@amresh_maverick look at first few terms of the sequence Amresh :)
You will get your mistake..
@amresh_maverick said:[] ----greatest Integer fun[5/2] + [5/2+1/100] +[5/2+2/100] ................[5/2+199/100] =???
OA:600
1> if inverse of f(x)= ax+b is f^-1 (x) = bx+a , find the vale of a and b
2> f(x+y) =f(x) +f(y) +f(x)*f(y) , if f(1) =3 , find f(3) ??
1> if inverse of f(x)= ax+b is f^-1 (x) = bx+a , find the vale of a and b
2> f(x+y) =f(x) +f(y) +f(x)*f(y) , if f(1) =3 , find f(3) ??
@iLoveTorres said:Q: 1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23 find the next termoptions a)24b)25c)26d)27
26? no prime factor is squared...
regards
scrabbler
@scrabbler said:26? no prime factor is squared...regardsscrabbler
bhai aap toh guru ho.. was waiting for you to see this question.. to be more precise it is called SQUARE FREE number series..
@bodhi_vriksha said:@amresh_maverick look at first few terms of the sequence Amresh You will get your mistake..
from [1^2/2013] ................... [44^2/2013] will give zero
from [45^2/2013] ................... [63^2/2013] will give 1
from [64^2/2013] ..............................will give 2
similarly
[2013^2/2013]. will give 2013
batti nahi jala , help
from [45^2/2013] ................... [63^2/2013] will give 1
from [64^2/2013] ..............................will give 2
similarly
[2013^2/2013]. will give 2013
batti nahi jala , help
@iLoveTorres said:bhai aap toh guru ho.. was waiting for you to see this question.. to be more precise it is called SQUARE FREE number series..
Lol...aaj main jyada nahin dekh paunga...am staying the night at a friend's place so can't hog his lappie 😉 Only on for 15 min...
regards
scrabbler
regards
scrabbler
@scrabbler said:Lol...aaj main jyada nahin dekh paunga...am staying the night at a friend's place so can't hog his lappie Only on for 15 min...regardsscrabbler
aapke 15 min humare 15 din ki padhai ke barabar hai.. :P
@amresh_maverick said:OA:600
2> f(x+y) =f(x) +f(y) +f(x)*f(y) , if f(1) =3 , find f(3) ??
63 ?
f(1 + 1) = f(1) + f(1) + f(1)*f(1) = 3 + 3 + 3*3 = 15
f(1 + 2) = f(1) + f(2) + f(1)*f(2) = 3 + 15 + 3*15 = 63
51^2003 mod 7^2003 = R
possible value of R ?
a) 7^550*3^2
b) 51^550
c) 3^550
d) 5^550
possible value of R ?
a) 7^550*3^2
b) 51^550
c) 3^550
d) 5^550
@amresh_maverick said:OA:6001> if inverse of f(x)= ax+b is f^-1 (x) = bx+a , find the vale of a and b2> f(x+y) =f(x) +f(y) +f(x)*f(y) , if f(1) =3 , find f(3) ??
1) If f(x) = ax + b, then f^-1(x) = x/a - b/a
Equating it with bx + a, we get 1/a = b i.e. ab = 1
AND a = -b/a i.e. a^2 = -b or a^3 = -ab = -1
i.e. a = -1 and also b = -1
2) Put x = y = 1, we get
f(2) = f(1) + f(1) + f(1)*f(1) = 3 + 3 + 3*3 = 15
Next put x = 2, y = 1 to get
f(3) = f(2) + f(1) + f(2)*f(1) = 15 + 3 + 15*3 = 63
Basically if you observe then f(x) = 2^2x - 1 :)
Team BV
@amresh_maverick said:from [1^2/2013] ................... [44^2/2013] will give zero from [45^2/2013] ................... [63^2/2013] will give 1 from [64^2/2013] ..............................will give 2similarly[2013^2/2013]. will give 2013 batti nahi jala , help
But look at 2012^2.....
