2011/111) Between 1000 and 2000 you can get all but one integers as the sum of positive consecutive integers. What is this number that you cannot get ?

It must have an odd factor it is proved in yahoo ans

first we know that each odd number can be expressed as sum of 2 consecutive numbers as

2n + 1 = n + (n+1)

Any even number is a power of 2 or can be expressed as power of 2 multiplied by an odd number >1

Say it is 2^p(2n+1) where p is highest power of 2 which is a factor.

Now 2n+1 can be written as n+ (n+1)

Now we have 2 cases 2^p <= n or 2^p > n

Case 1) 2^p < n

This can occur 2^p times but there shall be 2^p copies of n and 2^p copies on n + 1

In the kth copy subtract k-1 from n and add k-1 to n+ 1

So we get the values n-(2^p-1) to n+ 2^2p

The sum shall be 2^p ( n + 1 +n).

And all numbers are consecutive ( 1^st number >= 1) and positive

Case 2) 2^p > n

We take 2n+1 copies of 2^p

From the middle value at a distance of k we subtract k on from the left element and add k on the right element

So we have consecutive numbers 2^p-n to 2^p+ n

Sum = ½(2n+1)(2^p*2) = (2n+1) 2^p and as 2^p > n so all numbers are positive

Only power of 2 cannot be expressed as sum of consecutive number. Then the number is 1024 it has been proved in the link as above

2011/110) solve for a If a^n + a^(2n) = a^(3n) where n = 1+ log(base a)(cos(π/5)

a cannot be zero as base to a is taken

For a non trivial solution

a is not zero and dividing both sides by a^n we get

1+ a^n = a^2n

So a^2n = golden ration φ

n = 1+ log(base a)(cos(π/5)

so a^n = a * (cos(π/5)

= a * φ/2 = φ

Or a= 2

2011/109) How do you solve 1/x^2 + 1/y^2 = 1/z^2 in integers

1/x, 1/y . 1/z form numbers of (Pythagorean triplet assuming them to be integers and if they are not: which is generally true my multiplying by sutable nnumber ) so they are in the ratio

1/x : 1/y : 1/z = 2pq : p^2-q^2 : p^2 + q^2 (p and q are rational)

(knowing that (2pq)^2 + (p^2-q^2)^2 = (p^2+q^2) and you can find it in number theory book

or x: y : z : = 1/(2pq) : 1/ (p^2 - q^2) : 1/(p^2 + q^2)

or x: y: z = (p^2+q^2) (p^2-q^2) : 2pq(p^2+q^2) : 2pq(p^2-q^2)

putting p = m/r , q = n/r ( m,.n,.r are integers)

x:y:z = (m^2+n^2)(m^2-n^2) /r^4 : 2mn(m^2+n^2)/r^4 : 2mn(m^2-n^2)/r^ 4
or (m^2+n^2)(m^2-n^2) : 2mn(m^2+n^2) : 2mn(m^2-n^2)/


in a trivial case 1/x : 1/y: 1/z = 1/3: 1/4: 1/5 = 20:15: 12

2011/108) prove that sum of 2 sides of an euclidian triangle is greater than 3rd side

let the angles be A B C and opposite sides be a b c

we need to prove that a + b > c

we have A+B = 180-C

hence sin (A+B) = sin (180-C) = sin C

or sin A cos B + cos A sin B = sin C

as cos B < 1 and cos A < 1 we have (it is 1 at zero degree not possible)



sin A cos B + cos A sin B < sin A + sin B so sin C < sin A + sin B

as a/sin A = b/sin B = c/sin C we get c < a + b

proved

2011/107) Prove that... 1/sin(pi/7)=1/sin(2pi/7)+1/sin(3pi/7)

start from RHS as it is more complex

1/ (sin 2pi/ 7) + 1/(sin 3pi/7)
= 1/ ( sin 5pi/7) + 1/( sin 3pi/7) as sin 2pi/7 = sin (pi-2pi/7) = sin 5pi/7 and we can add with out a faction of the form pi/14)

= ( sin 3pi/7 + sin 5pi/7) / ( sin 5pi/7 sin 3pi/7)
= (2 sin 4pi/7 cos pi/7)/ ( ( sin 5pi/7 sin 3pi/7)
as sin 4pi/7 = sin 3pi/7 and one is in numerator and another in denominator
= (2 sin 3pi/7 cos pi/7)/( sin 5pi/7 sin 3pi/7)
= (2 cos pi/7)/ ( sin 5pi/7)
= (2 cos pi/7)/( sin 2pi/7) as sin 5pi/7 = sin 2pi/7
= ( 2 cospi/7 )/ ( 2 sin pi/7 cos pi/7)
= 1/ sin pi/7 = LHS

2011/106) A natural number n is chosen strictly between two consecutive perfect squares.

The smaller of these two squares is obtained by subtracting k from n an the larger is obtained by adding l to n. Prove that n-kl is a perfect square.

Proof:

Let the number n be between a^2 and (a+1)^2

As per given condition

n- a^2 = k …1

(a+1)^2 –n = l ….2

Adding (1) and (2)

k + l = 2a + 1

or l = 2a + 1 - k

now

n- kl = (a^2+k) – k(2a+1-k)

= a^2 + k – 2ak –k + k^2 = a^2 – 2ak + k^2 = (a-k)^2

Hence proved

that n-kl is a perfect square.

2011/105) Prove a^4 + b^4 + c^4 >= abc(a+b+c)

this can be done in 2 steps

we know by AM GM inequality

a^4+b^4 > = 2 a^2b^2
b^4+c^4 >= 2 b^2 c^2
c^4 + a^4 >= 2 a^2^2

adding the 3 above and deviding by 2 we get

a^4+b^4+c ^4 >= a^2b^2+b^2c^2 + c^2 a^2 ...1

now we repeat the process of a^2b^2 , b^2 c^2 and c^2 a^2 to get as below

a^2 b^2 + b^2 c^2 > = 2 b^2ac
b^2c^2 + c^2 a^2 >= 2 c^2ab
c^2a^2 + a^2 b^2 >= 2 a^2bc

adding the above and deviding by 2 we get

a^2b^2 + b^2 c^2 + c^2 a^2 >= (b^2ac+c^2ab+a^2bc) or abc(b+c+a) ,.2

from (1) and (2) it follows

a^4 + b^4 + c^4 >= abc(a+b+c)

2011/104) If a=(root 2+1)^ 1/3 - (root 2 -1 )^1/3 then value of a ^ 3 +3a-2 is zero. explain

a = -(root 2+1)^ 1/3-(root 2 -1)^ 1/3

so a + (-(root 2+1)^ 1/3) +(root 2 -1)^ 1/3 = 0

using x+y+z= 0 => x^3 +y^3 + z^3 = 3xyz

we get

a^3 + (-(root 2+1)) + ((root 2-1)) = 3a((-(root 2+1))((root 2-1)

or a^3 - 2 = - 3a and hence a^3+3a - 2 = 0

2011/103) Find sufficient condition on a, b, c that the roots of x³ + ax² + bx + c = 0 are in arithmetic progression.

Let the roots be m-d, m and m+d

then as per vieta's formula http://en.wikipedia.org/wiki/Vi%C3%A8te%…
1)
sum of roots = m-d + m + m+d = - a or 3m = - a
2)
b= m(m-d) + m(m+d) + (m-d)(m+d) = 3m^3 - d^2 or d^2 = 3m^2 - b or (a^2)/3- b
3)
product of roots = m(m-d)(m+d) = m(m^2 - d^2) = - c

or (-a/3)((a/3)^2 - (a^2/3 -b) = -c

or(- a/3)(2a^2/9 - b) = - c

or (a/3) (b - 2a^2/9) = c

or (a/27) (9b - 2a^2) = c
above condition is necessary and sufficient

2011/102) Prove z/ (1 + z^2) is real if and only if z is real or |z| =1.

if z is zero it is true so let us take for the case z not zero

so z/(1+z^2) = 1/(z + 1/z) is real

let z = r cos t + ir sin t

so 1/z = 1/r cos t - 1/r i sin t

z + 1/z imaginary part is zero

r sin t = - 1/r sin t
sin t = 0 or r = -1/r or |r| = 1 and hence |z| = 1


sin t = 0 means real

hence z is real or |z| = 1

2011/101) Prove this inequality? 4x^10+x^8+4x^2+1 ≥ 2x^9+4x^5+2x (x is a positive number)

we shall use the AM GM enaquality to prove it

we know (a+b)/2 > = sqrt(ab) or (a+b) > = 2 sqrt(ab)

a = 4x^10, b = x^8 gives 4x^10 + x^ 8 > = 4 x^9
a = 4x^10, b = 1 gives 4x^10 + 1 > = 4 x^5
a = x^8, b = 4x^2 gives x^ 8 +4x^2 > = 4x^5
a = 4x^2, b = 1 gives 4x^2 + 1 > = 4 x

adding we get

8x^10 + 2x^2+ 8x + 2 >=4x^9 + 8x^5 + 2

dividing by 2 we get 4x^10+x^8+4x^2+1 ≥ 2x^9+4x^5+2x

2011/100) Solve for x: (2x + 14) / (5x - 1) ≤ x + 3.

we need to get rid of denominator so multiplying by (5x-1) is problematic as we do not know whether it is positive or negative

so multiply by (5x-1)^2 which is positive to get

(2x+ 14) (5x -1 ) <= (5x-1)^2(x+3)

or (2x+14) (5x-1) - (5x-1)^2 (x+ 3) <= 0

or (5x-1) ((2x+14 - (5x- 1 )(x+3)) < = 0

or (5x-1)(2x+14 - (5x^2 + 14x - 3)) <= 0
or (5x-1)(-5x^2 - 12x + 17) < = 0
or (5x-1)(5x^2+ 12x - 17) >= 0
or (5x-1) (5x+17)(x-1) > = 0

there are 4 ranges - infinite to -17/5
- 17/5 to 1/5 (in this it is positive but x cannot be 1/5 as 5x-1 is in denominator)
1/5 to 1
and 1 to infinite (in this it is positive

x belongs to [- 17/5, 1/5) U [1, ∞).

