2016/121) Express $(1+a^2)(1+b^2)$ as sum of 2 squares

we have $(1+a^2)(1+b^2) = (1+ ia)(1-ia)(1+ib)(1-ib)$ factoring over complex
$=((1+ia)(1+ib))(1-ia)(1-ib))$
$=(( 1 - ab) + i(a+b))((1 - ab) - i(a+b)) = (1- ab)^2 + (a+b)^2$ multiplication over complex

2016/120) Solve in positive integers $p^x = y^4 + 4$ and p is prime

factor the RHS ro get $p^x = (y^2-2y+2)(y^2+2y+2)$
both are power of p and $y^2-2y+2)< y^2 + 2y + 2$ so $y^2-2y + 2$ is a factor of difference that is 4y
so $y^2-2y+2$ is a factor of $4y^2$ or $4y^2-4(y^2-2y+2) = 8(y-1)$
so $y^2-2y+2$ is a factor of 8 that is it is 1 or 2 or 4 or 8
if it is 1 then y = 1, p = 5 and x = 1
if it is 2 then y=2 giving $p^x = 20$ which is not possible and if it is 4 or 8 we do not have integer solution
so only solution is $p = 5, x = y= 1$

2016/119)show that if A, B,C are angles of triangle $\tan\, A \tan\, B \tan\, C>=3\sqrt3$

because A B and C are angles of a triangle we have
$A+B= (180^\circ-C)$
taking tan of both sides
$\tan (A + B) = -\tan \, C$
or $\frac{\tan\, A + \tan \, B}{1- \tan\, A \tan\, B} = -\tan \, C$
or $\tan \, A + \tan \, B = - \tan C + \tan\, A \tan\, B \tan \, C$
or  $\tan \, A + \tan \, B + \tan C =  \tan\, A \tan\, B \tan \, C$
so $\tan\, A \tan\, B \tan \, C = \tan \, A + \tan \, B + \tan C$
using AM GM inequalty we have
$\frac{\tan \, A + \tan \, B + \tan C}{3} >= \sqrt[3]{\tan \, A  \tan \, B + \tan C}$
or  $\frac{\tan \, A  \tan \, B  \tan C}{3} >= \sqrt[3]{\tan \, A  \tan \, B + \tan C}$
or  $\tan \, A  \tan \, B  \tan C >= 3 \sqrt[3]{\tan \, A  \tan \, B  \tan\, C}$
cube both sides to get $(\tan \, A  \tan \, B  \tan C)^3 >= 27 (\tan \, A  \tan \, B + \tan C)$
or $(\tan \, A  \tan \, B  \tan C)^2 >= 27 $
or $(\tan \, A  \tan \, B  \tan C) >= 3\sqrt{3}$

2016/118) Let X be set of +ve integers greater than or equal to 8 and f(x) X->X be a function such that f(x+y) = f(x)f(y).

we have f(9) = f( 4 + 5) = f(4 * 5) = f(20) = f( 4+ 16) = f(64) = f( 8 + 8) = f(16) = f( 4 * 4) = f(8) = 8
in the above each step is valid because x,y are >=4.

2016/117) If a,b,c are 3 positive numbers show that $\frac{a^2+1}{b+c} + \frac{b^2+1}{c+a} \frac{c^2+1}{a+b} >= 3$

we have $a^2+1 >= 2a$ hence $\frac{a^2+1}{b+c} >= 2\frac{a}{b+c}$
also $b^2+1 >= 2b$ hence $\frac{b^2+1}{c+a} >= 2\frac{b}{b+c}$
and $c^2+1 >= 2c$ hence $\frac{c^2+1}{a+b} >= 2\frac{c}{a+b}$
adding above 3 we get $\frac{a^2+1}{b+c} + \frac{b^2+1}{c+a} \frac{c^2+1}{a+b} >= 2(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b})$
now
If we can show that $2(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b})>=3$
we are through
we have $\frac{a}{b+c} + \frac{b}{c+a}$
= $\frac{ca + a^2 + b^2 + bc}{c+a}$
$>=\frac{ca + 2ab + b^2 + bc}{c+a}$ since $a^2+b^2>=2ab$
$= \frac{a(b+c) + b(c+a)}{(c+a)}$
Thus  $\frac{a}{b+c} + \frac{b}{c+a} >= \frac{a}{a+c} + frac{b}{b+c}$
Similarly, it can be proved that
$\frac{b}{c+a} + \frac{c}{a+b} >= \frac{b}{a+b} + frac{c}{c+a}$
and $\frac{c}{a+b} + \frac{a}{b+c} >= \frac{c}{b+c} + frac{a}{a+b}$
Adding corresponding sides of the above three inequalities, we get
$2(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b})>=3$

2016/116) Find the real roots of the equation

$\sqrt{x+3-4\sqrt{x-1}} + \sqrt{x+8-6\sqrt{x-1}} = 1$


Solution


let $x-1 = y^2$
so we get  $\sqrt{y^2+4-4y} + \sqrt{y^2+9-6y} = 1$
or $\sqrt{(y-2)^2} + \sqrt{(y-3)^2} = 1$
or $|y-2| + |y-3| = 1$
as 3-2 is 1 so $2 <=y <=3$ or $5  <= x <= 10$

2016/115) For $x,\,y$ and $z\in [0,\,1]$ such that $xy+yz+zx=1$, prove x,y.z can be sides of a triangle.

letting $x=\tan\,A$, $y=\tan\, B$, $z\tan\, C$ we have A, B, C, between 0 and $\frac{\pi}{4}$
also $\tan\, A \tan\,  B +  \tan\, B \tan\,  C + \tan\, C \tan\,  A = 1$
using $\tan(A+B+C)$ we get
$A+B+C=\frac{\pi}{2}$
so $B+C>=A$ from above and A between 0 and $\frac{\pi}{4}$
so $x <= y+z$
similarly $y <= z + x$ and $z <= x + y$ so x,y,z can be sides of a triangle

2016/114) Each coefficient of an equation $ax^2+bx+c=0$ is determined by throwing an ordinary die. find the probability that the equation shall have equal roots.

dice has to be thrown 3 times so number of ways = $6^3= 216$
for the equation to have equal roots we should have $b^2=4ac$
so b has to be even.
taking b=2 we have ac= 1 so a= 1, c = 1 one way
b=4 we have ac = 4 that in 3 ways (a = 1 , c= 4), ( a=2, c= 2), (a =4, c= 1)
b= 6 we have ac =9 in one way a=c=3
so number of ways = 5
hence probability is $\frac{5}{216}$

2016/113) Let a,b,c be respectively the sum of the first n, next n, and next n terms of a GP. Show that a,b,c are in GP.

