bhai it said two variables dont have same value and exactly one variable has a value less than or equal to 3
so if a=0 then b or c can attain values 1 or 29 i think rectify if iam wrong
@saurav205
@scrabbler said:Not ordered pairs. Not yet... With first var as 0, (0,4, 26) to (0, 14, 16) gives 11 sets. Similarly (1, 4, 25) to (1, 14, 15) gives 11 more and so on. Now each of these are assuming a min-to-max ordering which is not specified so since all are distinct just multiply by 3! to arrange in all possible ways.regardsscrabbler
bro but this is coming up to 252 only na
@sujamait said:yeah typo..forgot to add 2 wid ^ Find all pairs (x, y) of real numbers such that16^(x^2+y) + 16^(x+y^2)= 1.
16^(x^2+y) + 16^(x+y^2)= 1
Applying AM>= GM
16^(x^2+y) + 16^(x+y^2) >= 2 sqrt ( 16^(x^2+x+y^2+y) )
1 >= 2^[1+2(x^2+x+y^2+y)]
From here I could only find x=y= -1/2 
May be other are possible!
@scrabbler said:Not ordered pairs. Not yet... With first var as 0, (0,4, 26) to (0, 14, 16) gives 11 sets. Similarly (1, 4, 25) to (1, 14, 15) gives 11 more and so on. Now each of these are assuming a min-to-max ordering which is not specified so since all are distinct just multiply by 3! to arrange in all possible ways.regardsscrabbler
bro but this is coming up to 252 only na
@sujamait said:yeah typo..forgot to add 2 wid ^ Find all pairs (x, y) of real numbers such that16^(x^2+y) + 16^(x+y^2)= 1.
x=y=(-1/2)
AM>=GM
{16^(x^2 + y)+16^(x + y^2)}/2 >= sqrt(16^(x^2 + y)+16^(x + y^2)})
1/2 >= sqrt(16^(x^2 + y)+16^(x + y^2)})
squaring...
16^(x^2 + y)+16^(x + y^2) =
or 16^(x^2 + y)+16^(x + y^2) =
=> (x^2 + y)+(x + y^2) =
=>(x^2 + y)+(x + y^2) + 1/2 =
=> (x+1/2)^2 + (y+1/2)^2 =
this has to be (x+1/2)^2 + (y+1/2)^2 = 0
which is only possible if x=y=(-1/2)
@scrabbler said:Yeh kya tha?I repeat: exactly one variable has a value equal or less than three. To phir 3 + 25 kahaan se bhai? And uparwaale mein aur bhi extras hai...regardsscrabbler
hahahahahah.....bhai hassi aa gai srry... (apne upar aayi)... 
@raopradeep said:bhai post the solutioni am doing some mistake not getting answer
bhai i did it by using options
put n = 3
2*n + 1 = 2*3 + 1 = 7
so..7c1 + 7c2 + 7c3 = 63
@raopradeep said:@Logrhythm bhai how to find number of solutions from eqtn plz tell
bhai...counting is the best method i guess....but i counted wrong...
and finally count karne pe...22+22+20+20 = 84...
now wo 3c1 wala logic se 84*3 = 252 solutions aa rahe hai...
@scrabbler abb koi galti ??
@Logrhythm said:bhai...counting is the best method i guess....but i counted wrong...and finally count karne pe...22+22+20+20 = 84...now wo 3c1 wala logic se 84*3 = 252 solutions aa rahe hai...@scrabbler abb koi galti ??
Bas itna, ki OA kuch aur hai 😞 . I am convinced of 252...
regards
scrabbler
regards
scrabbler
@raopradeep said:@saurav205bhai it said two variables dont have same value and exactly one variable has a value less than or equal to 3 so if a=0 then b or c can attain values 1 or 29 i think rectify if iam wrong
it says exactly one variable has value less than or equal to 3
so you took a = 0
and now you say that b or c can attain values from 1 to 29..
so suppose if b = 2
so a = 0 and b =2..
so 2 values have value less than 3....but what does the question state ...exactly one variable has a value less than or equal to 3..
I hope you get the point...
@Logrhythm said:bhai...counting is the best method i guess....but i counted wrong...and finally count karne pe...22+22+20+20 = 84...now wo 3c1 wala logic se 84*3 = 252 solutions aa rahe hai...@scrabbler abb koi galti ??
I agree with 252 ...
I had done 84*3!..but got the mistake....
@Logrhythm said:bhai...counting is the best method i guess....but i counted wrong...and finally count karne pe...22+22+20+20 = 84...now wo 3c1 wala logic se 84*3 = 252 solutions aa rahe hai...@scrabbler abb koi galti ??
I agree with 252 ...
I had done 84*3!..but got the mistake....
@Logrhythm said:bhai...counting is the best method i guess....but i counted wrong...and finally count karne pe...22+22+20+20 = 84...now wo 3c1 wala logic se 84*3 = 252 solutions aa rahe hai...@scrabbler abb koi galti ??
bhai i am getting different values rectify where i am wrong
y+z=30
(1,29)
(2,28)
.
.
.
.
(14,16) so 14*2 =28 solutions check where iam going wrong 
@raopradeep said:bhai i am getting different values rectify where i am wrong y+z=30(1,29)(2,28)....(14,16) so 14*2 =28 solutions check where iam going wrong
Explained in 3 posts above by 2 different people...ab aur kya explain kare...
regards
scrabbler
@raopradeep said:bhai i am getting different values rectify where i am wrong y+z=30(1,29)(2,28)....(14,16) so 14*2 =28 solutions check where iam going wrong
let me explain 1 case...
x=0 wala..
here y+z = 30
subject to constrain -> y,z>3
y+z -> (4+26),(5+25),...,(26+4)
so total = (26-4)+1 = 23...but minus where both are equal, hence 22 solutions...
abb batao kuch samajh nahi aya toh??