2016/072) If $x=a(y+z), y= b(z+x), z=c(x+y)$ then prove that, $ab+bc+ca+2abc=1$

we have
$x+xa = x(1+a) = a(x+y+z)$ or $\frac{a}{1+a} = \frac{x}{x+y+z}$
similarly
$\frac{b}{1+b} = \frac{y}{x+y+z}$
$\frac{c}{1+c} = \frac{z}{x+y+z}$
add the above 3 to get
$\frac{a}{1+a} + \frac{b}{1+b}+ \frac{c}{1+c} = 1$
or $a(1+b)(1+c) + b(1+c)(1+a) + c(1+a)(1+b) = (1+a)(1+b)(1+c)$
or $a + ab + ac + abc + b + bc + ab + abc = (1+a)(1+b) = 1 + a + b + ab$
or $  ac + abc  + bc + ab + abc = 1$
or $ab + bc+ca + 2abc = 1$

2016/071) $\tan\, 1 + \tan\, 5 + \tan\, 9 + \cdots + \tan\, 177 =$ (angles in degrees)

we have $\tan\, 45 x  = 1$ has solutions $\tan\, 1 ,  \tan\, 5  \tan\, 9  \cdots  \tan\, 177$ there are 45 solutions
expand tan nx with n = 45 (odd) we get
$\tan\,45 x= \frac{\sum_{n=0}^{22} (-1)^{n}{45 \choose 2n+ 1}\tan^{2n+1}x }{\sum_{n=0}^{22}{45 \choose 2n}(-1)^{n}\tan^{2n}x}$
or $\frac{\sum_{n=0}^{22} (-1)^{n}{45 \choose 2n+ 1}\tan^{2n+1}x }{\sum_{n=0}^{22}{45 \choose 2n}(-1)^{n}\tan^{2n}x}= 1$
or $\sum_{n=0}^{22} (-1)^{n}{45 \choose 2n+ 1}\tan^{2n+1}x -\sum_{n=0}^{22}{45 \choose 2n}(-1)^{n}\tan^{2n}x= 0$
the sum of roots is -ve coefficient of $tan^{44}x$ which is 45
hece  $\tan\, 1 + \tan\, 5 + \tan\, 9 + \cdots + \tan\, 177 =45$

2016/070) If $\tan(x+y) = a + b$ and $\tan(x-y) = a - b$ show that $a\tan\,x - b\tan\,y = a^2 - b^2$

$\tan(x+y) = \frac{\tan\,x+\tan\,y}{1-\tan\,x\tan\,y}$ or $(a+b)(1-\tan\,x\tan\,y) = \tan\,x+\tan\,y \cdots(1)$
Similarly $(a-b)(1+\tan\,x\tan\,y) = \tan\,x-\tan\,y\cdots(2)$
multilying (1) by (a-b) and (2) by (a+b) and adding we get $(a^2-b^2) =  2a\tan\,x-2b\tan\,y$
devide both sides by 2 to get the result.

2016/069) Given $z'=1+i-\frac{2}{z}$. Let $z=x+iy$, prove that if $z'$ is a pure imaginary, then M moves on a circle

we have
$z'=1+i-\frac{2}{x+iy}$
multiply by conjugate to get
$z'= 1+i-\frac{2(x-iy)}{x^2+y^2}$
$=1- \frac{2x}{x^2+y^2} + i(1+ \frac{2y}{x^2+y^2})$
if it is imaginary real part is zero
or
$1- \frac{2x}{x^2+y^2} = 0$
or$x^2+y^2 - 2x = 0$
ort $x^2-2x+1 + y^2 =1$
or $(x-1)^2+y^2 = 1$
which is a circle with centre(1,0) radius 1

2016/068) Show that there exists 2016 consecutive numbers that contains exactly 100 primes.

we know that number of primes less than 1000  $= 168$
now let f(x) be number of  primes in a sequence of 2016 primes starting at x.
$f(1) > 100$.
now when we move to next number the number of primes increases/decreases by 1 or remains unchanged
$f(2017!+2) = 0$ as 2016 numbers starting from this number  all are composite
so from 1 going upto 2017!+2 the starting number ( $>100$) remains unchanged or increases by 1 or decreases by 1
going to 0.
hence at some point it is 100.