Let the first number be p and common ratio be q
so we have $a=p(1+q^2+\cdots+q^{n-1})$ , $b = pq^n(1+q^2+\cdots + q^{n-1})$, $c= pq^2(1+q^2+\cdots+q^{n-1})$
hence$\frac{b}{a} = q^n$ and $\frac{c}{b} = q^n$
or $\frac{b}{a} = \frac{c}{b}$ and as ratios are same so a,b,c are in GP

2016/112) Find the value of $ \displaystyle \sum_{k=1}^{n} \cot^2(\frac{\pi k}{2n+1})$

We have
$(\cos\, ma + i \sin\, ma) = (\cos\, a + i \sin\, a)^m$
= $\sum\limits_{k=0}^m {m \choose k } (i\sin\,a)^k (\cos\,a) ^{m-k} $
now compare imaginary parts on both sides to get
$\sin\, ma = {m \choose 1 }\cos^{m-1}a \sin\, a - {m \choose 3 }\cos^{m-3}a \sin ^3 a + ....  $
putting  m = 2n + 1 we get
$\sin(2n+1) a = {2n+ 1 \choose 1 }\cos^{2n}a \sin\, a - {2n+1 \choose 3 }\cos^{2n-2}a \sin ^3 a + ....  $
=  $\sin^{2n+1}a ({2n+ 1 \choose 1 }\cot^{2n}a - {2n+1 \choose 3 }\cot^{2n-2}a + ....) $
The LHS hand hence RHS  becomes zero for $a = \frac{k\pi}{2n+1}$
so for above t
${2n+ 1 \choose 1 }\cot^{2n}a - {2n+1 \choose 3 }\cot^{2n-2}a +... = 0 $
so $\cot^2\frac{\pi}{2n+1}$, $\cot^2\frac{2\pi}{2n+1}$ ... $\cot^2\frac{n\pi}{2n+1}$
are roots of the equation
$({2n+ 1 \choose 1 }x^n - {2n+1 \choose 3 }x^{n-1}  + ....=0$
so sum of roots
=  $\sum\limits_{k=1}^n \cot^2\frac{k}{2n+1} =\frac{{2n+1 \choose 3 }}{{2n+1 \choose 1 }} $
= $\frac{n(2n-1)}{3}$

2016/111) if $s+t+u+v = 0$

 show that $(s^3+t^3 + u^3+v^3)^2 = 9*(st-uv)(tu-sv)*(us-tv)$

Solution

From the above
$(s+t) = -(u+v)\cdots(1)$
also
$s+t+ u = -v \cdots(2)$
cube both sides of (1) to get
$(s+t)^3 = - (u+v)^3$
or $s^3+t^3 + 3st(s+t) = - (u^3+v^3 + 3uv(u+v)$
or $s^3+t^3+ u^3+v^3 = -(3st(s+t) + 3uv(u+v))= -3(st(s+t) - uv(s+t))$ using (1)
or $s^3+t^3+u^3+v^3 = -3(st-uv)(s+t)\cdots(3)$
by symmetry we can show that
 $s^3+t^3+u^3+v^3 = -3(su-tv)(s+u)\cdots(4)$
by multiplying (3) with (4) we get
$(s^3+t^3+u^3+v^3)^2 = 9(st-uv)(su-tv)((s+t)(s+u))$
= $9(st-uv)(su-tv)(s^2+ st + su + ut)$
= $9(st-uv)(su-tv)(s(s+t+u) + ut)$
= $9(st-uv)(su-tv)(-vs + ut)$ using (2)
= $9(st-uv)(ut-sv)(su-tv)$
Proved

2016/110) The sum of 3 numbers of GP is 42. if the 1st 2 number are increased by 2 and 3rd decreased by 4 the resultant of an AP. find the numbers

let the 1st number be a ratio be r
we have 3 numbers $a, ar, ar^2$
the sum = $a + ar + ar^2 = 42\cdots(1)$
from the given condition $(a+2), (ar+2)$ and $(ar^2-4)$ form ap or $2(ar+2) = a +2 + ar^2 - 4$
or $2ar + 4 = ar^2 + a - 2$
or $ar^2 -2ar + a - 6 = 0\cdots(2)$
from (2) we have $ar^2 = 2ar - a +  6$
putting in (1) we get
$a+ ar + 2ar - a +  6 = 42$
o $3ar = 36$
or $ar = 12$
putting above in (1) we get $2(1+r^2) = 5r$
solving this we get $r = 2$ or $r = \frac{1}{2}$
so a= 6(for r =2) or 24 for $r = \frac{1}{2}$
so the numbers are $(24,12,6)$ or $(6,12,24)$

2016/109) Let $-1 <=p <=1$. Show that the equation $4x^3-3x-p=0$ has a unique root in the interval $[\frac{1}{2}, 1]$ find it.

because p is between -1 and 1 so $cos^{-1}p$ is defined. let $x = \cos\,t$ so we get
$\cos\, 3t = p$
and hence $ 3t = \arccos p $ and so $x = \frac{1}{3}\cos^{-}p$
if $x > 1$ we get $4x^3-3x$ > 1 and if $x < \frac{1}{2}$ we get $4x^3 - 3x < -1$ and out side the
limit so x is between $\frac{1}{2}$ and 1

2016/108) If $S_n$ is sum of n terms of a GP show that $S_n(S_{3n} - S_{2n}) = (S_{2n} - S_{n})^2$

we have let 1st term be a and ratio be t
so $S_p = a\frac{t^{p}- t}{t-1}$
we have LHS
$= (a\frac{t^{n}- 1}{t-1})(\frac{a(t^{3n}- 1) -  a(t^{2n}- 1)}{t-1}$
$= \frac{(at^n-1)(a(t^{3n} - t^{2n}}{(t-1)}^2$
$= \frac{a^2t^2(t^n-1)^2}{(t-1)^2}$
$=(\frac{at^n(t^n-1)}{t-1})^2$
RHS = $(S_{2n} - S_{n})^2 = (\frac{at^{2n-1} - 1 - at^n + 1}{t-1})^2 =    (\frac{at^{2n}-at^n}{t-1})^2=LHS$

2016/107) Solve the system of equations

$xy+3y^2-x + 4y-7 = 0$

$2xy+y^2 -2x - 2y +1 = 0$

Solution
multiply the 1st one by 2 and subtract 2nd one to get
$5y^2 + 10y -15 = 0$
or $y^2+2y -3 = 0$ or $(y+3)(y-1)= 0$
y = -3 or y = 1
put y = -3 in 1st equation to get
$-3x + 27  - x - 12 -7 = 0$ or $ x= 2$
now check the 2nd one to get $ -12 + 9 -4 + 6 +1 = 0$ so 2nd one is satisfied and somution = $(x,y) = (2, -3)$
put y =1 in 1st one to get
$ -x + 3 -x + 4 -7$ or $x = 0$
put x= 0 and y = 1 in 2nd one and it is satisfied so solution $(x,y) = (0,1)$

2016/106) solve for x $(x^2+2)^2 + 8x^2 = 6x(x^2+2)$

taking the term on RHS to LHS
 $(x^2+2)(x^2 - 6x + 2) + 8(x^2)  = 0$
$(x^2-3x+2 + 3x )(x^2 - 3x + 2 - 3x) + 8x^2 = 0$
or $(x^2-3x+2)^2  - 9x^2 + 8x^2 = 0$
or $(x^2+ 3x +2)^2 - x^2 = 0$
or $(x^2+2x+ 2)(x^2+4x+2) = 0$
$x^2+2x+2=0$ gives $x= -1\pm i$ and $x^2+4x+2=0$ give $2\pm \sqrt{2}$

2016/105) Solve $x^4+4x^3+3x^2-14x+26=0$

As it is found that this does not have a rational factor and as we know that complex roots appear as conjugate pairs
so it can be factored to quadratic with real coefficient
so Let $x^4+4x^3+3x^2-14x+26$
$ = (x^2+px + q) (x^2 + sx + t)$ as coeffiefint of $x^4$  = 1
$= x^4 + x^3(s + p) + x^2(ps + q + t) + x(pt + qs) + qt$
Comparing coefficients
$s+ p = 4$
$ps +q + t = 3$
$pt+qs = - 14$
$qt = 26 => ( 1,26),(-1,-26),(2,13), (-2, - 13)$
it can be easily solved if we have rational coefficient (above 4 combinations)
by taking $q = 1, t = 26$ or $q = -1 t = -26$ you shall not find any solution(I have tried) but
if we take $q = 2, t = 13$ we get
$s+ p = 4 \cdots(1)$
$sp + 15 = 3$ OR  $ps = -12 \cdots(2)$
$13p + 2s = - 14 \cdots(3)$
From (1) and (3) you get p = - 2 and s = 6 and it satisfies (2)
So we get quadratic factor as $(x^2-2x+2)(x^2+6x+13) = 0$
So we are lucky to get factors with rational coefficients
$(x^2-2x+2) = 0$ gives 2 roots $1\pm i$
And $(x^2+6x+13) = 0$  gives 2 roots $-3\pm 2i$

