2014/116) What is the value of m in the quadratic equation $x^2-mx+4=0$ to have real and equal roots.

equal roots

discriminant = 0 ($b^2-4ac, a = 1 b = -m . c = 4$)

$m^2 - 4*4 = 0$ or m= 4 or -4

m = 4 means $^2-4x+4 = 0$   oots are 2 and 2
m = -4 means $^2+4x+4 = 0$r  roots 2 and -2

roots are real as well


refer to https://in.answers.yahoo.com/question/index?qid=20111021024215AACQbhH

2014/115) how many factors of 240 are therer which are of the form 4n+ 2

240 can be factored as $2^4 * 3 * 5$

now for a factor to be of the form it should be 2 multiplied by by an odd number,
so it is 1 * (1+1) * (1+1) ( one way of choosing 2, 2 ways of choosing power of 3 ( $3^0, 3^1$) and same way 2 ways of choosing 5) or it is 4

2014/114) solve the equation $(x-3)^4 + (x-7)^4 = 24832$

Before I provide the solution I would like to mention that generally a quartic polynomial is not easy to solve but this type of equation

$(x-a)^4 + (x-b)^4= c$ can be converted to a quadratic equation by transformation of

$y= \dfrac{(x-a) + (x-b)}{2}$

as below

we shall put

$y= \dfrac{(x-3) + (x-7)}{2}$ = x- 5

so we get
$(y+2)^4 + (y-2)^4 = 24832$
or $2(y^4 + 6 y^2 (-2)^2 + 16) =  24832$
or $y^4 + 24 y^2 = 12400$
$y^4 + 24 y^2 – 12400 = 0$
 or $(y^2 – 100)(y^2 + 124)  = 0$
so $y^2 = 100$
so $y = \pm 10$

or x = -5 or 15

2014/113) find the number of solutions of $\sin\,x = \dfrac{x}{100}$

first let us look at positive x and same number of solutions shall be for -ve x

as $\sin,x$ and $x$ both are odd functions.

as $\sin\,x$ is less than 1 so x shall be less than 100 and so if we draw a sin  curve there shall be $\dfrac{100}{2\pi}$ or 15.91 ( around 16) units so there shall be 16 parts( 1/2 oscilaltions) in positive side and 16 in -ve side.

for each on the curve y = x shall intersect the sin curve 2 times so 32 in positive side including zero) so there is 1 value at 0, 31 positive values and 31 -ve values or 63 values

2014/112) if x+y is divisble by 3 show that $x^3+y^3$ is divisible by 9

we have
$x^3+y^3= (x+y)^3 - 3xy(x+y)$
if $(x+y)$ is divisible by 3 then $(x+y)^3$ is divisible by 9 ( it is divisible by 27 but 9 is required) and $3xy(x+y)$ is divisible by 9 and hence the difference.

2014/111) Show that the function below has same remainder when divided by $x(x+1)$ and $x(x+1)^2$

$f(x)=2008+2007x+2006x^2+\cdots+3x^{2005}+2x^{2006}+x^{2007}$

1st we provide the premise

this shall have same remainder when divided by $x(x+1)$ and $x(x+1)^2$

provided this shall have same remainder when divided by $(x+1)$ and $(x+1)^2$

when we divide by (x+1)^2 we shall have a linear polynomial say m(x+1) + c and when we divide by (x+1) it shall be d and both are same if m= 0 and m+c = d

so if we convert

f(x) as a polynomial of (x+1) the coefficient of x should be zero

of f(x-1) should have coefficient of x to be zero

now we provide the solution based on premise

$f(x-1)=2008+2007(x-1)+2006(x-1)^2+\cdots$
   $+3(x-1)^{2005}+2(x-1)^{2006}+(x-1)^{2007}$

should have coefficient of x to be zero

the coefficient of x

$= 2007 + 2006 * (-2) + 2005 * 3 \cdots + 2 * 2006 - 2007 = 0$

hence proved

I have solved the problem at http://mathhelpboards.com/challenge-questions-puzzles-28/division-polynomial-13760.html#post65389 where you can find some more different solutions

