i too cant get ur approach.Dis approach is given in the solution also but cant get the step a998(x-1)(x-2)........(x-999)=xP(x)-1=0 if u can elaborate on it.Dis question is from past XAT Paper.
P(x)=a998x^998 +a997x^997 +.....+a1x^1+a0=1/x =>a998x^999 +a997x^998 + ......+a0x-1=0 Bro..I try to drive the point home .... The equation is satisfied by x=1,2,3,.......999 let us take for argument- if ax^2 +bx +c-1/x =0 has roots a1,a2,a3,then ax^2+bx+c=1/x too has roota at a1,a2,a3 and can be written as a(x-a1)(x-a2)(x-a3) on the same line- a998(x-1)(x-2)........(x-999)=xP(x)-1=0 =>product of roots=1/a998=1x2x3x4.....x999 putting x=1001 in a998(x-1)(x-2)........(x-999)=xP(x)-1=0 => a998x1000x999x998.........3x2 =1001P(1001)-1 =>1001 +1=1001P(1001) =>P(1001)=1 if still some doubt ... bother me pls
bro suppose there is function f(x)=a polynomial of n degree f(x)=f1(x)=1/x if F(x)=f(x)-f1(x)=/x(roots of this (n+1) order poly will be same as values for which f(x) is defined.... Note f(x) is polynomial of n order and it is defined for (n+1) values. therefore /x=/x=0 abhi bhi koi doubt ho to batao........
a998(x-1)(x-2)........(x-999)=xP(x)-1=0 brother yahan tak clear hain..... then where is problem simply put x=0 a998=1/1x2x3x4......x999 =>product of roots=1/a998=1x2x3x4.....x999
still didnt get ur idea..... (i think my basics are too weak )
what i dont understand is... u got ur equation a998(x-1)(x-2).......(x-999) = 0 - (1)
from the following P(x) = a998x^998 + a997x^997 ..............+a0 = 1/x which is true when 1so if x>999 , u dont know if P(x) = 1/x so u cannot get equation (1) if x>999 then ........ this is the most important line
"what is the meaning of putting an 'x' greater than 999 in an equation that we have obtained assuming 'x' is lesser than 999 in the first place"
actually i had this doubt all thru my school... aaj din aa gaya hai jab doodh ka doodh aur pani ka paniho jaye... so plzz. bear with me
Now here 20 = 5*4 , So 20^3 and greater will be divisible by 125 So we are left with
mod 125 or
mod 125 {Look for the last three digits}
(400 + 440 + 1) mod 125 = (25 + 65 + 1) mod 125 = 91
Approach looks Complex but it is Short & Simple....... ;)
Cant we do it like this : 125=25*5 Hence remainder of 21^22 when divided by 25 : E(25) =20 Hence 21^2/25 = 441^25 ...Remainder = 16 Also remainder when 21^22 is divided by 5 gives Remainder as 1
Hence using Chinese theorem, 25k + 16 = 5p +1 Hence p = 25k+15 / 5 = 5k+3 Putting k=1, p=8 Hence Remainder = 5*8 +1 =41
Cant we do it like this : 125=25*5 Hence remainder of 21^22 when divided by 25 : E(25) =20 Hence 21^2/25 = 441^25 ...Remainder = 16 Also remainder when 21^22 is divided by 5 gives Remainder as 1
Hence using Chinese theorem, 25k + 16 = 5p +1 Hence p = 25k+15 / 5 = 5k+3 Putting k=1, p=8 Hence Remainder = 5*8 +1 =41 Why there is a difference in the answer?
For Chinese Remainder Theorum it is Mandatory that the factors in which the Divisor is broken must be Coprime.
So here we can't break 125 as 25 and 5. As 125 = 5^3
Give remiander when 21^22 is divided by125. 21^22 mod 5^3 => (20+1)^22 mod 125 => (22C0 +22C1x20 +22C2x400)mod 125 => (1+22x20+11x21x400) mod 125 => (66+231x400) mod 125 =>(66+ 400) mod 125 => (466) mod 125 =>91 mod125 or (441)^11 mod 125 (66)^11 mod 125 6^11x11^11 mod 125 66x36^5x(125-4)^5 mod 125 - 66x36^5 x4^5 mod125 -14x36^5x6 mod 125 -14x36x46x6 mod125 -34 mod 125
will try to be regular from now onwards..(in this thrd)
okay one qstn from my side
5789^(1!+2!+3!-------+1000!)has the unit digit of ?
Hi Guys,
This question is posted on the second page of the thread but I have posted here for my doubt. My way for this question- 9^odd=9 and 9^even=1 so the expression can be written as- 9^!1+9!2............... =9+1+9+1+9+1...................... =9*500+1*500, as there will be 500 even and 500 odd nos =....oo so isnt the last digit zero?
x^998 +a997x^997 +.....+a1x^1+a0=1/x