2016/104) Find all real numbers a such that 3 $< a < 4$ and $a(a - 3\{a\})$ is an integer. where $\{x\}$ is fractional part

let a  = 3 + x
so we have (3+x) (3+x-3x) = (3+x)(3-2x) = n an integer
$9 - 3x -  2x^2 = $ integer
$2x^2 + 3x -k = 0$
the values can be 1 or 2 or 3 or 4
giving the value of x = $\frac{\sqrt{9+8k}-3}{4}$
putting k = 1 to 4 we get a = $3 + \frac{\sqrt{17}-3}{4}$,$ 3 + \frac{1}{2}$,$3 + \frac{\sqrt{33}-3}{4}$, $3 + \frac{\sqrt{41}-3}{4}$

2016/103) if p is natural number show that $p^{n+1} + (p+1)^{2n-1}$ is divisible by $p^2+p+1$

let $f(n) = p^{n+1} + (p+1)^{2n-1}$
we shall prove it principle of mathemetical induction
for n = 1 we have  $f(1) = p^2 + p+ 1 $ and obviously it is divsible by $p^2+p+1$
so we have shown it for base step
let it be dvisible for n = k so $p(k)$ is divisible by $p^2+p+1$
or $p^{k+1} + (p+1)^{2k-1} =  q(p^2+p+1)$
Now we need to show that it is true for k+ 1
$p^{k+2} + (p+1)^{2k+1} = p^(k+2) + (p+1)^2 (p+1)^{2k-1}$
$ = p^{k+2}+ (p^2 + 2p + 1) (p+1)^{2k-1}$
$= p^{k+2} + p((p+1)^{2k-1}) + (p^2 + 2p + 1)((p+1)^{2k-1})$
$=  p(p^{k+1} + (p+1)^{2k-1} + (p^2 + 2p + 1)((p+1)^{2k-1})$
$ = p(q(p^2+p+1)) + p^2 + 2p + 1)((p+1)^{2k-1})$
which is divisible by $p^2+p+1$ hence induction step is proved.
hence proved

2016/102) if${n \choose r-1} = 36$, ${n \choose r} = 84$, ${n \choose r+1} = 126$ find the values of n and r

we have
$\frac{n!}{(r-1)!(n-r+1)!} = 36$, ,
$\frac{n!}{(r!(n-r)!} = 84$
$\frac{n!}{(r+1)!(n-r-1)!} = 126$
by dividing we get
$\frac{r}{n-r+1} = \frac{3}{7}$ or $7r = 3n - 3r + 3 $ or $3n- 10r + 3 = 0$
and  $\frac{r+1}{n-r} = \frac{2}{3}$ or $3(r+1) = 2(n -r)$ or $5r -2n +3=0$
multiplying 2nd by 2 and adding gto 1st we get 9-n = 0 so n = 9 and putting in (1) we get r = 3

2016/101) The circles $x^2+y^2-10x + 16 = 0$ and $x^2+y^2 = r^2$ intersect each other in distinct points

1) $r < 2$ 2) $r > 8$ 3) $2 < r < 8$ 4) $2 <=r <= 8$

Solution
The circle $x^2+y^2 = r^2$ has centre at (0,0) and radius r
the circle $x^2+y^2 - 10x + 16 = 0$
$=>(x^2-10x + 25) + y^2 = 9$
$=>(x-5)^2 + y^2 = 3^2$
this has center at (5,0) and radius 3.
the distance between points = 5
one has radius 3 and another shall intersect if radius be between 5-3 and 5+ 3 or $2 < r < 8$ and hence(3)

2016/100) find $lim_{ x-> inf} (\frac{x+6}{x+1})^{x+4}$

we have $lim_{ x-> inf}(\frac{x+6}{x+1})^{x+4}$
= $lim_{ x-> inf}(1+ \frac{5}{x+1})^{x+4}$
= $lim_{ x-> inf}(1+ \frac{5}{x+1})^{x+1} (1+ \frac{5}{x+1})^3$
= $e^5 * 1 = e^5$

2016/099) Let $a\,b$ and $c$ be the sides of a triangle. Prove that $\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\ge 3$

if we put $x = b+c-a, y = a+c-b, z = a + b -c$ we get
given expression
$= \frac{1}{2}(\frac{y+z}{x} +\frac{z+x}{y} + \frac{x+y}{z})$
$= \frac{1}{2}((\frac{y}{x}+ \frac{x}{y})+( \frac{z}{y} + \frac{y}{z}) + ( \frac{z}{x} + \frac{x}{z}))$
$= \frac{1}{2}(((\sqrt{\frac{y}{x}}- \sqrt{\frac{x}{y}})^2+ 2) +((\sqrt{\frac{y}{z}}- \sqrt{\frac{z}{y}})^2+ 2) + ((\sqrt{\frac{z}{x}}- \sqrt{\frac{x}{z}})^2+ 2)$
$>= \frac{1}{2}(2+2+2)\,or\,3$

2016/098) Let $a = 5^{1000}\sin(1000\alpha)$ where $\sin(\alpha) = \frac{3}{4}$ prove that $a\in Z$

because $\sin \alpha = \frac{3}{5}$ we have $\cos \alpha = \frac{\pm 4}{5}$
we have $\sin(1000\alpha) = Im(\cos(1000\alpha) + i\sin (1000\alpha)) = Im(e^{1000\alpha i})$
$=Im((e^{\alpha i})^{1000})  = Im((\cos \alpha + i\sin\alpha)^{1000}) = Im ( \frac{\pm 4}{5} + i \frac{3}{5})^{1000}$
hence $a = Im (\pm 4 + 3i)^{1000} = Im(\sum_{n=0}^{1000}{1000 \choose b} (\pm 4)^n * (3i)^{1000-n}$
A is imaginary part of the above and as each element is integer we have a is integer

2016/097) Find all integers n such that $\{1,2,3,4, ...,n\}$ can be written as the disjoint union of the subsets, $A$, $B$ and $C$ -whose sums of elements are equal.

for 3 sets the sum of elements to be same sum upto n must be divisible by 3
or $\frac{n(n+1)}{2}$ should be divsible by 3
so n or n+ 1 must be divisible by 3
let us taken n divisible by 3
for n =3 we cannot divide into 3 equal groups
for 6 successive numbers say m+1 to m+ 6 we can devide into 3 groups $\{m+1,m+6\}, \{m+2,m+5\}.\{m+3,m+4\}$
so if n is multiple of 6 we can divide the numbers into groups as above
but if n is not divsible by 6 we can have a group of 9(1-9)  and multiple groups of 6
the multiple groups of 6 can be broken into 3 groups and we can combine them into 3 groups
and 9 can be grouped as $\{4,9,2\},\{3,5,7\}, \{(1,6,8\}$ and take the elements and add to the above groups to get the result.
Let us take n + 1 divisible by 3 so n is of the form 3p + 2 where p >= 1
for p = 1 or n= 5 we can group as $\{4,1\},\{2,3\}, \{5\}$
for p = 2 or n= 8 we can group is $\{4,8\},\{1,5,6\}, \{(2,3,7\}$
for p > 2 we have number of term 5 + 6k or 8 + 6k and we can choose A,B,C accordingly as above

2016/096) Consider the set of integers 1000,1001,1002,...1998,1999,20001000,1001,1002,...1998,1999,2000.