2014/110) Find closed form of

 $\cos\,1^\circ \cos\,2^\circ + \cos\,2^\circ \cos\,3^\circ\cdots \cos\,88^\circ \cos\,89^\circ$

we have $2 \cos(x) \cos (y) = \cos (x+y) + \cos (y-x)$

so $2 (\cos\,1^\circ \cos\,2^\circ ) = \cos\,3^\circ + \cos\,1^\circ$
 $2 (\cos\,2^\circ \cos\,3^\circ ) = \cos\,5^\circ + \cos\,1^\circ$
so on till
$2 (\cos\,44^\circ \cos\,45^\circ ) = \cos\,89^\circ + \cos\,1^\circ$

 $2 (\cos\,45^\circ\cos\,46^\circ ) = \cos\,91^\circ + \cos\,1^\circ$

or  $2 (\cos\,45^\circ \cos\,46^\circ ) = - \cos\,89^\circ + \cos\,1^\circ$


so on till

$2 (\cos\,88^\circ \cos\,89^\circ ) = - \cos\,3^\circ + \cos\,1^\circ$

on adding above for each positive term in 1st half for the first term there is a -ve term for the second half and we are left wth 88 times $\cos\,1^\circ$

so $2( \cos\,1^\circ \cos\,2^\circ + (\cos\,2^\circ \cos\,3^\circ +\cdots \cos\,88^\circ \cos\,89^\circ)  = 88 \cos\,1^\circ$

or  $( \cos\,1^\circ \cos\,2^\circ + (\cos\,2^\circ \cos\,3^\circ +\cdots \cos\,88^\circ \cos\,89^\circ)  = 44 \cos\,1^\circ$

problem 2014/109) prove that $\dfrac{\sin\,x - \cos\, x + 1}{\sin\, x + \cos\, x – 1} = \dfrac{1 + \tan \frac{x}{2}}{1- \tan\frac{x}{2}}$

Proof

convert it into $\dfrac{x}{2}$ form

numerator = $2 sin \dfrac{x}{2}\cos\dfrac{x}{2} + 2 \sin ^2\dfrac{x}{2}$ (as $\cos\, x= 1 - 2 \sin ^2 \dfrac{x}{2}$) = $2 \sin \dfrac{x}{2}( \cos \dfrac{x}{2} + \sin \dfrac{x}{2})$
denominator = $2 \sin \dfrac{x}{2} \cos \dfrac{x}{2} - 2 \cos^2 \dfrac{x}{2}$ (as $cos\, x= 2 \cos^2 \dfrac{x}{2} - 1$ = $2 \sin \dfrac{x}{2}(\cos \dfrac{x}{2}- \sin \dfrac{x}{2})$
dividing
we get
$\dfrac{\sin\,x - \cos\,x + 1}{\sin\,x + \cos\,x - 1} = \dfrac{\cos \dfrac{x}{2} + \sin \dfrac{x}{2}}{\cos\dfrac{x}{2}-\sin \dfrac{x}{2}}$
= $\dfrac{(1 + \tan \dfrac{x}{2})\cos\dfrac{x}{2}}{(1 - \tan \dfrac{x}{2})\cos\dfrac{x}{2}}$
= $\dfrac{1 + \tan \dfrac{x}{2}}{(1 - \tan \dfrac{x}{2}}$

2014/108) Find the remainder when $2^{1990}$ is divided by 1990

We have 1990 = 10*199 = 2* 5* 199

now we have

$2^4$ mod 5 = 1 as per fermat theorem
so $2^{1990}$ mod 5 = $2^2$ mod 5 = 4 mod 5

$2^{ 198} = 1$ mod 199

so $2^{1990}$ mod $199$ = $2^10$ mod 199 = 1024 mod 199 = 29 mod 199

$2^{1990}$ mod 2 = 0

so $2^{ 1990}$ mod 199 = 29
$2^{1990}$ mod 5 = 4

using Chinese Remainder Theorem you can proceed

continuing further

$2^{1990} = 0$ mod 2
$2^{1990} = 4$ mod 5

this gives $2^{1990} = 4$ mod 10

now using
$a = 4 (mod\, 10)$
$a = 29 (mod\, 199)$

Notice that 20*10+(-1)*199=1, thus 20*10≡1 (mod 199) and -199≡1 (mod 10)

let a= 29*(20•10)+4*(-199), then it is clear that
a = 29*(0)+4*(1)=4 (mod 10) and a=29•(1)+4•(0)=29 (mod 199) so this a works for what we want.