2011/099) Find cube root of 9 sqrt 3+11 sqrt 2

because there is sqrt(3) and sqrt(2) in the number the cube root is a combination of them
let it be x sqrt (3) + y sqrt (2)
so (x sqrt (3) + y sqrt (2))^3 = 9 sqrt 3+11 sqrt 2
or 3 x^3 sqrt (3) + 9x^2y sqrt 2 + 6xy^2 sqrt (2) + 2y^3 sqrt 2 = 9 sqrt 3+11 sqrt 2
comparing coefficient of sqrt (3) and sqrt(2) we get
3x^3 + 6x y^2 = 9 ..1
9x^2 y + 2y^3 = 11 .. 2
multiply (1) by 11 and 2nd one by 9 to get
33 x^3 + 66 xy^2 = 81 x^2y + 18y^3
divide both sides by 311 x^3 + 22xy^2 = 27x^2y + 6y^3
or 11x^3 - 27 x^2y + 22 xy^2 - 6y^3 = 0
putting x/y = t we get
11t^3-27 t^2 + 22t- 6 = 0
f(1) = 0 gives t = 1 is a sqrt ( as 11 + 22 = 27+ 6)
so t = 1 or 11t^2-16t + 6 = 0
or x = y is a solution
and
from (1) 9x^3 = 1 or x = 1 and y = 1
so result = sqrt(3) + sqrt (2)

2011/098) if a=0.5(5-root 21) then find a^3 + a^-3- 5a^2- 5a^-2+ a+ a^-1.?

a = .5(5-sqrt(21)

so 1/a = 2/(5- sqrt(21) = 2(5+ sqrt(21)/(5^2-21) = .5(5 + sqrt(21))

so a + 1/a = 5

a^3+ 1/a^3- 5 ( a^2+ 1/a^2) + (a+ 1/a)
= (a+1/a)^3 - 3 ( a + 1/a) - 5((a+1/a2^2) - 2) + (a+ 1/a)
= 5^3 - 3 * 5 - 5(5^2 - 2) + 5
= 125 - 15 - 125 + 10 + 5 = 0

2011/097) Prove the identity: (1+cosx+sinx)/(1+cosx-sinx) = secx+tanx?

regular solution
(1+ cos x + sin x)/(1+ cos x - sin x)
= (1+ cos x + sin x)^2/ (( 1+ cos x + sin x)(1+ cosx - sin x))
= (1+ cos x + sin x)^2/ ((1+ cos x) ^2 - sin ^2 x)
= (1+ cos x + sin x)^2/ ((1+ 2cos x + cos ^2 x - sin ^2 x)
= (1+ cos x + sin x)^2/ ((cos ^2 x+ 2cos x + cos ^2 x)
= (1+ cos x + sin x)^2/ (2(cos^2 x + 2 cos x))
= (1+ cos x + sin x)^2/ (2 cos x( 1+ cos x))
= ( 1 + 2 cos x + 2 sin x + 2 sin x cos x + cos ^2 x + sin ^2 x)/ (2 cos x( 1+ cos x))
= ( 2 + 2 cos x + 2 sin x + 2 cos x sin x)/(2 cos x( 1+ cos x))
= 2(1+ cos x)(1+ sin x)/ (2 cos x (1+ cos x))
= (1+ sin x)/cos x
= sec x + tan x


alternatively ( simpler solution)

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

using compnondo dividendo
we get
(1+cosx+sinx)/(1+cosx-sinx) = (cos x/2 + sin x/2)/( cos x/2 - sin x/2)
= ( cos x/2 + sin x/2)^2 / (( cos x/2 + sin x/2)(cos x/2 - sin x/2)
= ( cos ^2 x/ 2 + sin ^2 x/2 + 2 cos x/2 sin x/2)/(cos^2 x/2 - sin ^2 x/2)
= (1 + sin x)/ cos x
= sec x + tan x

2011/096) The equation P(x) = x^4 – 16x^3 + 94x^2 +px + q = 0 has two double roots. Solve

Because it has 2 double roots so it is square of a quadratic polynomial

Let it be (x^2+ax+b)^2

= x^4 + 2ax^3 + x^2(2b+a^2) + 2abx+ b^2

Comparing coefficients

2x = -16 or a = - 8

2b+a^2 = 94 or b = 15

p = 2ab = - 240

q = b^2 = 225

so equation

= x^4-16x^2 + 94x^2 – 240x + 225

It is (x^2 -8 x + 15) ^2 = (x-3)^2(x-5)^2 and roots are x = 3,3,5,5

2011/095) Prove: Let p and q be distinct primes, k be a natural number, and W be a natural number less than pq. Then W^(1+k(p-1)(q-1)) is congruent to W (mod pq

because p is a prime as per Formats Little theorem
now there are 3 cases and we deal as below
1) when w is coprime to p and q

w^(p-1) = 1 mod pwhen w is coprime to p

so w^(p-1)(q-1) = 1 mod p

similarly w^(p-1)(q-1) = 1 mod q

so w^(p-1)(q-1) = 1 mod pq

so w^k(p-1)(q-1) = 1 mod pq
or w^(1+k(p-1)(q-1)) = w mod pq

case 2
let w be a miultiple of p and not q

w^ t = w mod p as both are zero

so w ^(1+(k(p-1)(q-1)) = w mod p
as per argument in 1
w ^(1+(k(p-1)(q-1)) = w mod q

so w ^(1+(k(p-1)(q-1)) = w mod pq

3)
w is coprime to p and not q
same arguments as in 2

hence proved for all cases

2011/094) Show that the equation 2x+cosx=0 has exactly one real root?

we have
cos x + 2x = 0
so cos x = - 2x

as cos x is between -1 and + 1 so x has to be between -1/2 and + 1/2

now between -1/2 and 1/2

d/dx (cos x + 2x) = 2 - sin x > 0 so increasing

at 1/2 we have cos x + 2x > 0 (both cos x and x > 0)
at -1/2 we have cos x + 2x < 0 cos x < 1 and 2x = - 1

so it is increasing from -1 to 1 and hence exactly one real root between (-1/2 and 1/2)

2011/093) If a=4 root 6 /(root 2+ root 3) then find value of (a+2 root 2)/(a-2 root 2) + (a+ 2 root 3)/(a - 2 root 3)?

this is a perfect example of using componendo dividendo method

a=4 root 6 /(root 2+ root 3)

so a/(2 root 2) = 2 root3/(root 2 + root 3)

by componendo dividendo

( a+ 2 root 2)/(a- 2 root 2) = (2 root 3 + root 2 + root 3)/(root 3 - root 2)

= (3 root 3 + root 2) /(root 3 - root2) ...1

again for the second term

a=4 root 6 /(root 2+ root 3)
so a/(2 root 3) = 2 root2/(root 2 + root 3)

by componendo dividendo

( a+ 2 root 3)/(a- 2 root 3) = (2 root 2 + root 2 + root 3)/(root 2 - root 3)
= ( 3 root 2 + root 3)/ (root 2 - root 3) = - (- root 3 - 3 root 2)/(root 3 - root 2) .. 2

add (1) and (2) to get
( a+ 2 root 2)/(a- 2 root 2) + ( a+ 2 root 3)/(a- 2 root 3)
= 2(root 3 – root 2)/(root 3 – root 2) = 2


(comonendo and dividendo: for the persons who are not familiar

if a/b = c/d then (a+b)/(a-b) = (c+d)/(c-d))

for a proof of it (a+b)/(a-b) = (a/b +1)/(a/b-1)= (c/d + 1)/(c/d - 1) = (c+d)/(c-d))

2011/092) When is one of the roots of a cubic equal to the product of the other two roots?

Find, in terms of the coefficients, a necessary and sufficient condition for one of the roots of ax³ + bx² + cx + d = 0 (a ≠ 0) to be equal to the product of the other two roots.

Let the roots be p,q,pq

So a(x-p)(x-q)(x-pq)

= a x^3 – ax^2(p + q + pq) + ax(pq + pq^2 + p^2q) – ap^q^2

Comparing coefficients

b/a = -(p+q+pq) .. 1

c/a = pq + pq^2+ p^2q = pq(1+p+q) … 2

d/a = -p^2 q^2 ..3

subtract 1 from both sides of (1)

b/a - 1 = - (p+1)(q+1)

from (2 ) and (3)

c/a - d/a = pq(1+p+q+pq) = -pq(b/a- 1)

or c-d = - pq(b-a)

square both sides

(c-d)^2 = p^2q^2(b-a)^2 = -d(b-a)^2

Or (c-d)^2 + d(b-a)^2= 0

2011/91) What is the graph of xy = 3y

(A) one point
(B) two points
(C) one line
(D) two lines
(E) a right angle

this is a tricky question and ans is simple

xy = 3y => y(x-3) = 0 so y = 0 or x = 3

y= 0 is a line and x= 3 is another line and hence ans is D

2011/090) Prove that there do not exist four positive real numbers, all distinct from each other, such that a^3 + b ^3 = c^3 + d^3 and a+b = c+d

proof:
without loss of generality assume a>=b and c >= d

now as a+b = c+ d
cube both sides a^3+b^3 + 3ab(a+b)= c^3+d^3 + 3cd(c+d)

and a a^3+b^3 = c^3+d ^3 so 3ab(a+b) =3cd(c+d)
and hence ab = cd
(a-b)^2 = (a+b)^2 - 4ab = (c+d)^2 - 4cd = (c-d)^2 and hence a- b = c- d

as
a + b = c+ d
a- b = c - d
adding 2a = 2c = a = c and hence b = d
hence proved

2011/089) Evaluate the infinite product Π(n=2..∞)n^2/(n^2-1).

First we try to get a closed form for the product

We have t(n) = n^2/((n+1)(n-1))

Π(n=2) n^2/(n^2-1)= 2^2/(3)

Π(n=2..3)n^2/(n^2-1)= 2^2/(3) * 3^2 /(2*4) = 2 * 3 / 4

Π(n=2..4)n^2/(n^2-1)= 2*3/(4) * 4^2 /(3*5) = 2 * 4 / 5

From the pattern we feel that

Π(n=2..k)n^2/(n^2-1) = 2 * k/(k+1)

It is true for n = 2

If it is true for n =k then

Π(n=2..k+1)n^2/(n^2-1) = 2 * k/(k+1) * (k+1)^2/(k*(k+2) = 2 *(k+1)/(k+2)

So it is true for k+ 1

Π(n=2..k)n^2/(n^2-1) = 2k/(k+1)

Now that we have found the closed form as k goes to inifinite above product goes to 2 (converges to 2)

2011/088) Find a right angled triangle with integer sides whose all three sides are fibonacci numbers.

No solution

Proof:

Let the legs be x and y

X and y cannot be same then hypotenuse shall be x qrt(2) and it is not integer.