There are times when a pair of consecutive integers can be added without "carrying": 1213+12141213+1214 requires no carrying, whereas 1217+12181217+1218 does require carrying.

For how many pairs of consecutive integers is no carrying required when the two numbers are added?

Solution
Overflow shall not occur in the following cases
1) No digit is greater than 4 and next number digit  is < 5. The lower numbers are 1xxx number of numbers = 125
2) unit digit is 9 so next number unit digit is zetro and no other dight > 4. number of numbers = 25
3) unit and ten's digit is 9 so next number unit and tensdigit is zetro and no other dight > 4. number of numbers = 5
4) 1999 so next number 2000
this gives us 156 numbers

2016/095) If $\frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \cdots = \frac{\pi^4}{90}$

then  $\frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \cdots = ...$

Solution
Let $\frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \cdots = x$
$\frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \cdots = \frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \cdots + \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \cdots$
or $\frac{\pi^4}{90} = x + \frac{1}{2^4}(\frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \cdots)$
$\frac{\pi^4}{90} = x + \frac{1}{16}\frac{\pi^4}{90}$
or $y = \frac{\pi^4}{96}$

2016/094) Prove that in a sequence of numbers 49,4489, .... where each number is made by inserting 48 in the middle of previous number each number is a perfect square.

We have the $n^{th}$ number is n 4's followed by n 8's + 1
n 4's = $4 *(\frac{10^n-1}{9})$ n '8s = $8 *(\frac{10^n-1}{9})$
it may kindly be noted that the numerator  is divisible by 9
so the number = $4 *(\frac{10^-1}{9}) * 10^n + 8 *(\frac{10^-1}{9}) + 1$
$=\frac{4 * (10^n-1) * 10 ^n + 8(10^n-1) + 1}{9}$
$=\frac{ 4 * 10^{2n} + 4 * 10^n + 1}{9}$
$= \frac{(2 * 10^n)^2 + 2 * (2 * 20^n) + 1}{9}$
$= \frac{(2 * 10^n + 1)^2}{3^2}$
$= (\frac{2 * 10^n+1}{3})^2$
Beacuse numerator is divisible by 3 so it is a perfect square

2016/093) Suppose $a_1,a_2,...,a_n$ are poistive real numbers satisfying \(a_1\cdot a_2\cdots a_n=1\).

Show that $(a_1+1)(a_2+1)\cdots(a_n+1)>=2^n$

Solution 
We have using AM GM $\frac{a_k+1}{2} >= \sqrt{a_k}$ or $a_k + 1>=2\sqrt{a_k}$
multiplying over k from 1 to n we have
$\prod_{k=1}^n(a_k+1) >= 2^n \sqrt{\prod_{k=1}^na_k} = 2^n$
or $\prod_{k=1}^n(a_k+1) >= 2^n$

2016/092) a,b,c, are 1st 3 terms of a geometric series . if harmonic mean of a,b is 12 and of b,c is 36 find first 5 terms of the series.

let common ratio be t
then terms are a, at
the harmonic mean is 12
so $\frac{1}{a} + \frac{1}{at} = \frac{1}{6}\cdots(1)$
and
$\frac{1}{at} + \frac{1}{at} = \frac{1}{18}$
deiding (1) by (2) we get $t = 3$
now from (1) $\frac{1}{a} + \frac{1}{3a} = \frac{1}{6}$ or a = 8
giving 5 terms 8,24,72,216,648.

2016/91) if $a_1,a_2,\cdots,a_n$ are in HP show that $a_1a_2+a_2a_3+\cdots + a_{n-1}a_n = (n-1)a_1a_n$

we have
  $\frac{1}{a_1},\frac{1}{a_2}\cdots,\frac{1}{a_n}$ are in AP hence
or $(n-1)(\frac{1}{a_2} - \frac{1}{a_1})  = \frac{1}{a_n} - \frac{1}{a_1}$
or $(n-1)(a_1-a_2)(a_n a_1) = a_2a_1(a_1-a_n) \cdots(1)$
similarly
 $(n-1)(a_2-a_3)(a_n a_1) = a_3a_2(a_1-a_n) \cdots(2)$\
...
   $(n-1)(a_{n-1}-a_n)(a_n  a_1) = a_{n-1}a_m(a_1-a_n) \cdots(n-1)$\
adding above n eqautions we get
$(n-1)(a_1-a_n)(a_n a_1) = (a_1a_2+a_2a_3+\cdots + a_{n-1}a_n) (a_1-a_n)$
cancelling $a_1-a_n$ from both sides we get the result

2016/090) Find the value of $\frac{18^3+7^3+3.18 *7 *25}{3^6+6*243 *2 + 15 *81 *4+20 *27 *8 + 15 *9 *16 + 6*3*32 + 64}$

We have numerator = $18^3+7^3+3.18*7*25 = 18^3+7^3 + 3 *18 *7 (18+7) = (18+7)^3 = 25^3 = 5^6$
denominator = $3^6+6*243*2 + 15*81*4+20*27*8 + 15*9*16 + 6*3*32 + 64$*
$= 3^6 + 6 * 3^5 * 2 + 15 * 3^4 * 2^2  + 20 * 3^3 *2^3 + 15 * 3^2 * 2^4 + 6 * 3 * 2^5 + 2^6$
$= {6 \choose 0}3^6 + {6 \choose 1} * 3^5 * 2 + {6 \choose 2} * 3^4 * 2^2  + {6 \choose 3} * 3^3 *2^3 + {6 \choose 4} * 3^2 * 2^4 + {6 \choose 5} * 3 * 2^5 + {6 \choose 6}2^6$
$= ( 3+2)^6 = 5^6$
so given expression is 1

2016/089) Evaluate $\sin\,18^\circ$ and $\cos\,18^\circ$

Let, $A = 18^\circ$
Then $2A = 90^\circ - 3A$
Taking sine on both sides, we get
$\sin 2A = \sin (90^\circ - 3A) = \cos 3A$
$=> 2 \sin\, A \cos\, A = 4 \cos^3 A - 3 \cos\, A$
or $2 \sin\, A \cos\, A - 4 \cos^3 A + 3 \cos\, A = 0$
or $\cos\, A (2 \sin\, A - 4 \cos^2 A + 3) = 0$
Dividing both sides by $\cos\, A = \cos 18^\circ$ which is not zero we get
$2 \sin\, A - 4 (1 - \sin^2 A) + 3 = 0$
or $4 \sin^2 A + 2 \sin\ A - 1 = 0$ which is a quadratic in $\sin\ A$
hence $\sin\,A = \frac{-1\pm\sqrt{5}}{4}$
but as $\sin\, 18^\circ$ is positive we have $\sin 18^\circ = \frac{-1+\sqrt{5}}{4}$
now $\cos^2 18^\circ= 1- \sin ^2 18^\circ = 1 - (\frac{-1+\sqrt{5}}{4})^2$
$= 1 - \frac{5+1-2\sqrt{5}}{16} = \frac{10+2\sqrt{5}}{16}$
 $\cos\,18^\circ= \frac{\sqrt{10+2\sqrt{5}}}{4}$

2016/088) Which number is smaller $\sqrt{3} + \sqrt{5}$ or $\sqrt{2} + \sqrt{6}$

we have $\sqrt{3} - \sqrt{2}= \frac{1}{\sqrt{3} + \sqrt{2}}$
and  $\sqrt{6} - \sqrt{5}= \frac{1}{\sqrt{6} + \sqrt{5}}$
from the above $\sqrt{3} - \sqrt{2} > \sqrt{6} - \sqrt{5}$
or  $\sqrt{3} + \sqrt{5} > \sqrt{6} + \sqrt{2}$

2016/087) If $a\sin\,x=b\sin(x+\frac{2\pi}{3})=c\sin(x+\frac{4\pi}{3})$ prove that $ab+bc+ca=0$