a= 5004=1024 (mod 1990)

so the remainder is 1024

Q2014/107) find coefficient of a^2 in the following expression

 $(1-a^2)+(1-a^2)^2+(1-a^2)^3+(1-a^2)^4+(1-a^2)^5+(1-a^2)^6$
       $+(1-a^2)^7+(1-a^2)^8+(1-a^2)^9+(1-a^2)^{10}+(1-a^2)^{11}+(1-a^2)^{12}$

if we put $x= (1-a^2)$
we get given expression as
$x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^{10}+x^{11}+x^{12}$= $x\dfrac{1-x^{12}}{1-x}$
= $(1-a^2)\dfrac{1-(1-a^2)^{12}}{a^2}$
= $\dfrac{(1-a^2)-(1-a^2)^{13}}{a^2}$

now $\dfrac{1-a^2}{a^2}$ shall not contribute to $a^4$ so we need to find the coefficient of $a^6$ in $-(1-a^2)^{13}$ which is ${13\choose 3}$

that is the ans

2014/106) find sum of real roots of the following

$2x^8-9x^7+20x^6-33x^5+46x^4-66x^3+80x^2-72x+32=0$
we have
$2x^8-9x^7+20x^6-33x^5+46x^4-66x^3+80x^2-72x+32=0$
= $2(x^8+ 16) -9(x^7+8x) +20(x^6+ 4x^2) -33(x^5+2x^3) + 46x^4=0$
or deviding by $x^4$ as x = 0 is not a solution  we get
$2(x^4 + (\dfrac{2}{x})^4)-9(x^3 + (\dfrac{2}{x})^3)+20((x^2 + (\dfrac{2}{x})^2)-33(x (+\dfrac{2}{x}))+46 = 0$

now if we put  $x+\dfrac{2}{x}=t$
we get
$x^2+(\dfrac{2}{x})^2=t^2-4$
$x^3+(\dfrac{2}{x})^3=t^3-6t$
$x^4+(\dfrac{2}{x})^4=t^4-8t + 8$

so given relation reduces to
$2(t^2-8t^2 +8) -9(t^3-6t) +20(t^2-4) - 33t + 46= 0$
or $2t^4-9t^3+4t^2+21t-18=0$

now we see that t = 1 and t = 3 are solutions and hence we get
 $2t^4-9t^3+4t^2+21t-18=0$
= $2t^3(t-1) - 7t^2(t-1) -3t(t-1) + 18(t-1)=0$
or$(t-1)(2t^3-7t^2-3t+18) = 0$
gives a solution t = 1

or
$2t^3-7t^2- 3t + 18 = 0$ as 3 is a root we get
$2t^2(t-3) - t(t-3) - 6(t-3) = 0$
or $(t-3)(2t^2 - t^2-3) = 0$
so t = 3
or $2t^2 - t - 3 = 0 $
or $(2t-3)(t+1) = 0$

so t = 1 or 3 or - 1 or $-\dfrac{3}{2}$

now $t = x+ \dfrac{2}{x}$ and if x is positive then by AM GM inequality lowest value = $2\sqrt{2}$
or only possible value from above is

t = 3 (as t cannot be between  -$2\sqrt{2}$ and $2\sqrt{2}$)

t = 3 gives x = 1 or 2 and so sum of real roots = 3

I had the privilege to solve it at http://mathhelpboards.com/challenge-questions-puzzles-28/find-sum-real-roots-13522.html#post64344