Now let x < y and so hypotenuse >= (x+y)

as next Fibonacci number = (x+y) if x and y are consecutive Fibonacci numbers

and > (x+y) if they are not consecutive

So x^2+y^2 >= (x+y)^2 which is not possible unless x = 0

Hence no solution

2011/087) Prove Cos^4x =(1/8)(3+4cos2x+cos4x)

realizing that RHS has got cos of multiple of x

Cos^4x = 1/4(2 cos ^2 x )^2
= 1/4( cos 2x + 1)^2 knowing cos 2x = 2 cos^2 x - 1
= 1/4( cos^2 2x + 2 cos 2x + 1)
= 1/8( 2 cos ^2 2x + 4 cos 2x + 2)
= 1/8( cos 4x +1 + 4 cos 2x + 2) as 2 cos^2 2x - 1 = cos 4x)]
= 1/8( cos 4x + 4 cos 2x + 3)

2011/086) prove that the expression 2x+3y and 9x + 5y are divisible by 17 for same integral values of x and y

if 9x + 5y is divisible by 17 then
4(9x + 5y) is divisible by 17 (note below why multiply by 4)
or 36x + 20 y is divisible by 17
or 36x-34 x + 20y - 17 y or 2x + 3y is divisible by 17

multiplication by 4 was not magic

as 2 = 9m mod 17 gives m = 4

to find it

let us find inverse 0f 9 mod 17

we have 17 = 9 + 8
8 = 17-9
now 9= 8+1 = (17-9) + 1 or
2 * 9 = 17 + 1 so 2 is inverse of 9
so m = 4 as multiplying by 2 gives 1 so multiply by 4 to give 2

2011/085) The quadratic function f(x) is negative for x > 9/2 and x < -1, but for no other value of x. If f(1) = 28 find f(x)

As it is –ve for x > 9/2 and x < -1 it is of the form –A (x-9/2)(x+1) where A > 0

Or – B(2x-9)(x+1)(where B = A/2)

F(1) = -B(-7)(2) = 28A = 28 or B = 2

So we get

F(x) = – 2(2x-9)(x+1)

2011/084) If p q r are roots of equation x^3 – 3px^2 + 3q^2x – r^3 = 0 then prove that p = q = r

proof:

as sum of roots is 3p so p+ q + r = 3p

so there exists t such that

q = p- t, r = p+ t

coefficient of x = pq + pr + qr = 3q

or p(q+r) + qr = 3q^2

or p* 2p + (p – t)(p+t) = 3(p-t)^2

or 3p^2 – t^2 = 3p^2-6pt + 3t^2

or 4t^2-6pt = 0 or

t(2t-3p) = 0 ..1

constant = p(p+t)(p-t) = (p+t)^3

p+t = 0 or p(p-t) = (p+t)^2

or p^2-pt = p^2 +2 pt + t^2

or t^2 + 3pt =0

t(t+3p) = 0 ….2

fom (1) and (2) t = 0

or

2t-3p = 0 and t+ 3p = 0 which again gives t = 0 (also p = 0 this is redundant as a sub set of 1^st solution)

So t = 0

So p = q =r

hence proved

2011/083) Show that for p and q odd x^2 + 2px + 2q = 0 does not have rational solution

proof

the discriminant is

b^2-4ac = (2p)^2 – 8q = 4(p^2-2q)

as p and q are odd

p^2-2q = (2s+1)^2 – 2(2t+1) = (4s^2+ 4s+1) – (4t+2) =(4s^2 + 4s – 4t -1) which is 3 mod 4

so discriminant cannot be a perfect square so there is no rational root

2011/082) The equation x³ + px² + qx + r = 0 (where p, q, r are non zero) has roots α, β, γ such that 1/α , 1/β, 1/γ...?

are consecutive terms in an arithmetic sequence, show that β = -3r / q

f(x) = x^3+px^2+qx+r = 0 has the roots α, β, γ

so f(1/x) has the roots 1/α , 1/β, 1/γ. Which are in AP

f(1/x) = 1/x^3+p/x^2+q/x+ r = 0

or rx^3+qx^2+px+1 = 0

sum of roots = - q/r = 3/ β or β = -3r/q (sum of 3 terms of AP = 3 * middle term)

proved

2011/081) prove that cos(3π/14)sin(π/7)cos(π/14) + sin(π/14)cos(π/7)cos(3π/14) + sin(3π/14)sin(π/14)sin(π/7) = cos(π/7)sin(3π/14)cos(π/14)

proof

cos(3π/14)sin(π/7)cos(π/14) + sin(π/14)cos(π/7)cos(3π/14) + sin(3π/14)sin(π/14)sin(π/7)

(knowing that sin (3π/14) is on the right so combine the terms not having sin(3π/14)

= cos(3π/14)[sin(π/7)cos(π/14) + sin(π/14)cos(π/7)] + sin(3π/14)sin(π/14)sin(π/7)

(using sin A cos B + cos A sin B = sn (A+B) in next line

= cos(3π/14)[sin(π/7+π/14] + sin(3π/14)sin(π/14)sin(π/7)

= cos(3π/14)sin(3π/14) + sin(3π/14)sin(π/14)sin(π/7)

= sin(3π/14)[ cos(3π/14) + sin(π/14)sin(π/7)]

= sin(3π/14)[ cos(π/7+π/14)+ sin(π/14)sin(π/7)] (as cos (A+B) shall cancel sin A sin B by contributing –ve of it)

= sin(3π/14)[ cos(π/7) cos (π/14) - sin(π/14)sin(π/7+ sin(π/14)sin(π/7)]

= cos(π/7)sin(3π/14)cos(π/14)

2011/080) prove that Tan 55 = tan35+ 2tan20

20 = 55 - 35

take tan of both sides

tan 20 = (tan 55 - tan 35)/(1+ tan 55 tan 35)

but as 55 + 35 = 90 so tan 35 = cot(90-35) = cot 55

so tan 20 = (tan 55 - tan 35)/(1+ tan 55 tan 35) = tan 20 = (tan 55 - tan 35)/(1+ tan 55 cot 55)
= (tan 55 - tan 35)/(1+ 1) = (tan 55 - tan 35)/2
or 2 tan 20 = tan 55 - tan 35

or tan 55 = tan 35 + 2 tan 20

2011/079) Find the limit. lim[x->179] (x-179) / [ √(x+17) - 14]

(x-179) / [ √(x+17) - 14] (of the form 0/0)
= (x-179) [ √(x+17) + 14]/( [ √(x+17) + 14] [ √(x+17) - 14] ) rationalizing denomenator
= (x-179) [ √(x+17) + 14]/(x^1+17-196)
= (x-179) [ √(x+17) + 14]/(x^1-179)
= [ √(x+17) + 14]
and putting x= 179 we get 14+14 = 28

2011/078) If x^2+y^2=1,then (4x^3-3x)^2+(3y-4y^3)^2=

This can be solved both algebraically and also using trigonometry.

Algebra method

We putting y ^2 = (-1x^2) that is converting to terms of x

x² * (4x² - 3)² + y² * (3 - 4y²)²

=x² * (4x² - 3)² + (1 - x²) * [3 - (4 - 4x²)]²
= x² * (4x² - 3)² - (x² - 1) * ( 4x² - 1)²
= x² * (4x² - 3)² - x² * ( 4x² - 1)² + ( 4x² - 1)²
= x² [ (4x² - 3)² - ( 4x² - 1)² ]+ ( 4x² - 1)²
= x² [ (4x² - 3 + 4x² - 1)(4x² - 3 - 4x² + 1) ] + ( 4x² - 1)²
= x² * (8x² - 4)(-2) + ( 4x² - 1)²
= -16x^4 + 8x² + 16x^4 - 8x² + 1

= 1

Now for trigonometric method

Given : x² + y² = 1
then x can be considered as sin θ and y as cos θ
{Since sin² θ + cos² θ = 1 }

Now, 4x³ - 3x = 4 sin³ θ - 3 sin θ = sin 3θ
and 3y - 4y³ = 3 cos θ - 4 cos³ θ = cos 3θ

=> (4x³ - 3x)² + (3y - 4y³)²
= sin² 3θ + cos² 3θ
= 1

2011/077) factor (x+y+z)^5 -x^5-y^5-z^5.?

Note
factorization is base don understanding symmetric and finding some method based on it

Method
we see that (x+y+z)^5 - x^5 - (y^5+z^5)

(x+y+z)^5 - x^5 is divisible by y+z and y^5+z^5 by y+ z

so it is divisible by (y+z)

hence it is divisible by (z+x) and (x+y) by symmetry so (x+y)(x+z)(y+z)

(x+y+z)^5- x^5-y^5-z^5

= x^5+ 5x^4(y+z) + 10 x^3(y+z)^2 + 10 x^2(y+z)^3 + 5x(y+z)^4+ (y+z)^5 - - x^ 5- y^5 - z^5
= 5x^4(y+z) + 10 x^3(y+z)^2 + 10 x^2(y+z)^3 + 5x(y+z)^4 + (y+z)^5- (y^5+z^5)

Now (y+z)^5 = y^5+ 5y^4z + 10 y^3z^2 + 10y^2z^3+5yz^4 + z^5

Hence (y+z)^5- y^5-z^5 = 5y^4z + 10 y^3z^2 + 10y^2z^3+5y^4

= 5y^4z + 5 zy^4 + 10 y^3z^2 + 10y^2z^3
= 5yz(y^3+z^3) + 10 y^2z^2(y+z)
= 5yz(y+z)(y^2-yz+z^3) + 10 y^2z^2(y+z)
= 5yz[(y+z)(y^2-yz+z^2) + 2yz)
= 5yz(y+z)(y^2+yz + z^2)

So 5x^4(y+z) + 10 x^3(y+z)^2 + 10 x^2(y+z)^3 + 5x(y+z)^4 + (y+z)^5- (y^5+z^5)
= 5x^4(y+z) + 10 x^3(y+z)^2 + 10 x^2(y+z)^3 + 5x(y+z)^4 + 5yz(y+z)(y+yz+z^2)
= 5(y+z) (x^4+ 2x^3(y+z) + 2x^2(y+z)^2 + x(y+z)^3 + yz(y^2+z^2+ yz))

Now we need to factor

(x^4+ 2x^3(y+z) + 2x^2(y+z)^2 + x(y+z)^3 + yz(y^2+z^2+yz) …1

This has a factor x + y and also x+ z so we have a factor (x^2+(y+z)x + yz) so let other factor be (x^2+ax+c)

(x^2+(y+z)x + yz) (x^2+ax+c)

= x^4 + x^3(a+y+z) + x^2(c+a(y+z)+ yz) + x(ayz+c(y+z)) + cyz) …2

Comparing coefficients from 1 and 2
From x^3
a= y+z,
from constant
c= y^2+z^2+yz

so coefficient of x^2 in 2nd expression that is (2) = c+ a(y+z) + yz
= (y^2+z^2 + yz + (y+z)^2 + yz)
= (y^2 + z^2 + 2yz) + (y+z)^2 = 2(y2+z)^2 matches in (1)
Coefficient of x (in 2)
= ayz + c(y+z)
= (y+z)yz + (y^2+z^2+yz) (y+z) = (y+z)^3

So the coefficient of x^2 and x match in 1 and 2 and hence the value is consistent

So we have
(x+y+z)^5 - x^5 - (y^5+z^5)
= 5(y+z) (x^4+ 2x^3(y+z) + 2x^2(y+z)^2 + x(y+z)^3 + yz(y^2+z^2+ yz))
= 5(y+z)(x+z)(x+y)(x^2+ yx + zx + y^2+z^2+ yz)

This is the complete factorization

2011/076) Express (1-cot2010°)(1-cot1995°) as a single real number.

