2010/021) Prove that there is no a for which a^2-3a -19 is divisible by 289

Proof:
As a first step as we see that 289 = 17^2.

Now a^2-3a-19 = (a-10)(a+7) + 51

The 2nd term that is 51 is divisible by 17 and for the 1st term that is product to be divisible by 17 either (a-10) or (a+7) is divisible by 17. but if one of them is divisible by 17 then the 2nd one is divisible by 17.

So 1st term is divisible by 289 and 2nd one is not divisible by 289 so sum is not divisible by 289. Or the 2nd term is divisible by 17 and 1st term is not divisible by 17 so sum is not divisible by 17.

So the expression is not divisible by 289.

2010/020) Form a cubic equation with odd coefficients and rational root that fits x^3 + ax + b

because coefficient of x^3 is 1 so rational roots are integers say m, n , p

product is b odd so all roots (b= -mnp) are odd

coefficient of x^2 = -(m+n+p) = 0 which is not possible as m n and p are odd

so no solution

2010/019) Suppose two complex numbers z = a + ib and w = c + id

Suppose two complex numbers z = a + ib and w = c + id satisfy the equation ...?
(z + w)/z = w/(z + w).

Then,
a) both a and c are zero;
b) both b and d are zero;
c) both b and d must be non zero;
d) at least one of b and d is non-zero;
Kindly explain...

(z + w)/z = w/(z + w)
=> (z+w)^2 = wz
=> z^2+wz+w^2 = 0

let z/w = t so we get

t^2+t+1 = 0 and hence t is cube root of 1 that is cis 120 or cis 240

so z and w in complex plane and at angle 120

so at least one of b and d must be non zero because in case b and d both are zero then z/w is real

hence ans is d

2010/018) The set of all integers n for which sqrt(n^2 + n) is an integer is...?

The set of all integers n for which sqrt(n^2 + n) is an integer is...?
a) the set {0,-1};
b) a finite set with at least three elements;
c) an infinite set;
d) none of these sets;
Kindly explain...

Ans:
sqrt(n^2 + n) = sqrt(n(n+1)) is inetger

n and n+1 are coprimes so either n= 0 or n = - 1 or n and n+1 both squares

n = x^2 and n+1 = y^2 => 1 = (x+y)(y-x) => x+y = 1 and y-x = 1 => x = 0 y =1 => n = 0

or x+y = -1 and y-x = -1 => x = 0 y =-1 => n = 0

so only solution 0 and -1 so ans is a)

2010/017) Lim (n -> infinity) {(1 + 1/n)^n - (1 + 1/n)}^-n is

(a) 1; (b) 1/(e - 1); (c) 1 - e^-1; (d) 0;?

Ans:
take (1+1/n) out to get

(1+1/n)^-n((1+1/n)^(n-1)-1)^-n

as n->infinite (1+1/n)^-n = 1/e

((1+1/n)^(n-1)-1)^(-n) = (e-1)^(-n) = 0

so product = 0

hence d

2010/016) Let 1, a1, ... ,a6 denote the distinct roots of x^7 - 1.

Let 1, a1, ... ,a6 denote the distinct roots of x^7 - 1. Then the product (1 - a1)(1 - a2)(1 - a3) ...(1 - a6) is

is
a) 0;
b) 1;
c) 6;
d) 7;
Kindly explain...
x^7- 1 = (x-1)(x-a1)(x-a2)(x-a3)(x-a4)(x-a5)(x-a6… by definition of roots

dividing by (x-1) on both sides

x^6+x^5+x^4+x^3+x^2+x+1 = (x-a1)(x-a2)(x-a3)(x-a4)(x-a5)(x-a6)

putting x= 1 on both sides we get 7 = (1-a1)(1-a2)(1-a3)(1-a4)(1-a5)(1-a6)

hence d

2010/015) Suppose that one moves along the points (m,n) in the plane where m and n are integers in ...?

such a way that each move is a diagonal step, that is, consists of one unit to the right or left followed by one unit either up or down.
a) Which points (p,q) can be reached from the origin?
b) What is the minimum numbers of moves needed to reach such a point (p,q) ?

Ans:
) In each move x cordinate goes up/down by 1 and y cordinate goes up or down by 1 so x+y sum changes by an even number.

the starting point is (0,0) so sum of x and y is even so p+q should be even.

b) if p + q is odd then we cannot go

so let p+ q be even.

then number of steps required is maximum(|p|,|q|)

to explain say q and p positive and q> p

so we require q steps to wards y side up and out of it p + (q-p)/2 steps to the right and (q-p)/2 steps to left

note: if some term above is -ve then move that many positive steps in opposite direction

2010/014) Let a,b and c be the sides of a right angled triangle. Let theta be the smallest angle of this triangle.?

If 1/a, 1/b, 1/c are also the sides of a right angled triangle then show that Sin(theta) = (sqrt(5) - 1)/2;

proof
Let a < b < c


And theta (say t opposite to smaller side)

Sin t = a/c and cos t = b/c

Now 1/a > 1/b > 1/c and as it is right angled triangle we have

(1/a)^2 = (1/b)^2 + (1/c)^2

Or

(c/a)^2 = (c/b)^2 +1

(1/sin ^2 t) = (1/ cos^2 t ) +1

Or cos^2 t = sin ^2 t + sin ^2 t cos^2 t

Or 1- sin ^2t = sin ^2 t + sin ^2 t (1- sin ^2 t)
= 2 sin ^2 t – sin ^4 t
Or sin ^4 t – 3 sin ^2 t + 1 = 0

Sin^2 t = (3 +/- sqrt(9-4))/2 = (3 – sqrt(5))/2 , + cannot be taken as it shall be >1 not possible


So sin t = sqrt((3 – sqrt(5))/2) say sqrt(x) – sqrt(y) as there is sqrt(5) in square

So squaring we get

x+y – 2sqrt(ab) = (3 –sqrt(5))/2

so x+y = 3/2 and sqrt(xy) = sqrt(5)/4) or xy = 5/16

we can solve it as x = 5/4 and y= 1/4 or y = 1/4 and x = 5/4

so sin t = +/-(sqrt(5/4) - sqrt(1/4))

positive value to be taken as sin t > 0 so sin t = sqrt(5/4) – sqrt(1/4) = (sqrt(5)-1)/2