2014/105) show that all terms of $\dfrac{107811}{3}$, $\dfrac{110778111}{3}$ and $\dfrac{111077781111}{3}$ ... are pefect cubes

let us look at 1st few terms

1st term = $\dfrac{107811}{3} = \dfrac{1 * 10^5 + 77 * 10^2 + 111}{3}$ 
2nd term = $\dfrac{110778111}{3} = \dfrac{11 * 10^7 + 777 * 10* 3 + 1111}{3}$
3rd term = $\dfrac{111077781111}{3} = \dfrac{111 * 10^9 + 7777 * 10^4 + 11111}{3}$

so nth term = $\dfrac{(10^n- 1) * 10^{2n+3} + 7 * 10^{n+2} -1)* 10^{n+1} + (10^{n+2} - 1)}{3*9}$
=$\dfrac{ (10^n-1) * 1000 * 10^{2n} + 7 *( 100 * 10^n - 1)(10 * 10^n) + (100 * 10^n - 1)}{27}$
=$\dfrac{1000 * 10^{3n} - 1000 * 10^{2n} + 7 *1000 * 10^{2n} - 7 *10 * 10 ^n + 100 * 10^n - 1}{27}$
= $\dfrac{1000 * 10^3n - 300 * 10 ^2n + 30 * 10 ^n -1}{27}$
= $\dfrac{10^{3n+3} - 3 * 10^{2n+2} + 3 * 10^{n+1} -1}{27}
= \dfrac{(10^{n+1} - 1)^3}{27}$

so nth term = $(\dfrac{10^{n+1} -1}{3})^3$

which is a perfect cube


 

Q2014/104) find the degree of polynomial $(x+\sqrt{x^3-1})^7 + (x-\sqrt{x^3-1})^7$

We have

$(a+b)^7=a^7+7a^6b+21a^5b^2+35a^4b^3+35a^3b^4+21a^2b^5+7ab^6+b^7$
also  
$(a-b)^7=a^7-7a^6b+21a^5b^2-35a^4b^3+35a^3b^4-21a^2b^5+7ab^6-b^7$

so $(a+b)^7 + (a-b)^7= 2a^7+42a^5b^2+70a^3b^4+14ab^6$

So $(x+\sqrt{x^3-1})^7 + (x-\sqrt{x^3-1})^7$
= $2 x^7+42x^5(x^3-1)+70x^3(x^3-1)^2+14x(x^3-1)^3$

now the term with highest power of x is $x^{10}$ and so 10 is degree of polynomial 

2014/103) prove that $\cot 2x - \tan 2x = 2\cot 4x$

LHS = $\dfrac{\cos 2x }{\ sin 2x}- \dfrac{\sin 2x}{\cos 2x}$
= $\dfrac{\cos^2 2x - \sin^2 2x}{\sin 2x\cos 2x}$
= $\dfrac{\cos 4x}{\sin 2x\cos 2x}$
= $\dfrac{2\cos 4x}{2\sin 2x\cos 2x}$
= $\dfrac{2 \cos 4x}{\sin 4x}$
= $2 \cot 4x$

2014/102) show that $\lfloor(\sqrt{n} +\sqrt{n+1}\rfloor= \lfloor(\sqrt{4n+2}\rfloor$

we realise that $n(n+1) = (n+\dfrac{1}{2})^2 - \dfrac{1}{4}$
so $\sqrt{n(n+1)}\lt(n+\dfrac{1}{2})$
clearly $n\lt\sqrt{n(n+1)}$

so

we have
$(\sqrt{n} + \sqrt{n+1})^2 = n + n+ 1 + 2 \sqrt{n(n+1)}$
= $2n +1 + 2 \sqrt{n(n+1)}$
>$2n+ 1 + 2 n$ or > $4n+ 1$

and = $2n +1 + 2 \sqrt{n(n+1)}$ <  $2n +1 + 2 (n + \dfrac{1}{2})$
or <  (4n + 2)
so $(4n+1)\lt(\sqrt{n} + \sqrt{n+1})^2\lt(4n+2)$

because  4n+2 is not a perfect square

we have $\lfloor(\sqrt{4n+1}\rfloor = \lfloor(\sqrt{4n+2}\rfloor$
and as $ (\sqrt{n} + \sqrt{n+1})^2$ is between 4n + 1 and 4n +2 we have