Note that 2013^2 - 2012^2 will be 2012 + 2013....which when subtracted and divided by 2013 will give 2011...
regards
scrabbler
Note that 2013^2 - 2012^2 will be 2012 + 2013....which when subtracted and divided by 2013 will give 2011...
regards
scrabbler
@amresh_maverick said:from [1^2/2013] ................... [44^2/2013] will give zero from [45^2/2013] ................... [63^2/2013] will give 1 from [64^2/2013] ..............................will give 2similarly[2013^2/2013]. will give 2013 batti nahi jala , help
Amresh :)
There are total 2013 terms in the sequence and all are integers, right?
Now you have mentioned already that there are many terms which are zero, then some are 1 and so on.....So certainly there are less than 2013 distinct integers in the seires as many terms of the series yield in same integers.
I hope you follow what I just said :)
Team BV
@jain4444 said:51^2003 mod 7^2003 = R possible value of R ?a) 7^550*3^2 b) 51^550 c) 3^550 d) 5^550
Trying to solve orally....I would go with option c
Basically, 51^2003 mod 7 is 4, so mod any multiple of 7 it will be 7k+4....
Out of the options, first is divisible by 7, 2nd and 4th are 7k+2 and 3rd only is 7k+4
OA?
regards
scrabbler
Basically, 51^2003 mod 7 is 4, so mod any multiple of 7 it will be 7k+4....
Out of the options, first is divisible by 7, 2nd and 4th are 7k+2 and 3rd only is 7k+4
OA?
regards
scrabbler
@scrabbler said:Trying to solve orally....I would go with option cBasically, 51^2003 mod 7 is 4, so mod any multiple of 7 it will be 7k+4....Out of the options, first is divisible by 7, 2nd and 4th are 7k+2 and 3rd only is 7k+4OA?regardsscrabbler
bro whats the logic here? why are you not considering the power of 7 i.e 2003 while formulating the remainder?
@scrabbler said:Trying to solve orally....I would go with option cBasically, 51^2003 mod 7 is 4, so mod any multiple of 7 it will be 7k+4....Out of the options, first is divisible by 7, 2nd and 4th are 7k+2 and 3rd only is 7k+4OA?regardsscrabbler
E(7^2003) = 6*7^2002
51^2003 mod 6 = 3
answer should be 6k + 3 ki form
option A satisfies
51^2003 mod 6 = 3
answer should be 6k + 3 ki form
option A satisfies
btw options banane mei gadbadi kar dee 3^550 mod 6 = 3 also 
@jain4444 said:E(7^2003) = 6*7^200251^2003 mod 6 = 3answer should be 6k + 3 ki formoption A satisfiesbtw options banane mei gadbadi kar dee 3^550 mod 6 = 3 also
But option a is a multiple of 7...this can't happen. When we divide a non-multiple of 7 with a multiple of 7, remainder has to be a non-multiple of 7.
regards
scrabbler
regards
scrabbler
@iLoveTorres said:bro whats the logic here? why are you not considering the power of 7 i.e 2003 while formulating the remainder?
When we divide by 11 if remainder is say 3, then remainder with a multiple of 11, say 44, will be of the form 11k+3 (3, 14, 25, 36)
Similarly if we divide by 7 and find remainder 4, then on dividing by 7^2003 which is a multiple of 7, the remainder should be f the form 7k+4...
At least that is my feeling...
regards
scrabbler
Similarly if we divide by 7 and find remainder 4, then on dividing by 7^2003 which is a multiple of 7, the remainder should be f the form 7k+4...
At least that is my feeling...
regards
scrabbler
@scrabbler said:But option a is a multiple of 7...this can't happen. When we divide a non-multiple of 7 with a multiple of 7, remainder has to be a non-multiple of 7.regardsscrabbler
yeah my logic was wrong...bad question dont take it seriously.....mei chala sone you guys carry on
f(x)=x^2/(x^2-1) then 31f(30)*f(29).............*f(3)*f(2) equals to
??
@amresh_maverick said:f(x)=x^2/(x^2-1) then 31f(30)*f(29).............*f(3)*f(2) equals to ??
f(x) = [x/(x + 1)][x/(x - 1)]
i.e. 31f(30)*f(29).............*f(3)*f(2) = 31[30/31][30/29][29/30][29/28][28/29][28/27]...[2/3][2/1] = [30][2] = 60 :)
It seems that you have got Tathagat's function sheet from somewhere 😛