@bullseyes said:Remainder when 14! is divided by 17
8
use Wilson's method
@catter2011 said:ek statement se conclussion ni ho sakta kya .. ?eg: in this question conclustion 2, is nothing but a form of statement 1
I have never encountered a qstn where two statements are given & conclusion from any one statements is asked.
anyways other flaw is => nuns which is the middle term is not distributed in the statements.
this is against the rules !! middle term should be distributed atleast once !
anyways other flaw is => nuns which is the middle term is not distributed in the statements.
this is against the rules !! middle term should be distributed atleast once !
@maddy2807 said:8use Wilson's method
this is not for me buddy.. i try for others to learn. so someone who dsnt know Wilson's cannot do such qs. so next time try in some detail pls
@astute99 said:min and max value of 4^(sinx)+4^(cosx)??
y= 4^(sinx) + 4/4^(sinx)
=(2^sinx -2/2^sinx)^2 + 4
So minimum value of y is 4.
Maximum value 
@karl said:@catter2011 y= 4^(sinx) + 4/4^(sinx) =(2^sinx -2/2^sinx)^2 + 4 So minimum value of y is 4. Maximum value
put x=225 deg for min
and x=45 deg for max
and x=45 deg for max
Remainder when 14! is divided by 17
Wilson se kaise karte hai ye? btao koi, mere ko nai ata.
Wilson se kaise karte hai ye? btao koi, mere ko nai ata.
Two B schools IIT-Doms and IIM had a basketball game consisting of four rounds. Surprisingly both the teams scored equal points after first round . From second to last rounds the scores of iit were in A.P. and iim were in G.P. with the first round score. If iim scored 100 points and won the game by one point then what could be the possible first round score of both the teams??
option : a) 40 b) 38 c) 36 d) none of the above
@Torque024 said:Remainder when 14! is divided by 17Wilson se kaise karte hai ye? btao koi, mere ko nai ata.
Acc to Wilson,
(n-2)! mod n= 1 mod n
so in this case
15! mod 17= 1 mod 17
15*14! mod 17= 1 mod 17
-2*14! mod 17= -16 mod 17 [ divide 15 and 1 by 17 and write the respective remainders]
14! mod 17= 8 mod 17
@karl said:@catter2011 y= 4^(sinx) + 4/4^(sinx) =(2^sinx -2/2^sinx)^2 + 4 So minimum value of y is 4. Maximum value
Differentiate hi karna parega, no other approach.
@maddy2807 said:Acc to Wilson,(n-2)! mod n= 1 mod nso in this case15! mod 17= 1 mod 1715*14! mod 17= 1 mod 17-2*14! mod 17= -16 mod 17 [ divide 15 and 1 by 17 and write the respective remainders]14! mod 17= 8 mod 17
yeh wilson theorem se rem(4!/6) = 1 ana chahiye.. lekin basic calc karo tho.. 0 agya. ?
@bullseyes said:Remainder when 14! is divided by 17
16! mod 17 = -1
16*15*(14!) mod 17 = -1
(-1)(-2)*R = 17-1 => R = 8 ?
R = 14! mod 17
@catter2011 said:put x=225 deg for minand x=45 deg for max
Don't think max occurs at x=45 deg, as x=30 gives value of 15.xx while at x=45 gives a value of 14.xx
@karl bhai dekh sin or cos dono third quadrant mein negative hote hain toh mi. value wahin ayegi aur cos and sin sirf 1st mein positive hote hain toh max value wahin ayegi
@Torque024 said:Remainder when 14! is divided by 17Wilson se kaise karte hai ye? btao koi, mere ko nai ata.
16!/ 17 remainder--> -1;
15!/17 remainder ---> 1;
14! *15 / 17 remainder -->1,
8*15/17 remainder -->1 so remainder of 14!/17 will be 8. :P
@catter2011 said:yeh wilson theorem se rem(4!/6) = 1 ana chahiye.. lekin basic calc karo tho.. 0 agya. ?
Bhai...n is a PRIME NUMBER..mujhe lagta hai ye mention nahin hua hai..is liye confusion..
@albiesriram said:Two B schools IIT-Doms and IIM had a basketball game consisting of four rounds. Surprisingly both the teams scored equal points after first round . From second to last rounds the scores of iit were in A.P. and iim were in G.P. with the first round score. If iim scored 100 points and won the game by one point then what could be the possible first round score of both the teams??option : a) 40 b) 38 c) 36 d) none of the above
2(2a + 3d) = 99 ? hows this possible..
a(1+r+r^2+r^3)=100 = 2.2.5.5
D ?