$\lfloor(\sqrt{4n+1}\rfloor = \lfloor(\sqrt{4n+2}\rfloor= \lfloor(\sqrt{n} +\sqrt{n+1}\rfloor$

2014/101) If $\sin\theta = k \sin(\theta + 2\alpha)$ prove that $\tan (\theta + \alpha) = \dfrac{1+k}{1-k}\tan \alpha$

we have
$k = \dfrac{sin\theta}{\sin(\theta + 2\alpha)}$
using componendo dividendo
 $\dfrac{k+1}{k-1} = \dfrac{\sin\theta+\sin(\theta + 2\alpha)}{\sin\theta-\sin(\theta + 2\alpha)}$
=$\dfrac{ 2 \sin(\theta + \alpha ) \cos \theta}  {-2 \cos (\theta + \alpha) \sin \theta}$
= $\dfrac{ - \tan(\theta + \alpha )}{\tan \theta}$

hence $\dfrac{1 + k} {1- k} = \dfrac{\tan(\theta + \alpha )}{\tan \theta}$

or  $\dfrac{1 + k} {1- k}\tan \theta = \tan(\theta + \alpha) $
proved

Q2014/100) show that $\sqrt[3]{45 + 29\sqrt2} + \sqrt[3]{45 - 29\sqrt2}$ is rational

it is as below
 $x=\sqrt[3]{45 + 29\sqrt2} +  \sqrt[3]{45 - 29\sqrt2}$
or
$x-\sqrt[3]{45 + 29\sqrt2} -  \sqrt[3]{45 - 29\sqrt2}= 0$

using $a+b+c = 0 => a^3+b^3+ c^3 = 3abc$
we get
$x^3-(45 + 29\sqrt2) - (45 - 29\sqrt2)= 3 x\sqrt[3]{(45 + 29\sqrt2)(45 - 29\sqrt2)}$
or
$x^3-90= 3x\sqrt[3]{45^2 - 2 * 29^2}$
or $x^3-90 = 21x$
or $x^3 - 21x - 90 = 0$
or $(x-6)(x^2+6x+ 15) = 0$
has one real root = 6 and 2 complex roots

hence given expression = 6 which is real

2014/99 ) find the 50th smallest number which is coprime to 987

First we need to factor 987
987 = 3 * 329 = 3 * 7 * 47

to find a number which is coprime to 987 it should be co-prime to 3,7, and 47

now 1 is not coprime to any number so we need to find the 51st number which is  not divisible by 3 7 or 47

for x the numbers below or same as not divisible by 3 7 and 47 are

$f(x) =x-\lfloor\dfrac{x}{3}\rfloor-\lfloor\dfrac{x}{7}\rfloor-\lfloor\dfrac{x}{47}\rfloor+\lfloor\dfrac{x}{3*7}\rfloor+\lfloor\dfrac{x}{3* 47}\rfloor+\lfloor\dfrac{x}{7*47}\rfloor- \lfloor\dfrac{x}{3 * 7 * 47}\rfloor$

or $f(x) =x-\lfloor\dfrac{x}{3}\rfloor-\lfloor\dfrac{x}{7}\rfloor-\lfloor\dfrac{x}{47}\rfloor+\lfloor\dfrac{x}{21}\rfloor+\lfloor\dfrac{x}{141}\rfloor+\lfloor\dfrac{x}{329}\rfloor- \lfloor\dfrac{x}{987}\rfloor$

or $f(x) =x-\lfloor\dfrac{x}{3}\rfloor-\lfloor\dfrac{x}{7}\rfloor+\lfloor\dfrac{x}{21}\rfloor-\lfloor\dfrac{x}{47}\rfloor+\lfloor\dfrac{x}{141}\rfloor+\lfloor\dfrac{x}{329}\rfloor- \lfloor\dfrac{x}{987}\rfloor$


for estimating we take $x-\dfrac{x}{3}=\dfrac{2x}{3}=51$ or x = 76 ( rounded)

so f (x) = 76  - 25 - 10 + 3 = 44
we are falling short by 7

so we add 11 as it is 7 * 3/2 rounded

so we get x = 87 but as 87 is not coprime we take 88

f(88) = 88 - 29 - 12 + 4 - 1 = 50

so we take next number 89 which coprime

so x = 89  is the ans.

