@abhishekj1991 said:Find the remainder when (123123123123..................300digits) is divided by 37.
16?
(123*100) mod 37 =16
@viewpt said:@raopradeep said:what is the minimum value of the function |x-2|+|x-1|+|x+11| where x is real numbera)11B)13c)12d)1413
bhai approach kya hai..
@ishu1991 said:g(2)g(1)=g(2)+g(1)+g(2)-2 from here u get g(1)=8/9g(3)g(1)=g(3)+g(1)+g(3)-2 substitute the value of g(1) here u will get g(3)=10
g(1) = 8/9 kaise aaya ....i m getting 2..
thoda or explain karo...
@mailtoankit
sryy mine was wrng chck this nw
x-2,y-1 -> g(1) = 2
x-2,y-0 -> g(0) = 1
equation is
ax^2+bx+1
g(1)=2 -> a+b = 1
g(2) =5 -> 4a+2b=4-> b=0 -> a=1
eqn is x^2+1
so, g(3) is 10
sryy mine was wrng chck this nw
x-2,y-1 -> g(1) = 2
x-2,y-0 -> g(0) = 1
equation is
ax^2+bx+1
g(1)=2 -> a+b = 1
g(2) =5 -> 4a+2b=4-> b=0 -> a=1
eqn is x^2+1
so, g(3) is 10
Q: 37^41^57 div 67 find the R?
@viewpt said:Q: 37^41^57 div 67 find the R?
E(67)=66
so, we need to find 41^57 mod 66
E(66)=20
41^57mod66=41^17mod66
Also 41^20mod66=1
=> 41^17mod66 * 41^3 mod66=1
=> 41^17mod66 * 17 mod66 =1
=> 41^17mod66=35
so, now we need 37^35 mod 67
37^3mod67=1
So, 37^35mod67=37^2mod67=29
ANS=29
=> 41^17mod66=35
how did you get 35?
Ive always had trouble with this method
@VJ12
E(67) = 66
37^66k mod 67 = 1
41^37 mod 66 = 35
37^35 mod 67 = 29
PS : its a very lengthy solution...so puri calculation nahi daal sakta..
@viewpt said:@raopradeep said:what is the minimum value of the function |x-2|+|x-1|+|x+11| where x is real numbera)11B)13c)12d)1413
Minimum value of a function will occur at critical points or the average of the three. here it occurs at x=1. min value=13
@joyjitpal said:What is the remainder when the number 123123123.....123123 (300 digits) is divided by 99?a) 18 b) 27 c) 33 d) 36
33
123123(10^294+10^288.....1)/99
123123*50/99
@bs0409 said:The first and the last term of an arithmetic progression, having at least three terms, are 5 and 25 respectively. If all the terms of this arithmetic progression are integers, find the number of different values that the common difference of the arithmetic progression can take.
a=5 ; a+(n-1)d = 25 ==> (n-1)*d = 20 = 20*1 , 10*2, 5*4 ==> d can be 1,10,2,5,4... total 5??
@bs0409 said:OA=5..............A starts from a point P on a circular track ,at 6:00 am and runs around the track in clockwise direction .At 6:15 am ,B starts from the same point P and runs around the track in counter clockwise direction and meets A for the first time at 7:09 am.If B started at 6 am then he would have met A for the first time at 7 am.If both A & B start from P at 7:00 am and run in the opposite direction,then find the time at which they would meet at the starting point for the first time?
D / ( Sa + Sb ) = 3600 ; [D - Sa*900]/ Sa + Sb = 3240...dividing the two equations ===> d = Sa* 9000.
sbstitsting this eq (1) ==> 2Sb = 3 Sa .
Sa = D /9000 ; Sb = D / 6000 ;
Meeting for first time LCM { D/ Sa , D/sb } = LCM { 9000 , 6000 } = 180000 sec = 5 hours ==> they meet at 12 noon
sbstitsting this eq (1) ==> 2Sb = 3 Sa .
Sa = D /9000 ; Sb = D / 6000 ;
Meeting for first time LCM { D/ Sa , D/sb } = LCM { 9000 , 6000 } = 180000 sec = 5 hours ==> they meet at 12 noon
@ytstacks said:Find the remainder when (123123123123..................300digits) is divided by 37.Please explain along with the method?
123* 10^297 + 123*10^294 +.... 123*10^3 + 123 / 37
37*27 = 999
so., 123 * 1000^ 99 + 123 * 1000^98 +...........123 * 1000 + 123 / 37
==> [123 * (-1)^99 + 123 * (-1)^98 +...........123 * -1 + 123] / 37
==> [123 - 123 + 123 - 123 ........-123 + 123 ] / 17 total 99 terms..98 terms get cancelled..so its 123 /37 ==> rem = 16 :)
37*27 = 999
so., 123 * 1000^ 99 + 123 * 1000^98 +...........123 * 1000 + 123 / 37
==> [123 * (-1)^99 + 123 * (-1)^98 +...........123 * -1 + 123] / 37
==> [123 - 123 + 123 - 123 ........-123 + 123 ] / 17 total 99 terms..98 terms get cancelled..so its 123 /37 ==> rem = 16 :)
@raopradeep said:what is the minimum value of the function |x-2|+|x-1|+|x+11| where x is real numbera)11B)13c)12d)14
@bs0409 @Estallar12 bhai.. i guess der is a standard method for thhis kind of problems..ur views??
@The_Loser said:Q => N persons sitting around round table. Find the odd against two specified persons sitting next to each other.1) N+1/22) N-3/23) N+3/24) N+2/3
N-3/2