Let $asin\,x=b\sin(x+\frac{2\pi}{3})=c\sin(x+\frac{4\pi}{3})=k$
hence $\frac{k}{a} = \sin\,x\cdots(1)$
$\frac{k}{b} = \sin(x+\frac{2\pi}{3})\cdots(2)$
$\frac{k}{c} = \sin(x+\frac{4\pi}{3})$
or $\frac{k}{c} = \sin(x-\frac{2\pi}{3})\cdots(3)$

from (1),(2) and (3)
$\frac{k}{a} + \frac{k}{b} + \frac{k}{c} =\sin\,x +  \sin(x+\frac{2\pi}{3}) + \sin(x-\frac{2\pi}{3})$
$= \sin\,x +  \sin\,x\cos \frac{2\pi}{3} + \cos \,x\sin  \frac{2\pi}{3} + \sin\,x\cos \frac{2\pi}{3} - \cos \,x\sin  \frac{2\pi}{3}$
$= \sin\,x +  2\sin\,x\cos \frac{2\pi}{3}$
$= \sin\,x +  2\sin\,x( -\frac{1}{2})$
$= \sin\,x -  \sin\,x$
$= 0$

2016/086) If a,b,c are in AP find the fixed point wthough which line $ax+by+c= 0$ passes

a,b,c are in AP so $a+c = 2b$ or $c = 2b-a$
$ax+by+c=0$
$=>ax + by + (2b-a)=0$ or $(x-1) a + (y+2) b=0$
so the point through which the lines pass is (1,-2) as the above equation should be independent of (a,b)

2016/085)Let n be a positive integer. if $2^{nd}$ , $3^{rd}$ and $4^{th}$ term of expansion of $(1+x)^n$ are in AP find n

we are given
$ {n \choose 1} +  {n \choose 3}=  2{n \choose 2}$
or
$n + \frac{n(n-1)(n-2)}{6} = 2\frac{n(n-1)}{2}$
dividing both sides by $n$ we get
$ 1  + \frac{(n-1)(n-2)}{6} = (n-1)$
or $ (n-2)(1-\frac{n-1}{6} = 0$ giving n= 2 or 7
but $n > =3$ so n = 4

2016/084) If $\cos(x-y) = \frac{4}{5}$ and $\sin(x+y) = \frac{5}{13}$ and x,y are between 0 and $\frac{\pi}{4}$ then find $\tan (2x)$

 we have $\cos(x-y) = \frac{4}{5}=> \sin (x-y) = \sqrt{1-(\cos(x-y))^2} = \sqrt{1-\frac{16}{25}} = \frac{3}{5}$
 and  $\sin(x+y) = \frac{5}{13}=> \cos (x+y) = \sqrt{1-(\sin(x+y))^2} = \sqrt{1-\frac{144}{169}} = \frac{12}{13}$

giving $\tan (x-y) = \frac{\sin (x-y}{\cos(x-y)} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$
and $\tan (x+y) = \frac{\sin (x+y}{\cos(x+ y)} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12}$
hence
$\tan(2x) = \tan (x+y+x-y) = \frac{\tan (x+y) + \tan (x-y)}{1- \tan (x+y)\tan (x-y)} = \frac{\frac{5}{12} + \frac{3}{4}}{1- \frac{5}{12}\frac{3}{4}} = \frac{46}{33}$

2016/083) Find the smallest positive p such that $\cos(p\sin\, x) = \sin (p\cos\,x)$

we have
$\cos(p\sin\, x) = \sin (p\cos\,x)= \cos (\frac{\pi}{2} -  p\cos\,x)$
hence
$p\sin \,x  = \frac{\pi}{2} - p \cos \,x$ (other values shall given -ve / larger p)
hence
$p(\cos \,  x + \sin\, x) = \frac{\pi}{2}$
or $\sqrt{2}p( \sin \frac{\pi}{4} \cos \, x + \cos \frac{\pi}{4} \sin \, x) = \frac{\pi}{2}$
or $\sqrt{2}p( \sin ( x + \frac{\pi}{4}) = \frac{\pi}{2}$
the largest value of $\sin ( x + \frac{\pi}{4})$ is 1 hence smallest $p$ is $\frac{\pi}{2\sqrt{2}}$

2016/082) Let P(x) be a polynomial of x. Show that there exists a polynomial Q(x) such that P(x)Q(x) is a polynomial of $x^3$

We have
$P(x) = \sum_{n=0}^{\infty} a_nx^n$
this can be written as $A(x) + x B(x) + x^2C(x)$
where
$A(x) = \sum_{k=0}^{\infty} a_{3n}x^{3n}$
$B(x) = \sum_{k=0}^{\infty} a_{3n+1}x^{3n}$
$C(x) = \sum_{k=0}^{\infty} a_{3n+2}x^{3n}$
Now let
$R(x) =  A(x) + x B(x)w + x^2C(x)w^2$ where w is cube root of one
and
$S(x) = A(x) + x B(x)w^2 + x^2C(x)w$
using $(a+b+c)(a+bw^2+cw)(a+bw+cw^2) = a^3+b^3+c^3 - 3abc$
we get $P(x)R(x)S(x) = (\sum_{k=0}^{\infty} a_{3n}x^{3n})^3 + (\sum_{k=0}^{\infty} a_{3n+1}x^{3n})^3 x^3 + (\sum_{k=0}^{\infty} a_{3n+2}x^{3n})^3x^6$
$- 3(\sum_{k=0}^{\infty} a_{3n}x^{3n})(\sum_{k=0}^{\infty} a_{3n+1}x^{3n})(\sum_{k=0}^{\infty} a_{3n+2}x^{3n})x^3$
which is a polynomial of $x^3$
now using  $(a+bw+cw^2)( a+ bw^2 +cw) = (a^2+b^2+c^2-ab-bc-ca)$
we have
$R(x)S(x) = (A(x) + x B(x)w + x^2C(x)w^2)(A(x) + x B(x)w^2 + x^2C(x)w)$
$= (\sum_{k=0}^{\infty} a_{3n}x^{3n})^2 + (\sum_{k=0}^{\infty} a_{3n+1}x^{3n})^2)x^2 + (\sum_{k=0}^{\infty} a_{3n+1}x^{3n})^2x^4$
   $- (\sum_{k=0}^{\infty} a_{3n+1}x^{3n}) (\sum_{k=0}^{\infty} a_{3n+1}x^{3n})x$
  $- (\sum_{k=0}^{\infty} a_{3n+1}x^{3n}))(\sum_{k=0}^{\infty} a_{3n+2}x^{3n})x^3$
  $-(\sum_{k=0}^{\infty} a_{3n+2}x^{3n}\sum_{k=0}^{\infty} a_{3n}x^{3n})x^3$
multiplying by the above polynomial say Q(x) we get a polynomial of $x^3$

2016/081) find the number of integer-sided isosceles obtuse angled triangles with perimeter 2008.

let the sides by x,x,y.
so 2x + y = 2008
now 2x >y so y < 1004.
y has to be even.
x cannot be the longest side becuase if x is longest we have isosceles triangle with 2 large angles so cannot be obtuse
so the 3 sides are (y= 1004-2k, x= 502+k, x= 502 + k) and $k < 502$ and k is positive
because it is obtuse
$x^2+x^2 < y^2$ or $2(502+k)^2 < (1004-2k)^2$
solving these we get $k < 502(3-2\sqrt{2})$ or  $k > 502(3+2\sqrt{2})$
because k < 502 so we heve  $k < 502(3-2\sqrt{2})$ giving $k < 86.1432$
so there are 86 triangles