Q2014/098) Given $a^2+b^2=16$, $c^2+d^2=25$ find the maximium of ac

without loss of generality we can choose
$a=4\sin\,t$
$b=4\cos\,t$
$c=5\sin\,p$
$d=5\cos\,p$


so we get $ad-bc= 20\sin\, t \cos\, p - 20\sin\, p \cos\, t = 20\sin (t-p) = 20$
or $\sin(t-p) = 1$
so $t= p+ \dfrac{\pi}{2}$
hence
$ac = 20 \sin \, t \sin \ p$
= $20 \sin\, p + \sin (\dfrac{\pi}{2}+ p)$
= $-20 \cos \, p \sin\, p$
= $-10 \sin 2p$


clearly the largest value is 10 and smallest -10

2014/097) If $x^2-2x +4$ has roots a and b then show that $a^n + b^n = 2^{n+1} \cos\dfrac{\pi}{3}$ for n positive integer

We have as a and b are roots

$a+ b= 2$ and $ab = 4$

so $a +b = 2 = 2^2 \cos \dfrac{\pi}{3}$ true for 1

$(a^2+b^2) = (a+b)^2 – 2ba = 4 – 8 = -4 = 8 * \cos \dfrac{2\pi}{3}$ true for 2

Let it be true for n =k

We shall prove it by induction

$(a^k+b^k)(a+b) = a^{k+1} +b^{k+1} + ab(a^{k-1} +b^{k-1})$

Or 

$a^{k+1}+b^{k+1}= (a^k+b^k)(a+b)-ab(a^{k-1}+b^{k-1})$

we have

$a^{k+1}+b^{k+1}= (a^k+b^k)(a+b)-ab(a^{k-1}+b^{k-1})$

= $2^{n+1} \cos \dfrac{n\pi}{3} 2^2 \cos \dfrac{\pi}{3} – 4(2^n \cos\dfrac{(n-1)\pi}{3})$

= $2^{n+2}( 2\cos \dfrac{n\pi}{3} \cos \dfrac{\pi}{3}) – 4(2^n \cos\dfrac{(n-1)\pi}{3})$

= $2^{n+2}( \cos \dfrac{(n+1)\pi}{3} + \cos \dfrac{(n-1)\pi}{3}) – 2^{n+2} \cos\dfrac{(n-1)\pi}{3}$
= $2^{n+2}\cos \dfrac{(n+1)\pi}{3} +2^{n+2} \cos \dfrac{(n-1)\pi}{3} – 2^{n+2} \cos\dfrac{(n-1)\pi}{3}$

=  $2^{n+2}\cos \dfrac{(n+1)\pi}{3}$

We have proved the step for induction

Hence proved

2014/096) Evaluate $2\cos^3 \dfrac{\pi}{7}-\cos^2 \dfrac{\pi}{7}-\cos \dfrac{\pi}{7}$

$2\cos^3 \dfrac{\pi}{7}-\cos^2 \dfrac{\pi}{7}-\cos \dfrac{\pi}{7}$
=  $\cos \dfrac{\pi}{7}(2\cos^2 \dfrac{\pi}{7}-\cos \dfrac{\pi}{7}-1)$
= -  $\cos \dfrac{\pi}{7}(\cos \dfrac{\pi}{7} -  (2\cos^2 \dfrac{\pi}{7}-1))$
= -  $\cos \dfrac{\pi}{7}(\cos \dfrac{\pi}{7} -  \cos \dfrac{2\pi}{7})$
=  -$ 2 \cos \dfrac{\pi}{7}\sin \dfrac{3\pi}{14}  \sin\dfrac{\pi}{14}$

