 as a factor. 36=2*2*3*3
so 2 obvious solutions sambhav suggested, put k=0 and dance :D
about rest of them
let p-5 = q,
q(q+10)=36k
a number and ten added to it, multiply to give multiple of 36
so certainly assuming q=36x or q=36x-10, i.e 36,72 etc would be a nice solution i.e as kunal suggested, that would not be a complete solution
so q(q+10)=36k
RHS is multiple of 36,
now from q and q+10, either both will be even, or both odd, if both odd, then we will say get lost, we cant generate 36k, so both even
q=2z ,
so i get z(z+5)=9k
now 1 will be odd 1 will be even, but we want multiple of 9, so we make 2 cases,
1. either z or z+5 is multiple of 9,
2 both are multiples of 3 (even god knows this is not possible-dont think u need a 1 line proof for the same)
so z or z+5 is multiple of 9
so z=9k
or z=9k-5
q= 18 k or 18k-10
p =18k+5 or 18k-5
so here we do get infinite solutions
(hope i got the question right)
Bhaiya You Got the question right yaar.. har baar aisa nahi hota..
ReplyDeleteBut i checked on wolframalpha.. It is showing 4 solutions.. Ab kya karu ?
waise infinite solutions is no doubt a possibility yaar.. bt kya karu aap hi batao jab 4 hain toh hain.. :D
Solutions are 5 ,23 , 13, 31 no more.. :(( :(( :((
ReplyDeleteGotcha!
ReplyDeletehttp://www.tutorvista.com/math/linear-congruence
read solving a linear congruence part
u meant modulo 36, so we had to find solutions in region (0,35) (after that it had to be periodic), so the examiner was interested in this! thats why the problem arose
But still..
ReplyDeletethe solution is an infinite set of numbers
p=18k+-5
isn't it??
@devansh..
ReplyDeletecheck that 41,49,59 are also solutions..
it really is an infinite set..
Yea Yaar.. Sahi mein infinite hi honge.. :D
ReplyDeleteLag hi raha tha..
yeah solutions infinite hi honge obviously!
ReplyDeleteyaa.. missed some solutions..actually the thing was that getting infinite solutions i dint work on it much...mah mistake...
ReplyDeleteand yeah open this...
http://www.pocmania.com/friend.php?user=kunalsinghal1994