2016/080)Find the coefficient $x^3$ in $1+ (1+x) + (1+x)^2 + (1+x)^3\cdots(1+x)^n$

we have $1+ (1+x) + (1+x)^2 + (1+x)^3\cdots(1+x)^n= \frac{(1+x)^{n+1} -1 }{1+x-1}$
$= \frac{(1+x)^{n+1} -1 }{x}$
so coeffient $x^3$ in $1+ (1+x) + (1+x)^2 + (1+x)^3\cdots(1+x)^n$ is coefficient of $x^4$ in $(1+x)^{n+1} -1$
or ${n+1 \choose 4}$

2016/079) The lines $x^2-3xy+2y^2=0$ are shifted parallel to themselves so that their point of intersection comes to (1,1). find the combined equation of lines in the new position.

we have $x^2-3xy+y^2=0$ represent two lines $(x-y)(x - 2y)= 0$ giving $(x=y=0)$. if we need point of
 intersection 1,1
then we need to replace x by x-1 and y by y-1 in given equation giving
$(x-1)^2 -3(x-1)(y-1) + 2(y-1)^2 = 0$
or $x^2 - 2x +1 -3xy + 3x + 3y -3 + 2y^2-4y + 2= 0$
or $x^2+2y^2 - 3xy + x - y= 0$

2016/078)if $x=cy+bz$, $y= az + cx$ and $z=bx+ay$ then show that $a^2+b^2+c^2 + 2abc = 1$

we put the equations in standard form
$x-cy - bz = 0\cdots(1)$                    
$cx -y + az = 0\cdots(2)$    
$bx + ay - z = 0\cdots(3)$                            
multiply (1) by c and subtract it from (2) to get
$(c^2-1) y + (bc+ a) z = 0\cdots(4)$
mulltiply (1) by b and subtract from (3) to get
$(a+bc) y - (1-b^2) z = 0\cdots(5)$
as (4) and (5) are consistant we get
$(c^2-1)(b^2 - 1) -   (bc+ a)^2 = 0$
or $b^2c^2 - b^2 - c^2 + 1 -  b^2c^2 - a^2 - 2bca$
or $a^2+b^2+c^2 + 2abc = 1$

2016/077) Prove that $1+\cos 2x + \cos 4x + \cos 6x$$ = 4\cos\,x \cos 2x \cos 3x$

$1+\cos 2x + \cos 4x + \cos 6x$
= $(1+\cos 2x) + (\cos 4x + \cos 6x)$
= $(2\cos^2 x) + (\cos 4x + \cos 6x)$ using $cos 2x = 2cos^2 x - 1$
= $(2\cos^2 x) + 2\cos x  \cos 5x)$ using $\cos\, A + \cos \, B = 2(\cos \frac{A+B}{2} +  \cos \frac{B- A}{2})$
= $2\cos\,x(\cos \, x + \cos 5x)$
=$2 \cos\,x(2 \cos 2x  \cos 3x)$ using $\cos\, A + \cos\, B = 2(\cos \frac{A+B}{ 2} +  \cos \frac{B- A}{ 2})$
= $4\cos \, x\cos 2x \cos 3x$

2016/076) If a,,b,c are 3 unequal terms of AP then show that $\frac{b-c}{a-b}$ is rational.

proof
let $1^{st}$ term be p and difference d. Let a be $q^{th}$,  b be $r^{th}$,  c be $s^{th}$
so $a = p+(q-1)t$
$b= p+(r-1)t$
$c= p + (s-1)t$
hence $b-c = (r-s)t$
and $a-b = (q-r)t$
or  $\frac{b-c}{a-b} =  \frac{q-r}{r-s}$
as $q,r,s$ are intgers rhs is rational and hence the result

2016/075) derive the following given $u_n$ the $n^{th}$ fibonacci number $u_{n+1}^2 - 4u_nu_{n-1} = u_{n-2}^2$

we have
$u_{n+1}^2 -  u_{n-2}^2$
$= (u_n + u_{n-1})^2 -  (u_n - u_{n-1})^2$
$ =4u_nu_{n-1}$ using $(a+b)^2 - (a-b)^2 = 4ab$
hence $u_{n+1}^2 - 4u_nu_{n-1} = u_{n-2}^2$

2016/074) you have 25 horses and you have to pick fastest 3 out of the 25. In each race only 5 horses can run at the same time as there are only 5 tracks. what is the minimum number of races to pick the 3 horses without using a stopwatch.

you make the horses into 5 groups of 5 horses each say A,B,C,D,E . have 5 races. then pick the winner of the 5 races and have
a race that is race number 6.. Now pick the 3
winners out of the 5. Say the winner is from group A, the 2nd ranked is from Group B and 3rd one is from group C.
now let the 1st 3 postions in group A be A2,A2,A3. in group B be B1,B2,B3. and in group C be C1,C2, C3.
A1 is the 1st. B3 cannot be in top 3 beacuse A1, B1, B2 are faster.
So B3 is ruled out.
C2 and C3 cannot be in top 3 as A1,B1,C1 are faster. So have a race among A2,A3,B1,B2,C1 and choose the 2 fastest. the ranks
shall be 2nd and 3rd.
So we need 7 races.

2016/073) $\frac{1}{x} + \frac{1}{y} +\frac{1}{z}=1$ what is the value of $x,y$ and $z$ if $x>y>z$ and they are integers

$\frac{1}{x} + \frac{1}{y} +\frac{1}{z} > 3\frac{1}{z} = \frac{3}{z} = 1$

hece $ z < 3$ and as z cannot be 1 so $z = 2$
hence $\frac{1}{x} + \frac{1}{y} =\frac{1}{2}$
hence $2y+ 2x = xy$
hence $xy-2y-2x = 0$
or $xy-2x-2y + 4 = 4$
or $(x-2)(y-2) = 4$
as $x> y$ so we get $x = 2 = 4, y=2 =1$ giving $x = 6, y= 3$
hence $z=2,y=3,x = 6$

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 $N^6+N^3+1$

2016/062) If $x= 2 + \sqrt[3]{2^2} + \sqrt[3]{2}$ then the value of $x^3 - 6x^2+6x$ is

we have  $x-  2 =  \sqrt[3]{2^2} + \sqrt[3]{2}$
hence $(x-2)^3 = 4 + 2 + 3 * 2 *(\sqrt[3]{2^2} + \sqrt[3]{2}) = 6 + 3 * 2 (x-2) = 6x - 6$
Hence $x^3 - 6x^2 + 12x -8 = 6x-6 $ or $x^3-6x^2+6x = 2$

2016/061)Find the smallest positive integer m such that 5m is an exact 5th power, 6m is an exact 6th power, and 7m is an exact 7th power. (26th Irish)

The number has to be of the form $5^a6^b7^c$
now $5^{a+1}6^b7^c$ is a 5th power so  $a+1 \equiv 0 \pmod 5$ ,$b \equiv 0 \pmod 5$,$c \equiv 0 \pmod 5$
$5^a6^{b+1}7^c$ is a 6th power so  $a \equiv 0 \pmod 6$ ,$b+1  \equiv 0 \pmod 6$,$c \equiv 0 \pmod 6$
$5^a6^b7^{c+1}$ is a 7th power so  $a \equiv 0 \pmod 7$ ,$b \equiv 7 \pmod 5$,$c+1 \equiv 0 \pmod 7$
so we need to solve for
$a+1 \equiv 0 \pmod 5$ ,$a \equiv 0 \pmod 42$ giving a = 84 (taking multiples of 42 adding 1 to be divsible by 5)
$b+1 \equiv 0 \pmod 6$ ,$b \equiv 0 \pmod 35$ giving b = 35 (taking multiples of 35 adding 1 to be divsible by 6)
$c+1 \equiv 0 \pmod 7$ ,$c \equiv 0 \pmod 30$ giving c = 90 (taking multiples of 30 adding 1 to be divsible by 7)
so the number is $5^{84}* 6^{35}*c^{90}$