=  -$\dfrac{ 2 \cos \dfrac{\pi}{7}\sin \dfrac{3\pi}{14}  \sin\dfrac{\pi}{14} \cos \dfrac{\pi}{14}}{ \cos \dfrac{\pi}{14}}$
=  -$\dfrac{  \cos \dfrac{\pi}{7}\sin \dfrac{3\pi}{14}  \sin\dfrac{\pi}{7}}{ \cos \dfrac{\pi}{14}}$
=  -$\dfrac{  2\cos \dfrac{\pi}{7}\sin \dfrac{\pi}{7}  \sin\dfrac{3\pi}{14}}{2 \cos \dfrac{\pi}{14}}$
 = -$\dfrac{  \sin  \dfrac{2\pi}{7} \sin\dfrac{3\pi}{14}}{2  \cos \dfrac{\pi}{14}}$
 =- $\dfrac{  \sin  \dfrac{2\pi}{7} \cos (\dfrac{\pi}{2} - \dfrac{3\pi}{14})}{2  \cos \dfrac{\pi}{14}}$
= - $\dfrac{  \sin  \dfrac{2\pi}{7} \cos \dfrac{2\pi}{7}}{2  \cos \dfrac{\pi}{14}}$
= - $\dfrac{ 2 \sin  \dfrac{2\pi}{7} \cos \dfrac{2\pi}{7}}{4  \cos \dfrac{\pi}{14}}$
= -$\dfrac{  \sin  \dfrac{4\pi}{7}}{4  \cos \dfrac{\pi}{14}}$
= - $\dfrac{  \cos(  \dfrac{\pi}{2}- \dfrac{4\pi}{7})}{4  \cos \dfrac{\pi}{14}}$
= -$\dfrac{  \cos  \dfrac{-\pi}{14}}{4  \cos \dfrac{\pi}{14}}$
= - $\dfrac{  \cos  \dfrac{\pi}{14}}{4  \cos \dfrac{\pi}{14}}$
= - $\dfrac{1}{4}$

2014/095) Prove the sum of any 10 consecutive terms of the Fibonacci sequence is multiple of 11?

proof:

this can be proved as below
$F_3 = F_2 + F_1$
$F_4 = F_3 + F_2=2F_2+F_1$
$F_5 = F_4 + F_3=3F_2+2F_1$
$F_6 = F_5 + F_4=5F_2+3F_1$
$F_7 = F_6 + F_5=8F_2+5F_1$
$F_8 = F_7 + F_6=13F_2+8F_1$
$F_9 = F_8 + F_7=21F_2+13F_1$
$F_{10} = F_9 + F_8=34F_2+21F_1$

Adding we get $F_1+F_2+F_3+F_4+F_5+F_6+F_7+F_8+F_9+F_{10}$
=$88F_2+55F_1= 11(8F_2+5F_1) =11F_7$

This can be proved further below as(more fun)

$F_1+F_2+F_3+F_4+F_5+F_6+F_7+F_8+F_9+F_{10}$
=$(F_1+F_2)+(F_3+F_4)+(F_5+F_6)+F_7+F_8+F_9+F_{10}$
= $F_3+F_5+F_7+F_7+F_8+F_9+(F_8+F_9)$
= $F_3+F_5+2F_7+2F_8+2F_9$
= $F_3+F_5+2F_7+2F_8+2(F_7 + F_8)$
= $F_3+F_5+4F_7+4F_8$
=  $F_3+F_5+4F_7+4(F_6+F_7)$
=  $F_3+F_5+8F_7+4F_6$
=  $F_3+(F_5+F_6) +8F_7+3F_6$
= $F_3+9F_7+3F_6$
= $F_3+F_6+ 9F_7+2F_6$
=$F_3+(F_4+F_5)+ 9F_7+2F_6$
= $(F_3+F_4)+F_5+ 9F_7+2F_6$ 
= $F_5+F_5+ 9F_7+2F_6$
= $2 F_5+ 9F_7+2F_6$
= $2(F_5+F_6)+ 9F_7$
= $2F_7 + 9F_7$
= $11F_7$
Done

2014/094) If $a+b+c=0$ then prove that $\dfrac{a^2}{a^2-bc} +\dfrac{b^2}{b^2-ca} +\dfrac{c^2}{c^2-ab} =2$