2010° = [(11) * (180) + 30]° and
1995° = [(11) * (180) + 15]°
= (1 - cot2010°) (1 - cot1995°)
= (1 - cot30°) (1 - cot15°)
= (1 - cot30° - cot15° + cot30° cot15°)
= (1 - (cot30° + cot15° - cot30° cot15°) ...1

now as 30° and 15° add to give 45°

and sum and product have come above let us see if we can get combination

45° = 30° + 15°

so cot 45° = cot(30°+15°) = (cot 30° cot 15° -1 )/(cot 30°+ cot 15°)
or 1 = (cot 30° cot 15° -1 )/(cot 30°+ cot 15°)

or (cot 30°+ cot 15°) = (cot 30° cot 15° -1 )
or (cot 30°+ cot 15°) - (cot 30° cot 15°) = - 1
from 1 and above we get sum as 1+ 1 = 2

2011/075) 2+4+8+16+32+... equals -1

for a proof of it refer to

adding past infinity

of course it is not rue.

where is the fallacy

the problem is in the line

2+ 4 +8+ 16 ...
-1 -2 -4 -8 -16 + ...

if we realise that in the 1st line there is one extra term and that is not infinitesimal to be ignored( the sum diverges and does not converge we see that one term that should not be left out is ignored

2011/074) If the ratio of GM and HM of 2 numbers is 5: 4 find the ratio of numbers?

without loss of generality let the numbers be a and at

as GM^2 = AM * HM
GM/HM = AM/GM = (a+at)/(2a sqrt(t)) = 5/4

so (1+t)/( 2sqrt(t) = 5/4 and putting sqrt(t) = x

(1+x^2) / (2x)= 5/4

or 4(1+x^2) = 10 x
o (4x^2-10x + 4) = 0 or (2(2x-1)(x-2) = 0 or x = 1/2 or 2 or t = 1/4 or 4

so ratio = 1 : 4 (or 4:1)

2011/073) Solve the system of quadratic equations

5x^2 + 4xy + 5y^2 + 3x + 3y = 74
x^2 + 2xy + y^2 - 6x - 6y = -8

5x^2 + 4xy + 5y^2 + 3x + 3y = 74

the coefficent of x^ and y^2 are same and so is of x and y

so we combine accordingly

5(x+y)^2 - 6xy + 3(x+y) = 74 ... 1


from 2nd equation by same anology

(x+y)^2 - 6(x+y) = -8

putting x + y = t

t^2 - 6t +8 = 0 or (t-4)(t-2) = 0 so t = 4 or 2

5(x+y)^2 - 6xy + 3(x+y) = 74
80-6xy + 12 = 74
or 6xy = 18
or xy - 3

x+ y= 4 .. 2
xy = 3 ... 3

from 2 and 3 we get (x-y)^2 = (x+y)^2 - 4xy = 16-12 = 4 or x-y = 2 or -2

x+y = 4 and x- y = 2 give x = 3 y = 1
x+y = 4 x-y = -2 give x = 1 and y = 3

putting x+ y =2 in 1 we get
5(x+y)^2 - 6xy + 3(x+y) = 74
20 - 6xy + 6 = 74
6xy = 48
xy = -8

so x = 4 y = -2 or x= -2 y = 4 (we can clove by above method)

so we get
(1,3), (3,1)(4,-2),(-2,4) 4 set of solutions

2011/072) Prove that ( 1+ cos π/8)( 1+ cos 3π/8)( 1+ cos 5π/8)( 1+ cos 7π/8) =1/8?

(1 + cos π/8)(1 + cos 3π/8)(1 + cos 5π/8)(1 + cos 7π/8)
= (1 + cos π/8)(1 + cos 3π/8)(1 - cos 3π/8)(1 - cos π/8), since cos(π - t) = -cos t
= (1 - cos² π/8)(1 - cos² 3π/8)
= (1 - cos² π/8)(1 - sin² π/8), via cos(π/2 - t) = sin t
= (1 - cos² π/8)(1 - sin² π/8)
= sin ^2 π/8 cos ^2 π/8
= (1/4) (2 sin π/8 cos π/8)²
= (1/4) (sin 2π/8)², by double angle formula
= (1/4) (sin π/4)²
= (1/4) (1/√2)²
= (1/4)(1/2)
= 1/8.

2011/071) Show that 2 tan 20˚ + 4 tan 40˚ + 8 tan 80˚ = 9 (cot 10˚ - tan 10˚)

we first prove one relationship
cot x - tan x = cos x/ sin x - sin x/ cos x = (cos^2 x - sin ^2 x) /( sinx cos x) = 2 cot 2x

so

cot x - tan x = 2 cot 2x ...1

also
tan x = cot x - 2 cot 2x ...2

from 2 we generate following 3 relationships

2 tan 20= 2 cot 20 - 4 cot 40 ... 3
4 tan 40 = 4 cot 40 - 8 cot 40... 4
8 tan 80 = 8 cot 80 - 16 cot 160... 5
adding we get
2 tan 20˚ + 4 tan 40˚ + 8 tan 80 = 2 cot 20 - 16 cot 160
= 2 cot 20 + 16 cot(180-160)as cot x = - cot(180-x)
= 2 cot 20 + 16 cot 20
= 18 cot 20
= 9( 2 cot 20)
= 9( cot 10 - tan 10)from 1
proved

20111/070) If a+b+c=1, ab+bc+ca=2, abc=3, then find the value of 1/(a+bc)+1/(b+ca)+1/(c+ab).

we have
a+b+c=1 ...1
ab+bc+ca=2 .. 2
abc=3 ...3

so a b c are roots of equation

x^3-x^2 + 2x - 3 = 0

further

as a+ b+ c = 1

so a = 1- (b+c)
so a + bc = 1-(b+c) + bc = (1-b)(1-c)

similarly b+ ca = (1-c)(1-a)
c + ba = (1-b)(1-a)

we need to find 1/(a+bc)+1/(b+ca)+1/(c+ab) = 1/ ((1-b)(1-c)) + 1/ ((1-c)(1-a)) + 1/ ((1-b)(1-a))

= (1-a) + (1-b) + (1-c)/(1-a)(1-b)(1-c) = ((a-1) + (b-1) + (c-1))/(a-1)(b-1)(c-1)



so we need to form an equation whose roots are a-1 , b- 1 and c- 1

as a b and c are roots of f(x) = x^3-x^2 + 2x - 3

so a-1 b- 1 and c- 1 are roots of

f(x+1) = (x+1)^3 - (x+1)^2 + 2(x+1) - 3 = x^3 + 2 x^2 + 3x -1

so (a-1) + (b-1) + (c-1) = - 2

(a-1)(b-1)(c-1) = 1

so given expression = ((a-1) + (b-1) + (c-1))/((a-1)(b-1)(c-1)) = - 2

2011/069 ) Factor the following?

(x - 9753)(x -7531) + 234321

this can be factored by multiplying out and solving as quadratic however let is take x –a = t where a = (9753+7531)/2 = 8642

now 9753-8642 = 8642-7531 = 1111

so (x - 9753)(x -7531) + 234321
= (x-8642-1111)(x-8642 + 1111) + 234321
= (t-1111)(t+1111) + 234321 taking t = x- 8642
= t^2 – 1111^2 + 234321
= t^2- 1234321 + 234321
= t^2- 1000000
= t^2- 1000^2
= (t+1000)(t-1000)
= (x-8642+1000)(x-8642-1000)
= (x-7642)(x-9642)

This trick helps in keeping expression simple

2011/068) Find all ordered pairs of integers (x,y) which satisfy x^3 + y^3 - 3x^2 + 6y^2 + 3x + 12y + 6 = 0

Rewrite this using x terms and y terms in groups

(x^3 - 3x^2 + 3x) + (y^3 + 6y^2 + 12y) = -6
==> (x^3 - 3x^2 + 3x - 1) + (y^3 + 6y^2 + 12y + 8) = -6 - 1 + 8
==> (x - 1)^3 + (y + 2)^3 = 1
as difference of 2 cubes cannot be 1 so one of them is zero and another is 1 and there is no other choice

x - 1 = 1 and y + 2 = 0 ==> (x, y) = (2, -2), or
x - 1 = 0 and y + 2 = 1 ==> (x, y) = (1, -1).

so solution sets are (2,-2) and (1,-1)

2011/067) factor $a^6 + 18a^3 + 125$

the above is not difficult though it looks so

we realize $a^6 = (a^2)^3$ and $125 = 5^3$

now if we can split (is it possible) 18 so that $18 = x^3 + y$ and  $-y = 15 x$  (which is true x = 3 and y = -45)

we get $a^6 + 18a^3 + 125 = a^6 + 27 a^3 - 45 a^3 + 125$
$= (a^2)^3 + (3a)^2 + 5^3 - 3(a^2)(3a) 5$

and using $x^3 + y^3 + z^3 = (x+y+z)(x^2+y^2+z^2- xy-yz-zx)$
we get
$(a² - 3a + 5)((a^4) + 9a^2 + 25 + 3(a^3) + 15a - 5a²)$

$= (a² - 3a + 5)((a^4) + 3(a^3) + 4a² + 15a + 25)$

2011/066) find near issoceles triangles

that is the 2 legs ar of length x, x+ 1
let legs be x and x+1 and hypotenuse y

x^2+ (x+1)^2 = y^2
or 2x^2 + 2x + 1 = y^2
or doubling

(4x^2+4x+2) = 2y^2
or (2x+1)^2 = 2y^2 -1
putting 2x +1 = m
we get 2y^2-m ^2= 1

now by trial we see that y = 5 and m = 7 is a solution
if (a,b) is a solution then (2a^2-b^2) = 1

so (a sqrt(2) – b) (a sqrt(2) + b) = 1
so | (a sqrt(2) – b) | = 1
cubing both sides ( sqaring shall not help as we get ( a+b- 2ab(sqrt(2)) this arises as |1| = |-1| = 1
(2a^3 + 3ab^2) sqrt(2) –(12a^2 + b^3) (the sqrt(2) term is positive and the non sqrt(2) is – ve)
So it is a solution
Starting with (5,7) we get (985,1393) or the triple is ( 696,697,845)
So we can get the terms as far as possible.
But It shall not give all solutions

2011/065) If sides of length a b c form a triangle prove sqrt(a), sqrt(b), sqrt(c) form a triangle

Let's assume a >= b >= c > 0. As a, b, c are sides of a triangle, they fit the triangle inequality a < b + c.

