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}$

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