2016/067) Solve $\cos^{ -1}\frac{x^2-1}{x^2+1} + \frac{1}{2}\ tan^{-1}\frac{-2x}{1-x^2}=\frac{2\pi}{3}$

Let $x=\tan \,t$
we know $\cos^{-1}-x = \pi - \cos^{-1}x$
hence $\cos^{-1}\frac{x^2 - 1}{x^2 + 1} = \pi - \cos^{-1}\frac{1 - x^2}{x^2 + 1}$
$= \pi - \cos^{-1}\frac{1 - \tan 2 t}{1 + \tan  ^2 t}= \pi - 2t$

Similarly, $\tan^{-1}\frac{-2x}{1 - x^2} = -\tan^{-1}\frac{2x}{1 - x^2} = - 2t$

The given equation reduces to:
$\pi - 2t - \frac{1}{2}2t= \frac{2\pi}{3}$
$3tan^-1(x) = \frac{\pi}{3}$
or $\tan^{-1}x = \frac{\pi}{9}$
Thus $x = \tan\frac{\pi}{9}$

2016/066) Find the smallest m such that 18 is a factor of m!

$18 = 2 * 3 ^2$
so in n! there should be at least one multiple of 2 and 2 multiples of 3( 3 square counting as 2) so smallest is 6.

2016/065) For which positive integer n does 2n devide the sum of $1^{st}$ n numbers, does 2n+1 devide the sum of $1^{st}$ n numbers

we have sum of $1^{st}$ n numbers =$\frac{n(n+1)}{2}$
for this to be divisible by 2n we should have $4n$ should devide$n(n+1)$ or 4 should devide n+1 or n should be of the form $4k-1$

as (2n+1) is co-prime to n and n+1 so for no n 2n+1 devides the sum

2016/064)Find all solutions in integers $x,y$ of the equation $y^2+2y= x^4+20x^3+104x^2 + 40x + 2003$(Irish 2003 paper 2)

Add 1 to both sides to get
$(y+1)^2 = x^4+20x^3+104x^2+40x+2004 = x^4+20x^3+ 104x^2+40x+ 4 + 2000 $
$= (x^2+10x+2)^2 +2000$
or $(y+1)^2 - (x^2+10x+2)^2 = 2000$
for the above to have solution we need to have $(y+1)$ and $(x^2+10x+2)$ both should be positive and as $(x^2+10x+2) = (x+5)^2-23$
so the lower number need to be 23 less than a perfect square.
so let us find (t,z) which are $(\pm501,\pm499),(\pm252,\pm248),(\pm129,\pm121),(\pm105,\pm95),(\pm60,\pm40),(\pm45,\pm5)$ out of which
only z = 121 which 23 less than is a perfect square
so we get
$y+1= \pm 129, (x+5) = \pm 12$ giving 4 solutions ($y=-130, x = - 17$), ($y= -130,x= 7$), ($y=128,x=7$),($y= 128, x= -17$)

2016/063) Prove for all integers $N>1$ $(N^2)^{2014}- (N^{11})^{106}$ is divisible by $N^6+ N^3+1$

$(N^2)^{2014}- (N^{11})^{106}$
$=N^{4028} - N^{1166}$
$=N^{1166}(N^{2862}-1) = N^{1166}((N^9)^{318}-1)$ is divisible by $N^9-1$
as $N^9-1 = (N^3)^3 -1 = (N^3-1)(N^6+N^3+1)$
as $(N^2)^{2014}- (N^{11})^{106}$ is divsible by $N^9-1$ which is divisible by $N^6+N^3+1$ hence  $(N^2)^{2014}- (N^{11})^{106}$
is divisible by $N^6+N^3+1$