(1) (sqrt(b) + sqrt(c))^2 = b + 2*sqrt(b)*sqrt(c) + c

Value of sqrt(x) is positive for any positive x, so

(2) (sqrt(b) + sqrt(c))^2 > b + c > a

As square root is an increasing function, (2) follows

(3) sqrt(b) + sqrt(c) > sqrt(a)

Hence proved

2011/064) Solve 6x^5-41x^4+97x^3-97x^2 + 41x – 6 = 0

This as the coefficients are afternates signs and symmetric so they can be combined as below

6(x^5-1) – 41(x^4- x) + 97(x^3-1)
= 6(x-1)(x^4 + x^3 + x^2 + x + 1) – 41x(x-1)(x^2+x + 1) + 97x^2(x-1)
= (x-1)( 6x^4 + 6x^3 + 6x ^2 + 6x + 1 – 41x^3- 41 x^2 – 41 + 97 x^2 )
= (x-1)(6x^4 – 35x^3 + 62x^3 – 35 + 6)
So x= 1 is a solution and other factor is
(6x^4 – 35x^3 + 62x^2 – 35 + 6)
As the coefficients as symmetric we have combining similarly

6(x^4+1) – 35x(x^2+1) + 62 x^2 ...1
Taking x^2 out we get symmetric

X^2(6(x^2+ 1/x^2) - 35(x+1/x) + 62)

Put x+ 1/x = t o get
X^2(6(t^2-2) - 35 t + 62) = 0 and as x is not zero we get
(6(t^2-2) - 35 t + 62 = 0
Or 6t^2 - 35 t + 50 = 0
Ot (2t-5) (3t – 10) = 0
t= 5/2 or 10/3
now x + 1/x = 5/2 gives x ^2 + 1- 5/2 x= 0 or (4x^2 + 4x- 5) = 0 or (2x-1)(x-2) or x = ½ or 2
x + 1/x = 10/3 gives x ^2 + 1- 10/3 x= 0 or (3x^2 + 3x- 10) = 0 or (3x-1)(x-3) or x = 1/3 or 3

so solution = ½ , 1/3 , 1 , 2 , 3

you can also proceed from 1 putting x^2+1 = tx

2011/063) factor x^5 + x^4 + 1

This is a qunitic polynomial for which there is no direct method
(
one of the methods commonly used is given below at the end but a short cut can be used

(since using the Rational Root Theorem implies that the only possible rational roots are -1 and 1, and neither (x - 1) nor (x + 1) are factors).

as it is of the form x^(3n+2) + x^(3m+1) + 1 so
x = w and x = w^2 are zeros ( w is cube root of 1)
(x-w)(x-w^2) or (x^2 + x + 1) is a factor and by division

x^5 + x^4 + 1 = (x^2+x+1)(x³ - x + 1)


The common method
It's easy to check that x^5 + x^4 + 1 has no linear factors (since using the Rational Root Theorem implies that the only possible rational roots are -1 and 1, and neither (x - 1) nor (x + 1) are factors).

So, if x^5 + x^4 + 1 factors, then it must be the product of a quadratic factor and a cubic factor.
Hence, we assume that x^5 + x^4 + 1 = (x² + Ax + B)(x³ + Cx² + Dx + E).

Expanding yields
x^5 + x^4 + 1 = x^5 + (A + C)x^4 + (B + AC + D)x³ + (AD + BC + E)x² + (BD + AE)x + BE

Equate like coefficients:
A + C = 1
B + AC + D = 0
AD + BC + E = 0
BD + AE = 0
BE = 1.

Note that C = 1 - A and E = 1/B.
We can say more; since these variables should be integers, we know that B = E = ±1.

Thus, we obtain
±1 + A(1 - A) + D = 0
AD ± (1 - A) ± 1 = 0
±D ± A = 0

Let's look at the +1 case (it turns out that we don't need the -1 case...)
1 + A(1 - A) + D = 0
AD + (1 - A) + 1 = 0
D + A = 0

Since D = -A, we have
1 + A(1 - A) - A = 0 ==> A^2 - 1 = 0
-A^2 + (1 - A) + 1 = 0 ==> A^2 + A - 2 = 0.
These two equations are both true when A = 1.

Hence assuming that B = E = 1, we have A = 1, D = -1 ==> C = 0.
------------------
So, we have the factorization x^5 + x^4 + 1 = (x² + x + 1)(x³ - x + 1).)

2011/062) The sum of two natural numbers when added to their LCM, gives a total of 143.? How many such pairs of numbers exis

let the numbers be mx and my with x and y co prime

so LCM = mxy

so mx + my + mxy = 143

or m(x+y+xy) = 143

add m on both sides

m(x+y+xy+1) = 143+m

m(x+1)(y+1) = 143 +m or (x+1)(y+1) = 143/m + 1

so m is factor or 143 can be 1,11,13, 143

m = 1=> (x+1)(y+1) = 144
144 = 2 * 72( x=1 y = 71)
= 3 * 48 (2,47)
= 4 * 36(3,35)
= 6 * 24( 5,23)
= 8 * 18 (7,17)
= 9 * 16(8,15)
= 12 * 12 (11,11)
all except (11,11) are coprimes so solution

m = 11 => (x+1)(y+1) = 14 , x + 1 = 2, x = 1 and y+1 = 7 y = 6 so 11 and 66
m = 13 => (x+1)(y+1) = 12 (x=1, y = 5), (x=2, y = 3) so 13, 65 or 26,39
m = 143 => (x+1)(y+1) = 2 so no solution

so 9 solutions (1,71),(5,23), (8,15), (7,17),(3,35), (2,47) (11,66), (13,65), (26,39)

2011/061) if x = 0.666... (base 8) and y = 0.555... (base 6), what is the value of y - x in base 7?

we need to compute x and y say in base 10

x = .6666 base 8
= (6/8)( 1 + (1/8)^2 + (1/8)^2 + ....
= (6/8) /(1-(1/8)) = (6/7)

y = .55555 (base 6) = 5/6( 1+ 1/6 + (1/6)^2 ....) = (5/6)/(1-1/6) = 1
hence
y - x = 1- 6/7 = 1/7 and hence .1 in base 7

2011/060)If a^2=5a-3, b^2=5b-3, then equation having a/b and b/a as its roots is___?

a is not equal to b
A. 3x^2+19x+3
B. 3x^2-19x+3
C. 3x^2-19x-3
D. x^2-16x+1
Ans:
since a^2=5a-3, b^2=5b-3
a,b are soln's for the eqn. x^2=5x-3
ie x^2-5x+3=0
and they are not same so they are 2 roots of x^2-5x+3=0

hence sum of roots =a+b= -(-5/1) =5
and product of roots=ab=+( 3/1)=3

now if a/b and b/a are roots then

sum of the roots is
a/b + b/a = a^2 + b^2 /ab =(a+b)^2 -2ab /ab =5^2 -2*3 /3 =25-6 /3=19/3

and product is a/b *b/a =1

so eqn wth roots a/b&*b/a is
x^2 -(19/3)x +1 =0
=>3x^2-19x+3=0

hence the ans is B

2011/059) For all n>= 1 , prove that:- (1)+(1/1+2) + (1/1+2+3)+ ....... + (1/1+2+3+......+n)=(2n/n+1)

this type of sum come under telescopic sum

k th term = 1/(1+2 +3 + ... k) = 2/(k(k+i) = 2 * ( 1/k - 1/(k+1))

1st term = 2(1/1 - 1/2)
2nd term = 2(1/2 - 1/3)

nth term = 2( 1/n - 1/(n+1))

sum (all terms except 1st and last cancel) = 2/(1/1- 1/(n+1)) = 2n/(n+1)

2011/058)find the highest power of 2 that divides a number from n to m

for example from 12 to 14 the highest power of 2 that divides is 2^2 that divides 12.

this can be done by factoring all numbers from n to m in form 2^k * l and then finding the highest k however that can be a very slow process. and this can be improved signifcantly

let n = 2^p * q where q is odd( p = 0 in case n is odd)

next multiple of 2^p shall be n + 2^p

so we have algorithm

let n = 2 ^p* q

set r = 2 ^ p + q (1st number)
loop step:
if r + 2^p > m exit
add 2 ^p to r and 1 to q( r = r + 2^p , q = q + 1)
keep deviding q by 2 adding 1 to p until q is odd
go to loop step:
r is the value and p is the highest power

suppose we need to find between 21 and 31

21 = 2^0 * 21
add 1 22 = 2 ^ 0 * 22 or 2^1 * 11
add 2^1 or 2, 24 = 2^1 * 12 or 2^3 *3
add 2^3 or 8 we get 32 > 31

so and = 24 = 2^3 * 3

2011/057) Given |a | < 1 and |b| < 1

If $1+a+a^2+\cdots =x$ and $1+b+b^2+\cdots.=y$ then find $1+ab+a^2.b^2+.... $ in terms of x and y

Solution
we have
$x = \dfrac{1}{1-a}$
or $(1-a) = \dfrac{1}{x}$
or $a = 1 – \dfrac{1}{x} = \dfrac{x-1}{x}$
similarly
$b = \dfrac{y-1}{y}$
now as |a | and |b| both < 1 so |ab| < 1 and hence
$1+ ab + a^2b^2 + a^3b^3 + \cdots= \dfrac{1}{1-ab} = \dfrac{1}{1- \frac{x-1}{x} * \frac{y-1}{y}}$
= $\dfrac{xy}{xy - (x - 1)(y - 1)}$
= $\dfrac{xy}{x + y - 1}$ 

2011/056) Find lim n-> inf (n/(n+2))^n

(n/(n+2))^n
= (1/(1+2/n)^n
= ((1/(1+2/n)^n/2)^2

As (1+1/x) ^x as x -> infinite is e so /(1+2/n)^n/2 = e so given limit = 1/e^2

2011/055) Find the sum of rational terms of (2^(1/2) + 3^(1/5))^ 10

Let a rational term be (10Cn) (2^1/2)^n (3(^1/5)^(10-n)

Now n/2 and (10-n)/5 need to be integers

n = 0 and n =1 10 satisfy the criteria and for n = 0 we get 9 and for n = 10 we get 32 so sum of rational terms = 41

2011/054) If a = e^(2ipi/7) and f(x) = A0 + sum ( k = 1 to 20) Ak x^k find the value of f(a) + f(ax) + f(a^2x) + f(a^3x) + + f(a^4x)+ + f(a^5x) + f(a^6x)