2016/060) Find all integers x such that $x(x+1)(x+7)(x+8)$ is square of an integer (21st Irish math olympiad)

we see that x = 0, x = -1, x = -7 and x = -8 gives the answer zero so a perfect square
let us look for other values
we have
$x(x+1)(x+7)(x+8)$
= $x(x+8)(x+1)(x+7)$
= $(x^2+8x)(x^2+8x+7)$
$= y(y+7)$ where y is $x^2+8x$
for it to be a perfect square we see that $(GCD(y,y+7) = GCD(y,7)$
y cannot be a multiple of if 7 beacuse then y and y + 7 are consecutive multiples of 7 and as y is not zero product cannot
be a perfect square.
so y and y + 7 are coprimes and hence perfect square or -ve of perfect square
let $y = n^2$ and $y+7 = m^2$
giving $n^2+7=m^2$
or $m^2-n^2 = 7$
or $(m+n)(n-n) = 7 * 1$ hence $m+n = 7, m-n= 1$ or $m= 4,n= 3$
so $y = 16$
hence $x^2+8x-9=0$ giving $x = 1,=9$
taking -ve values we have $y= - 16 , y + 7 = - 9$
or $x^2+8x+ 16= 0 => x = - 4$
so we have x is one of -9,-8,-7, -1,0,1$

2016/059) Show that in 10 consecutive numbers there is at least one number which is co-prime to other 9 numbers

Out of 10 consecutive numbers 5 numbers are divisible by 2 and not more that 4 numbers are divisible by 3 out of which maximum 2 numbers are odd and divisible by by 3 that makes 7, 2 numbers are divisible by 5 out of which is even so there are maximum one number is divisible by 5 and neither 2 nor 3 and that nakes 8 and maximum 2 numbers are divisible by 7 out of which is only one is odd maximum one number is divisible by 7 and neither 2 nor 3 nor 5 and that makes maximum 9 numbers that are divisible by one of 2,3,5,7. so there is at least one number which is not divisible by 2,3,5 or 7 so the lowest prime factor of the same is 11 and it cannot devide any other number of the set. so this number is co-prime to rest 9.

2016/058) Show that $\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\cdots\frac{1}{2n-1}$

=$1-\frac{1}{2} + \frac{1}{3} + \frac{1}{4}\cdots -\frac{1}{2n-1}$
 We have RHS = $\sum_{k=1}^{n}\frac{1}{2k-1} - \sum_{k=1}^{n}\frac{1}{2n}$ $=\sum_{k=1}^{n}\frac{1}{2n-1} + \sum_{k=1}^{n}\frac{1}{2k} - 2 \sum_{k=1}^{n-1}\frac{1}{2k}$ $=\sum_{k=1}^{2n-1}\frac{1}{k}- \sum_{k=1}^{n-1}\frac{1}{k}$ $=\sum_{k=n}^{2n-1}\frac{1}{k}=LHS$

2016/057)If a,b,c and d are in G.P then show that $(b-c)^2+(c-a)^2+(d-b)^2=(a-d)^2$

a,b,c,d are in GP let common ratio be x $b = ax,c = ax^2,, d = ax^3$ LHS = $(ax-ax^2)^2 + (ax^2-a)^2 + (ax^3- ax)^2$ $= a^2(x^2 (1-x)^2) + (x^2-1)^2 + x^2(x^2-1)^2)$ $= a^2(x^2(1-2x+x^4) + ( 1- 2x^2 + x^ 4) + x^2(x^4-2x^2+ 1)$ $= a^2(1- 2x^3+x^6)$ $= a^2(1-x^3)^2$ $= ( a- (ax^3))^2$ $= (a-d)^2$

2016/056) If $n^4+an^3+bn^2-8n+1 $ is a perfect square for all integer values of n. find a,b

$n^4+an^3+bn^2-8n+1 $ is a perfect square for all integer values of n so this is square of a polynomial.
hence
$n^4+an^3+bn^2-8n+1 = (n ^2 + mn + c)^2 $
comparing the constant term we have $1=c^2$ or $c = \pm 1$
so we need to consider 2 cases
c = 1 and c= -1
case 1
c = -1
this give $(n^2 + mn + 1)^2 = n^4 + mn^3 + n^2(m^2+2) + mn +1 = n^4+an^3+bn^2-8n+1$
comparing coeffcients on both sides $m = a, b = m^2 + 2, - m = -8$ giving $a= 8, b = 66$
case 2
c = -1
this give $(n^2 + mn - 1)^2 = n^4 + mn^3 + n^2(m^2-2) - mn +1 = n^4+an^3+bn^2-8n+1$
comparing coefficients on both sides $m = a, b = m^2 - 2, - m = -8$ giving $a= -8, b = 62$
so we have 2 sets $a=8, b= 62$ and $a= -8,b=66$

2016/055) what is the smallest number with 101 factors

a number of the form $p_1^{q_1} * p_2^{q_2} * p_3 ^{q_3}\cdots$ has $(1+q_1)(1+q_2)(1 + q_3) \cdots$ factors
now 101 is prime so smallest number = $2^{100}$

2016/054) Let $a,b,c$ be rational and one of the roots of $ax^3+bx+c=0$ is equal to product of other two roots. Prove that this root is rational.

we can devide by a to get $x^3+dx+e=0$ where $d=\frac{b}{a},d=\frac{c}{a}$ $d,e$ are rational.
Let $\alpha,\beta,\alpha\beta$ be the three roots
so we get using vieta's relations
$\alpha+\beta+\alpha\beta= 0\cdots(1)$
$\alpha\beta+\alpha \alpha\beta + \beta\alpha\beta  = d\cdots(2)$
$\alpha\beta\alpha\beta= \alpha^2\beta^2= -e\cdots(3)$
from (2)
$\alpha\beta+ \alpha\beta(\alpha+\beta)  = d$
or $\alpha\beta+ \alpha\beta(\alpha+\beta)  = d$
or $\alpha\beta+ \alpha\beta(-\alpha\beta)  = d$ (Using (1)
or $\alpha\beta - (\alpha\beta)^2  = d$
or $\alpha\beta + e  = d$
or $\alpha\beta = d - e$
hence the root  $\alpha\beta$ is rational

2016/053) Solve the system of equations in real

$4x^2+25y^2 +9z^2 - 10xy -15yz - 6xz = 0\cdots(1)$
$x+y+z=5\cdots(2)$

Solution
from (1) we have
$8x^2+50y^2 +18z^2 - 20xy -30yz - 12xz = 0$
or  $(4x^2 - 20xy + 25y^2) + (25y^2 - 30yz + 9z^2) + (9z^2 - 12xz + 4x^2)= 0$
or $(2x-5y)^2 + (5y-3z)^2 + (3z-2x)^2=0$
above is sum of 3 squares and hence each of them is zero ior $2x = 5y = 3z=k$ (say)
so $x= \frac{k}{2}$, $y= \frac{k}{5}$,$z= \frac{k}{3}$
putting in (2) we get
$\frac{k}{2} + \frac{k}{5}+  \frac{k}{3} = 5$
or $\frac{31k}{30} = 5$
or  $k = \frac{150}{31}$
so $x= \frac{75}{31}$, $y= \frac{30}{31}$,$z= \frac{50}{31}$