Solution

We are given $a + b+ c = 0\cdots(1)$

$(a+b+c)^2 = a^2 +b^2 + c^2 +2ab + 2bc + 2ca$

 so 

$a^2 +b^2 + c^2 = -2(ab +bc +ca)\cdots(2)$

 From (1)
$a(a+b+c) = 0$

$a^2 + ab + ca = 0$

or $a^2 - bc = -(ab + bc+ ca)\cdots(3)$

Similarly  

$b^2 - ca = -(ab + bc+ ca)\cdots(4)$

$c^2 - ab = -(ab + bc+ ca)\cdots(5)$

Thus, using (3), (4) & (5), we can rearrange the LHS as follows;

LHS = $\dfrac{a^2}{a^2-bc} +\dfrac{b^2}{b^2-ca} +\dfrac{c^2}{c^2-ab}$

 = $\dfrac{a^2 +b^2+ c^2}{-ab-bc-ca}$

= -$\dfrac{a^2 +b^2+ c^2}{ab+bc+ca}$

= -(- 2)$\dfrac{ab+bc+ca}{ab+bc+ca}$

= 2 

 

2014/093) Find the number of non zero integral solutions of $(1-i)^x=2^x$

We have

 $|(1-i)|=2$

and 

 $|2|=2$

 So no power other than zero shall meet modulus and x = 0 is the only solution as both sides are 1

Q2014/092) If a, b, c ϵ R, $x = a² - bc$, $y = b² - ca$, $z = c² - ab$, then prove that, $x³ + y³ + z³ - 3xyz$ is a perfect square

we have

$x^3 + y^3 + z^3 - 3xyz$
= $(x + y + z) (x^2 + y^2 + z^2 - xy - yz - zx)$
= $\dfrac{1}{2} (x + y + z) [(x - y)^2 + (y - z)^2 + (z - x)^2] \cdots(1)$... ( 1 )

$x + y + z$
= $a^2 - bc + b^2 - ca + c^2 - ab$
= $\dfrac{1}{2}[(a - b)^2 + (b -c)^2 + (c - a)^2] \cdots (2)$

further

$x - y$
= $(a^2 - bc) - (b^2 - ca)$
= $a^2 - b^2 + ca - bc$
= $(a - b) (a + b) + c (a - b)$
= (a + b + c) (a - b)

Similarly,
$y - z = (a + b + c) (b - c)$ and
$z - x = (a + b + c) (c - a)$

=> $(x - y)^2 + (y - z)^2 + (z - x)^2$
= $(a + b+ c)^2 ((a - b)^2 + (b - c)^2 + (c - a)^2)\cdots ( 3 )$

Plugging results ( 2 ) and ( 3 ) into ( 1),

$x^3 + y^3 + z^3 - 3xyz$
= $\dfrac{1}{2} * \dfrac{1}{2}((a - b)^2 + (b -c)^2 + (c - a)^2) *$

            $(a + b+ c)^2 ((a - b)^2 + (b - c)^2 + (c - a)^2)$
= $(\dfrac{1}{2} (a + b + c) ((a - b)^2 + (b - c)^2 + (c - a)^2))^2$

which is a perfect square.

2014/091) The LCM and GCD of two composite numbers total 111. What are the numbers?

Let the gcd be t
the 2 numbers are at and bt with a < b where a and b are coprimes or a  =1

Therefore, the LCM = abt

 so abt + t = 111 or 111 = t(1+ab) = 3 * 37

then we have following cases
t = 1 , ab = 110 then a = 1, b= 110 giving 2 numbers 1 and 110
                                 a = 2, b= 55 giving 2 numbers 2 and 55
                         or a = 5 , b = 22 giving 2 numbers 5 and 22
                         or a = 10, b = 11 giving 2 numbers 10 and 11
t=3, ab = 36 then a = 1, b= 36 giving 2 numbers 3 and 108
                             a = 4, b= 9 giving 2 numbers 12 and 27
t = 37, ab =2 then a = 1, b=2 giving 37 and 74