We have a = e^(2ipi/7)

So a^7 = e^(2ipi) = 1

or a^(7 t) = 1 or

or (a^7t) – 1 = 0

or (a^t-1)(a^t + a^2t + a^3t + a^4 t + a^5 t + a^6 t) = 0

if t is multiple of 7 then a^k = 1

and if t is not multiple of 7 then a^t-1 is not zero so a^t + a^2t + a^3t + a^4 t + a^5 t + a^6 t) = 0

so in f(a) + f(ax) + f(a^2x) + f(a^3x) + + f(a^4x)+ + f(a^5x) + f(a^6x)

for k multiple of 7 sum is 7 and for k not multiple of 7 sum of x^k = 0

so sum = 7 A0 + 7A7 x^7 + 7 A14 x^14

2011/053) find the relation between a and b when a^3+b^3 = 8 - 6 ab

take 8 to the left

a^3+b^3-8 = -6 ab
or a^3+b^3 + (-2)^3 = 3(-2) a b

as x^3 +y^3 + z^3 = 3xyz then x = y = z or x + y+ z = 0

comparing we get a = b = -2 or a + b- 2 = 0

that is a= b = -2 or a + b = 2

2011/052) prove that x^2+3x+3 is a factor of (x+1)^(n+1) + (x+2)^(2n-1)

for integer n > = 1


This can be proved by induction

n =1 gives (x+1)^(n+1) + (x+2)^(2n-1) = (x+1)^2 + (x+2) = x^2 + 2x + 1 + x + 2 = x^2 + 3x + 3 divisible

so the 1st step is proved

let it be true for n = k.
we need to prove for n= k+ 1

let f(k) = (x+1)^(k+1) + (x+ 2)^(2k- 1)

so f(k+1) = (x+1)^(k+2) + (x+2)^(2k+ 1)
= (x+1)(x+1)^k + (x+2)^2 *(x+2)^(2k-1)
= (x+1)(x+1)^k + (x^2+4x + 4) *(x+2)^(2k-1)
= (x+1)(x+1)^k + (x + 1 + x^2+3x + 3) *(x+2)^(2k-1)
= (x+1)(x+1)^k + (x + 1)(x+2)^(2k-1) + (x^2+3x + 3) *(x+2)^(2k-1)
= (x+1)[(x+1)^k + (x+2)^(2k-1)] + (x^2+3x + 3) *(x+2)^(2k-1)
= (x+1) f(k) + (x^2+3x + 3) *(x+2)^(2k-1)
as both terms are divisible so f(k+1) is divisible

so step of induction is proved

hence proved

2011/051) if a+ib is root of equation x^3+px + q = 0 the form an equation whose root is 2a

and p and q are real

a + ib is a root and coefficients are real so another root = a - ib

sum of roots = 0 (coefficient of x^2 =0) so 3rd root = -2a

now -2a is root of f(x) = x^3+ px + q = 0

so 2a is root of f(-x) = (-x)^3 - px + q = 0 or x^3+px -q = 0

2011/050) If a and b are roots of the equation x^2 – 2x cos t +1 = 0 form the equation whose roots are a^n and b^n

As a and b are roots so

a + b = 2 cos t … 1
ab = 1 …2

we have (a-b)^2 = (a+b)^2 – 4ab = 4 cos^2 t – 4 = - 4 sin ^2 t

or a-b = 2 i sin t … 3

from 1 and 3 adding 2a = 2 (cos t + i sin t) = 2 e^it or a= e^it
subtracting 2b = 2 (cos t - i sin t) = 2 e^it or b = e^-it

now a^n + b^n = (e^int + e^-int) = 2 cos nt

and a^nb^n = (ab)^n = 1

so the equation = (x^2- 2x cos nt + 1) = 0

2011/049) the value of a for which the equation

(a^2+4a+3) x^2 + (a^2-a-2) x + (a+1)a = 0 has more than 2 solutions is


ans:
if above is quadratic equation then it has 2 solutions and hence it must be identity so
a^2+4a + 3 = a(a+1) = 0
(a^2-a-2) = (a+1)(a-2) = 0
a(a+1) = 0

hence a+ 1 = 0 or a = -1

2011/048) show that (a+bw+cw^2) ^3 + (a + bw^2_+cw)^3 = (2a-b-c)(2b-a-c)(2c-a-b) when w is cube root of unity

let x = (a+bw+cw^2) ...1
and y = (a + bw^2+cw) ...2

we need to factor x^3 + y^3

x^3+y^3 = (x+y) ((x+y)^2 - 3 xy) ...3

add (2) and (1) to get

x+y = (2a+b(w+w^2)+ c(w+w^2))
= (2a -b - c) ... 4

now xy = (a+bw+cw^2)(a + bw^2+cw)
= (a^2 + b^2 + c^2 + ab(w+w^2) + bc(w+w^2) + ca(w+w^2))
= (a^2+b^2 + c^2 - ab - bc - ca)

so (x+y)^2 - 3xy
= (2a-b -c)^2 - 3(a^2+b^2 + c^2 - ab - bc - ca)
= (4a^2 + b^2 + c^2 - 4ab + 2bc - 4ac)- 3(a^2+b^2 + c^2 - ab - bc - ca)
= a^2 -2b^2 - 2c ^2 - ab - ac + 5 bc
= a^2 - a(b+c) - (2b^2 + 2c^2 - 5bc)
= a^2 -(a (2b-c) + (2c-b)) - (2b-c)(b-2c))
= a^2 -(a (2b-c) + (2c-b)) + (2b-c)(2c-b))
= (a - 2b + c )(a-2c + b)= (2b-a -c)(2 c-a -b )... 5


from 3 4 and 5 we get the result

2011/047) the value of the expression

2(1+w)(1+w^2) + 3(2w+1)(2w^1+1) + ... (n+1)(nw+1)(nw^2+1) when w is cube root of unit is

1) n^2 (n + 1)^2/4
2) n^2 (n + 1)^2/4 -n
3) n^2 (n + 1)^2/4 + n
4_ no of these

ans 3) because

kth term = (k+1)(kw+1)(kw^2+1) = (k+1) (k^2w^3+ k(w+w^2) + 1)
= (k+1)(k^2 -k +1) as w^3 = 1 and w + w^2- 1
= (k^3 + 1)

so sum = (sum of cubes upto n) + n
= (n(n+1)/2)^2 + n
= n^2(n+1)^2 /4 + n

2011/046) let n be any positive integer >1. then show that n^4+4^n is a composite number

there are 2 cases

1) n is even.

n^4+4^n > 2 and even so composite

2) n is odd

n^4 + 4^n
= n^4 + 2n^2*2^n + 4^n - 2n^2*2^n
= (n^2 + 2^n)^2 - 2n^2*2^n
= (n^2 + 2^n)^2 - n^2*2^(n+1)
= (n^2 + 2^n - n * 2^((n+1)/2)) * (n^2 + 2^n + n * 2^((n+1)/2)). Therefore,

if n is odd it has 2 factors so composite if 1st term is not 1 as second term > 1

for n > 1 1st term is > 1 and hence the given expression is composite

2011/045) What is the remainder when x^81 + x^49 + x^25 + x^9 + x is divided by x^3 - 1

when we divide by x^3-1 it shall be a maximal quadratic polynomial

ax^2 + bx + c

x^81 - 1 is divisible by x^3 -1 so x^81 divided by x^3-1 leaves 1

x^48 -1 divided by x^3 -1 leaves 0

so x(x^48-1) divided by x^3-1 leaves 0
x^49 divided by x^3-1 leaves x

x^25 divided by x^3-1 leaves x( similarly)
x^9 divided by x^3-1 leaves 1
so adding all the above we get remainder 1 + x + x + 1+ x = 3x + 2

2011/044) show that $\sin \frac{\pi}{14} \sin \frac{3\pi}{14} \sin\frac{5\pi}{14} = \frac{1}{8}$

LHS
$= \sin \frac{\pi}{14} \cos (\frac{\pi}{2} -\frac{3\pi}{14}) \cos(\frac{\pi}{2} -\frac{5\pi}{14})$

$= \sin \frac{\pi}{14} \cos (\frac{4\pi}{14})  \cos (\frac{2 \pi}{14})$

$= \sin \frac{\pi}{14} \cos (\frac{2\pi}{14})  \cos (\frac{4 \pi}{14})$
$= \dfrac{\cos \frac{\pi}{14} \sin \frac{\pi}{14} \cos (\frac{2\pi}{14})  \cos (\frac{4 \pi}{14})}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{2}\frac{\sin \frac{2\pi}{14} \cos (\frac{2\pi}{14})  \cos (\frac{4 \pi}{14})}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{4}\dfrac{\sin \frac{4\pi}{14}  \cos (\frac{4 \pi}{14})}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{8}\dfrac{\sin \frac{8\pi}{14}}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{8}\dfrac{\sin(\pi -  \frac{8\pi}{14})}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{8}\dfrac{\sin(\frac{6\pi}{14}}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{8}\dfrac{\cos(\frac{\pi}{2} -  \frac{6\pi}{14})}{\cos \frac{\pi}{14}} $
$=\dfrac{1}{8}\dfrac{\cos(\frac{\pi}{2} -  \frac{6\pi}{14})}{\cos \frac{\pi}{14}} $
 $=\dfrac{1}{8}\frac{\cos \frac{\pi}{14}}{\cos \frac{\pi}{14}} $
$=\frac{1}{8}$

edited the above based on the comment to keep the flow

2011/043) find limit $\lim_{n\to\infty} \sqrt[n]{3^n+4^n}$

We can we take 3 or 4 out but taking 3 it diverges

we get $3 \sqrt[n]{1^n+(\frac{4}{3})^n}$ and reach no where

taking 4 we get
$4 \sqrt[n]{1^n+(\frac{3}{4})^n}$

now $1 < 1+(\frac{3}{4})^n < 2$

so  $\sqrt[n]{1} < \sqrt[n]{1+(\frac{3}{4})^n} < \sqrt[n]{2} $ and as $\sqrt[n]{2}$ as n goes to infinite goes to 1

so result = 4 * 1 or 4.

2011/042) Given $x^2+y^2 = 1$ and $r^2 + s^2 =1$ maximise $rx + sy$

solution:
this can be done by AM GM inequality also but here I show a different way

as $x^2+y^2 = 1 $so let $x = \sin\, t , y= \cos\, t$
and $r^2+s^2 = 1$ so let $r = \sin\, a , s = \cos\, a$

$xr+ sy = \sin\, t \sin\, a+ \cos\, t \cos\, a = cos (t-a)$

maximum when $t = a$ that is 1 ( it is possible when x = r = 1 and y = s = 0)
so maximum 1 is possible
so 1 is the result

2011/041) factoring by diminishing power $6bc - 9c² - 12cd - 8be + 12ce +16de$

one of the factoring that I saw in the net was factor

$6bc - 9c² - 12cd - 8be + 12ce +16de$

...the "clue" you need to recognize is that the coefficients of the first three terms have a common factor...and the coefficients of the last three terms have a common factor...and, the ratios are constant...so, group accordingly...

