21\\\textup{Q3) } \textup{let } N=1992^{1998} - 1955^{1998} - 1938^{1998} + 1901^{1998}\\ \textup{we know } a^n-b^n \textup{ is always divisible by (a-b) for any +ve integer n}\\ \therefore \; 1992^n - 1938^n \textup{ is divisible by (1992-1938)i.e 54.}\\ \textup{similarly } 1955^n- 1901^n \textup{ is divisible by (1955-1901)=54} \\\therefore \textup{ N is divisible by 54}\\ \textup{Now Similarly } 1992^n - 1955^n \textup{ is divisible by (1992-1955)= 37}\\ \textup{ and}\\ 1938^n - 1901^n \textup{ is divisible by (1938-1901)= 37 }\\ \textup{therefore N is divisible by 37}\\ \textup{Now since N is divisible by 54 and 37 it has to be divisible by LCM(54,37)=54.37=1998}
106to the last question .... zero (provided a,b,c....,h,k, A,B,C,....,H are real)
2305Answer 5:
Number of terms in (x1+x2+x3.....+xk)n is -n+r-1Cr-1
Answer - n+3C3
2305
Shaswata Roy
·2012-09-25 06:05:18
Answer 2:
32 ≡ 2 (mod 3)
3232 ≡ (26)5*4 ≡ 645*4 ≡ 4≡1(mod 3)
Let 3232 = 3k + 1
323232 ≡ (323)k*32 ≡ 64k * 32 ≡ 32 ≡ 4(mod 7)
Answer - 4
2305
Shaswata Roy
·2012-09-25 05:35:07
Answer 1:
1111 ≡ 5 (mod 7)
2222 ≡ 10 (mod 7)
22225555 ≡ 1000001111 (mod 7)
5555 ≡ 25 (mod 7)
55552222 ≡ 6251111 (mod 7)
Therefore, 22225555 + 55552222≡(100000+625)≡0( mod 7)
21
eragon24 _Retired
·2010-03-29 21:11:31
ok for 2nd here is teh link-
http://www.targetiit.com/iit-jee-forum/posts/remainder-11788.html
1sorry asish there was a typo got it edited.
SORRY AGAIN [3][3]
106
Asish Mahapatra
·2010-03-29 18:49:09
Q6. i think there is a typo
Q2. something missing at end of question
1z= ( a - i b ) ^3
x + i y = ( a - i b ) ^3
on cubing r.h.s and equating real and imaginary terms we get
x = a^3 - 3a b^2
y = b^3 - 3a^2 b
\frac{x}{a} = a^2 - 3b^2
\frac{y}{b} = b^2 - 3a^2
\frac{x}{a} -\frac{y}{b} = a^2 - 3b^2 - ( b^2 -3a^2)
{\color{red} \frac{x}{a} - \frac{y}{b}= 4( a^2 - b^2 )}
1
Manmay kumar Mohanty
·2010-03-29 09:45:16
Q6) added and can any one solve Q2)
39Q. If alpha, beta, gamma are cube roots of p where p < 0, for any x, y, z find the value of
\frac{x\alpha +y\beta + z\gamma }{x\beta +y\gamma +z\alpha}
39
Pritish Chakraborty
·2010-03-29 06:28:30
On request of gr8dreams, I'm posting some more conceptual complex number problems.
Q. If aeix, beiy, ceiz are three collinear points, find a trigonometric relation which satisfies x,y,z.
Q. If we are given that
z = x + iy
z1/3 = a - ib
Prove that
\frac{x}{a} - \frac{y}{b} = 4(a^2 - b^2)
39
Pritish Chakraborty
·2010-03-29 06:25:07
Yes Asish..spot on. You can hide your ans now :P
21
eragon24 _Retired
·2010-03-25 07:30:38
well these probs hav been hashed, re hashed, re-re hashed many a times here...
(i heard hsbhatt sir saying this to someone on goiit :P)
Q1) http://targetiit.com/iit-jee-forum/posts/reasoning-type-9087.html
Q2) this has also been done....though not finding the link rit now
39
Pritish Chakraborty
·2010-03-28 07:03:25
Here's an easy one....
Q. Find the number of imaginary roots in the equation -:
\frac {A^2}{x-a} + \frac{B^2}{x-b} + \frac{C^2}{x-c} +....+\frac{H^2}{x-h} = k
106
Asish Mahapatra
·2010-03-28 02:18:59
no. of terms in (x+y+z+w)n
=> Let power of x be a, y be b, z be c and w be d
Then acc. to multinomial theorem,
a+b+c+d =n n≥a,b,c,d≥0
=> coeff of xn in (1+x+x2+...)4
= (1-x)-4
= n+4-1Cn which is the answer
21
eragon24 _Retired
·2010-03-26 22:27:45
@manmay i guess i posted the ans for Q3 only
1mistake its e0=1 ,thanks.
39
Pritish Chakraborty
·2010-03-26 10:14:26
No ut10...your approach is right but answer is not...try again :)
1
ut10
·2010-03-26 08:24:37
using
a3+b3+c3-3abc=(a+b+c)(a+bw+cw2)(a+bw2+cw) the answer is 0.
39
Pritish Chakraborty
·2010-03-26 05:53:49
Try this one...an addition to your algebra thread lol
Q. If
a = 1 + \frac{x^3}{3!} + \frac{x^6}{6!} + ......\infty
b = x + \frac{x^4}{4!} + \frac{x^7}{7!} + ......\infty
c = \frac{x^2}{2!} + \frac{x^5}{5!} + \frac{x^8}{8!} + ......\infty
Find the value of
a^3 + b^3 + c^3 - 3abc
