@ScareCrow28 said:Nothing Saw a good question after a long time on QT
mazaak udaing?! 
PS sorry for spamming
mazaak udaing?! PS sorry for spamming
@scrabbler said:Square. Sec^2 - tan^2 = 1regardsscrabbler
yeah typo..forgot to add 2 wid ^
Find all pairs (x, y) of real numbers such that
16^(x^2+y) + 16^(x+y^2)= 1.
16^(x^2+y) + 16^(x+y^2)= 1.
@saurav205 said:@scrabbler Bhai eska approach please...i am getting 504 or something..cause I multiplied 84*3! (3! cause number of solutions mein a,b and c can be interchanged..approach galat hai kya??
No idea. My answer was 42 (not 84...don't arrange here as you are anyway multiplying bny 3! later) sets so 42*6 = 252 arrangements.
@raopradeep Is there also an explanation? I don't see where the extra 42 = 7*6 are coming from.regards
scrabbler
@scrabbler said:No idea. My answer was 42 (not 84...don't arrange here as you are anyway multiplying bny 3! later) sets so 42*6 = 252 arrangements.@raopradeep Is there also an explanation? I don't see where the extra 42 = 7*6 are coming from.regardsscrabbler
nahi bhai i dont have any solutions anyway how you approached?
@raopradeep said:nahi bhai i dont have any solutions anyway how you approached?
Cases, took the smallest as 0, 1, 2, 3...tried to find combos of the other 2 s.t they are diff, both more than 3, and add up to 30...got 11 + 11 + 10 + 10 = 42...now into 3!.
regards
scrabbler
regards
scrabbler
for a scholarship , at the most n candidates out of 2n+1 candidates can be selected . if the number of different ways of selection of atleast 1 candidate is 63 then find the maximum number of candidates that can be selected
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@raopradeep said:three variables x,y,z have a sum of 30. all three of them are non-negative integers . if any two integers dont have same value and exactly one variable has a value equal or less than three .find number of possible solutiins9828568294
i am getting 96...
x+y+z = 30
let's take x=0
y+z=30 has 31 solutions but minus when either y=0 or z=0 and when y=z=15...
hence 28 solutions..
x=1
y+z=29 -> 26 solutions
x=2
y+z=28 -> 22 solutions
x=3
y+z=27 -> 20 solutions
total = 28+26+22+20 = 96...
-------------------------
is it that we need to select the variable whose value is less than or equal to three in 3c1 ways??
then 96*3 = 288 solutions fir bhi.... 294 nahi aa raha mera....check karo thoda pls kahan galti hai.. :(
@Logrhythm said:@saurav205@scrabbleri am getting 96... x+y+z = 30let's take x=0y+z=30 has 31 solutions but minus when either y=0 or z=0 and when y=z=15...hence 28 solutions..
exactly one variable has a value equal or less than three...so remove cases when any other is 1, 2, 3...(same for later cases too)
regards
scrabbler
@scrabbler said:Cases, took the smallest as 0, 1, 2, 3...tried to find combos of the other 2 s.t they are diff, both more than 3, and add up to 30...got 11 + 11 + 10 + 10 = 42...now into 3!.regardsscrabbler
bhai how come 11 ,11,10 ,10??
for eg if a = 0
then b+c =30
now b,c both will be >= 4
so varies from 4 to 26..
so i guess 21 ways aayega..
similarly for others...
kuch galat???
@Logrhythm said:@saurav205@scrabbleri am getting 96... x+y+z = 30let's take x=0y+z=30 has 31 solutions but minus when either y=0 or z=0 and when y=z=15...hence 28 solutions..x=1y+z=29 -> 26 solutionsx=2y+z=28 -> 22 solutionsx=3y+z=27 -> 20 solutionstotal = 28+26+22+20 = 96...-------------------------is it that we need to select the variable whose value is less than or equal to three in 3c1 ways??then 96*3 = 288 solutions fir bhi.... 294 nahi aa raha mera....check karo thoda pls kahan galti hai..
bhai you are very close to answer just check one more time might be you will get it
i have a doubt how do you get number of solutions from eqtns like y+z=29 kindly tell the approach
@scrabbler said:exactly one variable has a value equal or less than three...so remove cases when any other is 1, 2, 3...(same for later cases too)regardsscrabbler
i have removed those na...tabhi toh utne aa rahe hai..
x=2 wale ka example lo...
y+z=28
3+25
4+24
5+23
.
.
.
25+3
so total solutions -> (25-3)+1 = 23 solutions...
but isme ek case hai where y=z=14....so hata do toh 22 cases...
that is what i have done...
@saurav205 said:bhai how come 11 ,11,10 ,10??for eg if a = 0then b+c =30now b,c both will be >= 4so varies from 4 to 26..so i guess 21 ways aayega..similarly for others...kuch galat???
i think baki do ke liye >=4 nahi bola hai kahi
@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.
(-1/2 , -1/2) ??
@raopradeep said:bhai you are very close to answer just check one more time might be you will get iti have a doubt how do you get number of solutions from eqtns like y+z=29 kindly tell the approach
bhai i have even counted manually...mere toh itne hi aa rahe hai...shayad i am missing out on something...pata nahi chal raha... :(
@saurav205 said:bhai how come 11 ,11,10 ,10??for eg if a = 0then b+c =30now b,c both will be >= 4so varies from 4 to 26..so i guess 21 ways aayega..similarly for others...kuch galat???
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.
regards
scrabbler
regards
scrabbler
@raopradeep said:i think baki do ke liye >=4 nahi bola hai kahi
bhai if only one number is greater than or equal to 3 ..toh baaki dono ko toh hona padega na 4 ya usse bada??
@raopradeep said:for a scholarship , at the most n candidates out of 2n+1 candidates can be selected . if the number of different ways of selection of atleast 1 candidate is 63 then find the maximum number of candidates that can be selected3245
@raopradeep said:for a scholarship , at the most n candidates out of 2n+1 candidates can be selected . if the number of different ways of selection of atleast 1 candidate is 63 then find the maximum number of candidates that can be selected3245
3 ?
@Logrhythm said:i have removed those na...tabhi toh utne aa rahe hai..x=2 wale ka example lo...y+z=283+254+245+23...25+3so total solutions -> (25-3)+1 = 23 solutions...but isme ek case hai where y=z=14....so hata do toh 22 cases...that is what i have done...
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...
regards
scrabbler
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...
regards
scrabbler