$(6bc - 9c^2 - 12cd) - (8be - 12ce - 16de) = 3c(2b - 3c - 4d) - 4e(2b - 3c - 4d) =$
$(2b - 3c - 4d)(3c - 4e)$

the above is correct and it was luck that grouping was there but what is luck is not there
then see that highest power of c is 2 and keep them in descending order

$ 6bc - 9c² - 12cd - 8be + 12ce +16de$
$= - 9c² + 6bc - 12cd + 12ce - 8be + 16de$
now you can factor as quadratic in c eliminating the luck factor
$= - 9c² - + 6c( b - 2d + 2e) -8e(b- 2d)$
$= - 9c² - + 6c(( b - 2d) + 2e)) -8e(b- 2d)$
letting b - 2d = a we get
$= - 9c^2 + 6c(a + 2e) - 8ea$
$= - 9c^2 + 6ca + 12ec - 8ea$
$= - 3c(3c - 2a) + 4e(3c-2a)$
$= (3c-2a)(4e - 3c)$
$= (3c-2b+4d)(4e-3c)$
which is same as 1st one

I do not mean to say that 2nd one is preferable to 1st but 2nd one can be used when 1st one does not work

2011/040) Boolean algebra simplify A'(A+B) + (B+AA) (A+B')

A'(A+B)
= A'A + A'B
= A'B as A'A = 0 for any A
(B+AA) (A+B')
= ( B + A) (A + B') as AA = A
= ( B A + AA + BB' + AB'
= BA + A + AB' as AA = A and BB' = 0
= A + BA as A + AB' = A
so A'(A+B) + (B+AA) (A+B')
= A'B + A + BA
= A + B(A'+A)
= A + B as A' + A = 1

2011/039) prove that (-1)* (-1) = 1

a lot of proof of the same is there including the proof below

we know
(-1) + 1 = 0

multiply both sides by -1 to get
(-1)*(-1) + 1*(-1) = 0
add 1* (1) on both sides

to get (-1)*(-1) + 1*(-1) + 1* (1) = 1
or (-1)*(-1) + (1*(-1) + 1* (1))= 1
or(-1)*(-1) + 0= 1

or (-1)*(-1) = 1

proved

However all the proofs including the above come from back ward calculation from LDMA
The LDMA shall fail unless we define it the above way which is a big sacrifice and hence it need to be defined this way to maintain the consistency of LDMA.

( similarly 0! is defined to be 1 so that law of combinatorial holds

2011/038) Find lim n-> inf (n+1)^1/2 – n ^(1/2)

This is of the form inf- inf

We can solve it in 2 ways

1) binomial expansion
(n+1)^(1/2) – n ^(1/2) = n^(1/2) + ½(1/n(^(1/2)) + … - n^(/12)
= ½(1/n(^(1/2)) = 0 as n -> inf

2) by rationalising the numerator
(n+1)^(1/2) – n ^(1/2) = ((n+1)^(1/2) – n ^(1/2) ((n+1)^(1/2) + n ^(1/2) / ((n+1)^(1/2) + n ^(1/2)
= 1/(n+1)^(1/2) + n ^(1/2))
= 1/(inf + inf) = 0

The above method shows rationalising the numerator instead of denominator which is generally used can be used as a tool

2011/037) If α is a root of equation x² + x + 1 = 0 then show that α is a root of equation x^(3p) + x^(3q+1) + x^(3m+ 2) = 0

Proof:
α is a root of equation x² + x + 1 = 0

so α^2 + α + 1 = 0 and hence α^3 =1

now there are 2 approaches

1)

α ^(3p) + α ^(3q+1) + α ^(3m+ 2)
= (α ^3)^p + (α ^3)^3q* α +( α ^3)3m* α ^2
= 1 + α+ α^2
= 0
Hence α is a root

2)

α ^(3p) + α ^(3q+1) + α ^(3m+ 2)
= α ^(3p) + α ^(3p+1) + α ^(3p+ 2) + α ^(3q+1) - α ^(3p+ 1) + α ^(3m+ 2) - α ^(3p+ 2)
= α ^(3p)( 1+ α + α ^ 2) + α (α ^(3q) - α ^(3p)) + + α^2 (α ^(3m) - α ^(3p))
= 0 + 0+ 0 = 0 as each part is zero
Hence α is a root


as a corollary we see that x² + x + 1 devides x^(3p) + x^(3q+1) + x^(3m+ 2)

2011/036) Find z^2010 where The complex number z is such that z^9 = zz', where z and z' are complex conjugates.

let z = r e^(it)

then z^9 = r^9 e^(9it)

zz' = r^2 = r^9 e^(9it)

so e^(9it) = r^7 or r = 1 and hence zz' = 1

so z^9 = 1 so z^2007 = 1 and so z^2010 = z^3 = cube root of 1

z = r = 0 is also another solution

so solutions are 3 cube roots of 1 and zero

2011/035) Prove that if the sides of a triangle are prime numbers its surface can not be whole number.

proof:
Let the sides are a,b c, and a <=b <= c

now there are 2 cases

a = 2 or all are odd

now A^2 = ( a+b-c)(a+b+c)(a-b+c)(b+c-a)/ 4
if all are odd then all 4 terms on the RHS are odd then A^2 cannot be integer so A cannot be whole number

case 2:
for a= 2 and b = 2 or a= 2 and b != 2

a = 2 then b =2 => c = 2 or 3

a =2 b = 2 c = 2 => A^2 = 6*2^3/4 = 12 so A is not integer

a =2 , b= 2 c = 3 => A^2 = 7 * 1 * 3 * 3/4 so A is not integer


if b != 2 then b= c because if c >b then c>= b+2 or a+b= c

so we get A^2 = (2+2b)* (2b-2)* b^2 /4 = b^2(b^2-1)/4 cannot be a perfect square


so no solution

2011/034) show that 3^2n+ 3^n + 1 is divisible by 13 if n is not multiple of 3

proof:

we know 3^3 = 27 = 1 mod 13

so we proceed

if n is of the form 3k + 2 then 2n of form 3m + 1 and if n is of the form 3k + 1 the 2n of the form 3m + 2

so we have
3^2n+ 3^n + 1 = 3^ (3k + 2) + 3^(3m + 1) + 1

= 27^k * 9 + 27^m * 3 + 1
= 9 + 3 +1 mod 13 = 13 mod 13 = 0 mod 13 do divisible by 13

2011/033) Prove: If p and q are distinct primes, show that p^q + q^p ≅ (p + q) mod pq

q is prime

so p^(q-1) mod q = 1 as per FLT


so p^(q-1) = mq + 1

multiply by p on both sides p^q = mpq + p = p mod pq

similarly as q is prime q^p = p mod pq

adding we get (p^q + q^p) mod pq = (p+q)

proved

2011/032) factorize x^4+2x^3y-3x^2y^2-4xy^3-y^4?

seeing coefficient of x^4 as 1 and y^4 is -1 and this is homogeneous in x y

that is power of x and y sum is 4 that is same

we see that if shall be (x^2+axy + y^2)(x^2+bxy-y^2)

= x^2 + x^3y(a+b) + x^2y^2(ab) + (b-a) xy^2 - y^4

so a + b = 2, ab = -3 and a-b = 4

a + b = 2, and a-b = 4 give a = 3 and b = -1 and this satisfies
ab = -3

so we get (x^2+axy + y^2)(x^2+bxy-y^2)
= (x^2+3xy + y^2)(x^2-xy-y^2)

2011/031) If f(x) + 2f(2002/x) = 3x then find f(2)?

ans: this type of problem we need to eliminate f(2002/x)and put x = 2

(x) + 2f(2002/x) = 3x --- 1
we realize that x and 2002/x are reciprocal so putting 2002/x for x we get

f(2002/x) + 2 f(x) = 3(2002/x) ... 2

multiply 2nd by 2 and subtract (1) fom it

3 f(x) = 6(2002/x) - 3 x

or f(x) = 4004/ x - x

putting x = 2 we get

f(2) = 2000

2011/030) prove that

(1....1) = (22....2)+ (3333....3)^2
2n times ntimes + n times

Proof:

LHS = (99999...9)/9 = (10^2n -1 )/ 9
RHS = 2/9( 10^n-1)+ (10^n-1)/9

so we need to prove that (RHS = LHS as RHS is more complex)

(10^n-1 )^2 + 2(10^n-1) = (10^2n- 1)

LHS = 10^2n - 2* 10^n +1 + 2*10^n - 2 = (10^2n- 1)

2011/029) can x^y be rational for both irrational x and y

we know that sum and product of irrational x and y can be rational

(a + sqrt(b)) + (a- sqrt(b))= 2a

and product = a^2 - b^2

so taking a and b rational such that sqrt(b) is irrational we can get easily


but for power

say x = sqrt(2) and y = sqrt(2)


is x^y rational

we do noy know

but if it is then we are through

and if it is not then (x^y)^y = 2 whhich is rational

hence x^y can be rational for both irrational x and y

2011/028) to prove that sqrt(2) is irrational

before proving it we need to prove a lemma

lemma: if 3 / (a^2+ b^2) then 3 | a and 3| b

proof:

we have a mod 3 = 0 or 1 or 2

so a^2 mod 3 = 0 or 1
and b^2 mod 3 = 0 or 1

so a^2 + b^2 mod 3 = 0 if a mod 3 = 0 and b mod 3 = 0
or 1 or 2 otherwise

hence lemma is proved

now for the proof of sqrt(2) is irrational

let sqrt(2) = a/b where GCD(a,b) = 1 ( we can always reduce it in this form)

so a^2 = 2b^2 or a^2 + b^2 = 3b^2

so 3 / (a^2+b^2) or hence as per above lemma 3 | a and 3 | b

so GCD(a,b) is not 1 so a contradiction

so sqrt(2) is irrational

2011/027) obtain the sum of

1/(x+1) + 2/(x^2+1) + 4/(x^4+1) + (2^2n/(x^2n + 1)

we realise that 1/(x+1) - 1/(x-1) = -2 /(x^2-1)

so add and subtract 1/(x-1) to get

1/(x-1) + (1/(x+1) - 1/(x- 1) + 2/(x^2+1) + 4/(x^4+1) +.... (2^2n/(x^2n + 1)))
= 1/(x-1) + (-2/(x^2-1)+ 2/(x^2+1) + 4/(x^4+1) + (2^2n/(x^2n + 1)))
= 1/(x-1) + (-4/(x^4-1)+ ... + (2^2n/(x^2n + 1)))

applying repeatedly we get 2^(4n)/(x^4n+1) + 1/(x-1)

2011/026) Factor by^2-b^2y-ay^2+a^2y+ab^2-ba^2

rewrite in decreasing power of any one say y and then proceed

by^2-b^2y-ay^2+a^2y+ab^2-ba^2

= by^2-ay^2- b^2y+a^2y+ab^2-ba^2
= y^2(b-a) +y (a^2-b^2) + ab(b-a)
= (b-a) (y^2 + y(a-b)(a+b) + ab)
= (b-a) (y^-y(a+b) + ab)
= (b-a) (y-a)(y-b)

thus it becomes easy

2011/025) solve x = a/(b+c) = b/(c+a) = c/(a+b)

x cannot be zero as if x= 0 then a = b =c so a/(b+c) = b/(c+a) = c/(a+b) are indeterminate form

so we take reciprocals

1/x = (b+c)/ a= (c+a)/b = (a+b)/c
add 1 on all 2 sides

1/x +1 = (a+b+c)/a = (a+b+c)/b = (a+b+c)/ c

so a+b+c = 0 = 1/x + 1 or a= b = c

a+b+c = 0 = 1/x + 1 => x= -1 => b+ c = - a,if a = 1 then b= k and c = -1 - k

this satisfies all 1/x = (b+c)/ a= (c+a)/b = (a+b)/c
add 1 on all 2 sides


if a=b= c then we get x = 1/2 so solution (x=1/2, a= k. b= k,c =k ) for any k

if x = -1 then (x= -1, a = 1 , b= k, c = - 1 -k) or any permutation or any multiple of the same.

2011/024) if p and p^2 + 8 are both primes then prove p^3+4 is a prime

proof:
p is either 2 or 3 or of the form (6n+/-1)

if p is of the form (6n+/-1)

then p^2 + 8 = 36n^2-12n + 9 = 3 ( 12n^2 - 4n + 3) divisible by 3 and ( 12n^2 - 4n + 3) > 1

so not a prime

if p = 2 then p^2 + 8 is even and not a prime

so only p =3 is left and p^2+ 8 = 17 is a prime and p =3 is the only number satisfies the criteria and p^3+ 4 = 31 is a prime as well

2011/023) the number of real solutions of the equation

sin e^x = 5^x + 1/5^x

is

a) 0
b) 1
c) 2
d) many

ans:

we know sin e^x is between -1 and + 1

and
5^x + 1/5^x is cannot lie between - 2 and 2

so zero solutions

2011/022) To show that 1 = 1/2+1/4+ 1/8 + 1/16+ ….

We can prove from the RHS that

1/2+1/4+ 1/8 + 1/16+ .. is a GP with a= 1/2 and r = 1/2 and the GP converges to

1/2/(1-/12) = 1/2/(1/2) = 1

but beauty lies in expanding from LHS

1 = 1/2 + 1/2
= 1/2 + 1/4 + 1/4 ( take the last term and make into 2 halves)
= 1/2 + 1/4 + 1/8 + 1/8 and so on to get the result

2011/021) Pythagorean triplet with Gaussian Integer

Here we discuss the Pyhagorean triplet using Gaussian integers

As I have discussed in the previous blog entry about Gaussian integers that Gaussian integers are complex number of the form α = u + vi, where u and v are ordinary integers d i is the square root of minus one.

A primitive Pythagorean triple is one in which a and b are coprimes

That is integers (a,b,c) form a primitive Pythagorean triple where a^2+b^2= c^2 and

a and b are coprime

as a and b are co primes both cannot be even and only one can be odd. Both cannot be odd as

a mod 2 = 1 and b mod 2 = 1 shall give a^2 + b^2 mod 4 = 2 which is not a perfect square.

Now c^2 = (a^2+ b^2) = (a+ib)(a-ib)

Now both factors a + ib and a-ib cannot share any common factor that is they most be coprime to each other

So both of them must be perfect squares
Let
(a+ib) = (u+iv)^2 (and so (a-ib) = (u-iv)^2)

squaring RHS we get

a = u^2-v^2
b= 2uv

and c = (u+iv)(u-iv) = (u^2+v^2)

so we get

(u^2-v^2, 2uv, u^2+v^2) as Pythagorean triplet

2011/020) Gaussian Integer

Here i discuss Gaussian integer and property of Gaussian integer

x+iy in complex number is a Gaussian integer if x and y are integers

they are closed under addition subtraction and multiplication

A Gaussian integer a + bi is prime if and only if:

* one of a, b is zero and the other is a prime of the form 4n + 3 or its negative − (4n + 3) (where n \geq 0)
* or both are nonzero and a2 + b2 is prime.


13 is not a Gaussian prime because 13 = 9+4 = (3^2+2^2) = (3+2i)(3-2i)

any prime number of the form 4n+1 can be expressed as sum of 2 squares and cannot be a Gaussian prime.

but a prime of the form 4n+3 cannot be expressed as sum of 2 squares and hence Gaussian prime.

2 is a special prime as it is not odd and 2 = 1^2+1^2 = (1+i)(1-i)

2011/019) Prove that Sin (a +b ) sin (a-b) = sin ^2 a – sin ^2b

The above is an interesting identify in trigonometry and this can be done wrongly with out understanding trigonometry

As we do not know trigonometry we take sin as a multiplier

So sin (a+b) = sin a + sin b

And sin (a-b) = sin a – sin b

Multiply them to get

Sin (a+b) sin (a-b) = (sin a + sin b) (sin a - sin b)
= sin ^2 a – sin ^2 b

however above is wrong but the ans is right

now to the correct approach

Sin (a+b) sin (a-b)
= (sin a cos b + cos a sin b)(sin a cos b – cos a sin b)
= sin ^2 a cos^2b – cos^2 a sin ^2b
= sin ^2 a(1- sin ^2 b) – cos^2 a sin ^2 b
= sin ^2 a – sin ^2 a sin ^2 b – cos^2 a sin^2 b
= sin ^2 a – sin ^2 b(sin ^2 a + cos^2 a)
= sin ^2 a – sin ^2 b


The above example is to demonstrate that we get right result through erroneous approach and one need to be careful about it

2011/018) Prove that in the product

(1-x+x^2-x^3+ ..... - x^99 + x^100) ( 1 +x +x^2 + .. x^99 + x^100) after multiplying there does not appear a term of x odd degree.

this problem is solved by taking coefficient in a couple of books and I try to solve in a different approach

LHS = (1-x^101)/(1-x) * (1+ x^101)/(1+x)
= (1-x^202)/(1-x^2)

if we define f(x) = (1-x^101)/(1-x) then f(x^2) is the given expression and as it is division of p(x^2) by q(x^2) so cannot have any term other than (x^2)^2k or odd power of x

2011/017) if ( p and q) are roots of equation (x-a)(x-b) = c then

-->

Roots of equation (x-p)(x-q) + c are

a) a c

b) bc

c) a,b

d) (a+c), (b+c)

solution

-->

we have

(x-a)(x-b)) - c = x^2- (a+b) x+ (ab-c)

As p and q are roots so p+ q = a + b …1

and pq = ab – c or pq + c = ab …2

Now

(x-p)(x-q) + c = x^2 - (p+ q) x + (pq + c)

= x^2-(a+b)x + ab = (x-a)(x-b) ( from 1 and 2)

So roots are a and b

hence ans is c)

2011/016) a problem with repunit

A number is called repuint if all the digits of the number are 1 that is it is 1 , 11 , 111 so on

in a repunit number with 50 1's the 26th digit from the left is modified to some digit so that is is divisible by 13. what is the digit

Ans:
method 1
say the digit x
the number is 1(25 times)x1(24 times)
= (10^50-1)/9 + (x-1)(10^25-1)/9

we need to find x such that
(10^50-1)/9 + (x-1)(10^25-1)/9 mod 13 = 0

as 9 is coprime to 13

so (10^50-1)+ (x-1) )(10^25 -1)mod 13 = 0


as 10 and 13 are coprime so as per format's little theorem


10^12 = 1 mod 13

so 10^48 = 1 mod 13

so 10^50 = 100 mod 13

so (10^50-1)+ (x-1) )(10 -1) =

(100-1) + 9(x-1) mod 13 = 0

99 + 9x - 9 or 90+9x = 0 mod 13 or 10 +x = 0 mod 13 or x= 3

so ans = 3

Method 2


sequence of 12 1's is divisible by 13 so we subtract sequence of 12 1's and divide by 10^12 and do it 2 times

we get 1...(25times)x

as seqeunece of sequence of 12 1's is divisible by 13 so multiplying by any number so multiply by 10^14 and subtract

we get 1..(13 times) x

as sequence of 12 1's is divisible by 13 so multiplying by any number so multiply by 10^2 and subtract

1x is the number and divisible by 13 so x = 3

2011/015) Let p be a prime number > 5. Show that p divides infinitely many of the numbers 1, 11, 111 ...

as p > 5 p and 10 are coprimes and so are p and 9

so if p devides x <=> p devides 9x

so p devides 9, 99, 999, so on

so 10^n mod p = 1

now as p and 10 are co-primes then

10^n(p-1) is 1 mod p(for n = 1 ... )

or 10^n(p-1) -1 is divsible by p

or (10^n(p-1)-1)/9 that is 1 repeated n(p -1) is dvisible by p

so there are many numbers divisible by p

2011/014) Proof of existence of infinite prime numbers.

There are infininite number of prime numbers and there are a couple of proofs these are in the linked list below

http://primes.utm.edu/notes/proofs/infinite/index.html

I propose a proof that is simpler

If I prove that if for any n there is a prime > n then I am through

Let us consider n!+1

This does not have any factor from 2 to n and hence it is a prime or in case it is not a prime then prime factor > n.

So for any n there is a prime number> n

Hence proved

2011/013) factorization using short cuts.

1) (a+b+c)^3 - a^3-b^3-c^3

we know that (a+b+c)^3 - a^3 is divisible by (a+b+c) – a that is b+ c

and b^3+c^3 by (b+c)

hence b+c is a factor

by symmetry (a+b) and (c+a) are also factors

2) so it is (a+b+c)^3 - a^3-b^3-c^3 = m(a+b)(b+c)(c+a)

now as LHS does not contain a^3 or b^3 or c^3 (as they cancel out) so m has to be a constant

putting a =1 b = 1 and c = 1 we get LHS = 24 and RHS = 8m or m = 3

hence (a+b+c)^3 - a^3-b^3-c^3 = 3(a+b)(b+c)(c+a)