2016/052) if $n\sin \theta =\sin (\theta+ 2a)$ then $\tan (\theta+a) = ?$

$\frac{\sin (\theta+2a)}{\sin\theta} = n$
using componendo dividendo we have
$\frac{\sin (\theta+2a) +  \sin \theta}{\sin (\theta+2a) -  \sin \theta} = \frac{n+1}{n-1}$
Or $\frac{2 \sin (\theta+a)  \cos\,  a}{2 \cos (\theta+a)  \sin\, a} = \frac{n+1}{n-1}$
Or $\frac{tan  (\theta+a)}{\tan\,   a} = \frac{n+1}{n-1}$

Or  $\tan  (\theta+a) =    \frac{n+1}{n-1} \tan\, a$

2016/051) If $1, w, w^2$ are cube roots of unity and $a+b+c=0$ then prove that $(a+bw+cw^2)^3+(a+bw^2+cw)^3=27abc$

let $X = a+bw+cw^2$
$Y = a+bw^2+cw$
$X+Y = 2a + b(w+w^2) + c(w^2+w) = 2a - b -c = 3a - (a+b+c) = 3a$
let $Z = -3a$
so $X+Y+Z=0$
so $X^3 + Y^3 + Z^3 = 3XYZ = -9a(a+bw+cw^2)(a+bw^2+cw)$
           $ = -9a(a^2 + b^2 + c^2 - ab - ac - bc)$
or $X^3 + Y^3 + (-3a)^3 = -9a((b+c)^2 + b^2 + c^2 - a(b +c) - bc)$
or $X^3 + Y^3 + (-3a)^3 = -9a((b+c)^2 + (b+c)^2 - 2bc + (b +c)^3 - bc)$
or $X^3 + Y^3 = 27a^3 -9a(3(b+c)^2 + 3 bc)$
$ = 27a(a^2 -  (b+c)^2  + bc)$
$= 27a((b+c)^2 - (b+c)^2 + bc) = 27abc$
hence  $(a+bw+cw^2)^3+(a+bw^2+cw)^3=27abc$

2016/050) Let a,b be roots of $x^2-3x + A =0$ and c,d be roots of $x^2-12x+B=0$ if $a,b,c,d$ form an increasing GP then find A and B.

because a,b,c,d are in GP let common ratio be t then we have $b=at,c=at^2,d=at^3$
a,b are roots of $x^2-3x + A =0$ so $a+at= 3$ or $a(1+t) = 3\cdots(1)$
c,d are roots of $x^2-12x+C=0$ so $c+d = 12$ or $at^2+bt^3= 12$ or $at^2(1+t) = 12\cdots(2)$
from (1) and (2) $t^2 = 4$ ot $t=2$ because t has to be positive otherwise $a,b,c,d$ cannot be increasing
from (1) $a = 1$ so $a=1,b=2,c=4,d=8$ and hence $A=ab= 2,C=cd = 32$

2016/049) Find the equation of the normal to the curve $x^2= 4y$ that passes through the pont(1,2) (IIT 1984)

equation of the curve $4y = x^2$
so slope of the tangent = $\frac{dy}{dx} = \frac{x}{2}$
so slope of the normal  = $- \frac{2}{x_1}$
the line passes through   $x_1,\frac{x_1^2}{4}$ and $(1,2)$ so slope of line
$\frac{frac{x_1^2}{4} - 2}{x_1-1} = - \frac{2}{x_1}$
$x_1^3 - 8x_1 = - 8x_1 + 8$ or $x_1 = 2$
giving slope of normal = - 1
so equation of normal $= (y-2) = - (x-1)$ or $x+y=3$

2016/048) Let $f(x) = Ax^2 + Bx +C$ where A,B,C are real numbers. prove that if (x) is integer for all integers x then 2A, A + B, C are integers. prove the converse as well.

we have $f(x) = A x^2 + Bx + C = A (x^2-x) + (A+B) x + C=2A\frac{x(x-1)}{2} + (A+B) x + C$
now x and $\frac{x(x-1)}{2}$ are integers for integer x. so if (A+B),2A and C are integers the $f(x)$ is integer for integer x
if f(x) is integer for all x then $f(0) = C$ is integer.
$f(1) = (A+B) 1 + C$ is integer so $A+B$ is integer
$f(2) = 2A + 2(A +B) + C$ is integer so 2A is integer

2016/047) if $\alpha+\beta+\gamma = \pi$ then prove that $\sin ^2\alpha + \sin ^2 \beta - \sin ^2 \gamma = 2 \sin\,\alpha\sin\,\beta \cos\,\gamma$

$\sin ^2\alpha + \sin ^2 \beta - \sin ^2 \gamma$
$= \sin ^2\alpha + \sin^2  \beta - \sin ^2 \gamma(\sin ^2 \beta + \cos  ^2 \beta)$
$= \sin ^2\alpha + \sin^2  \beta(1 - \sin ^2 \gamma) - \sin ^2 \gamma \cos  ^2 \beta$
$= \sin ^2\alpha + \sin^2  \beta\cos ^2 \gamma - \sin ^2 \gamma \cos  ^2 \beta$
$= \sin ^2\alpha + (\sin\, \beta\cos\, \gamma + \sin\, \gamma \cos\, \beta)(\sin\, \beta\cos\, \gamma - \sin\, \gamma \cos\, \beta)$
$= \sin ^2\alpha + (\sin(\beta + \gamma)\sin(\beta- \gamma)$
$=\sin\,\alpha \sin (\sin(\beta + \gamma) + \sin\,\alpha\sin(\beta - \gamma)$ as $\sin\,\alpha= \sin(\beta + \gamma)$
$=\sin\,\alpha (\sin(\beta + \gamma)+\sin(\beta - \gamma))$
$=\sin\,\alpha (2\sin\,\beta \sin\,\gamma)$
$=2 \sin\,\alpha \sin\,\beta \sin\,\gamma$

2016/046) Show that there is no combination of integers a,b,c for which $f(n) = n^3 + an^2 + bn+ c$ is perfect square for all n

If we can show that it cannot be for n = 1,2,3,4 then we are through
for n = 1 we have $f(1) = 1 + a + b+ c\cdots(1)$
for n= 2 we have $f(2) = 8 + 4a + 2b + c\cdots(2)$
$f(3) = 27 + 9a + 3b + +c\cdots(3)$
$f(4) = 64 + 16a + 4b + c\cdots(4)$
if f(1) and f(3) are perfect squares then from (1) and (3) we get $26+ 8a+ 2b \equiv 0 \pmod 4$ or $2b \equiv 2 \pmod 4\cdots(5)$
if f(2) and f(4) are perfect squares then from (2) and (4) we get $56+ 12a+ 2b \equiv 0 \pmod 4$ or $2b \equiv 0 \pmod 4\cdots(6)$
(5) and (6) cannot be satisfied at the same time so for n = 1 to 4 it cannot be a perfect square

2016/045) Show that there exists infinitely many pairs of coprime integers (a,b) such that both $x^2+ax+b=0$ and $x^2+2ax+b=0$ have integer solution

from the above b can be expressed as product of 2 numbers in 2 different ways that is at least 3 integers say uvw.
now a = uv + w
and 2a = u + vw
so 2(uv+w) = u + vw
or u(2v-1) = w (v- 2)
the above holds if we choose (49) w = 2v- 1 and u = v-2.
that gives
$a = v(v-2) + 2v-1 = v^2 - 1 = (v-1)(v+1)$
$b= v(v-2)(2v-1)$
for a b to be coprime (v-1) should be co prime to v, (v-2) and (2v-1) which is true for any v
and (v+1) should be co prime to v, (v-2) and (2v-1) which is true unless v+1 is divisible by 3.
so we can choose v not divisible by 3 and get corresponding a,b for the same.
hence there are inifinite of them. for example v = 7 giving a = 48 and b= 13 * 35 = 455
$x^2+48x + 455 = 0$ gives solution x = -33,-35
$x^2+96x + 455 = 0$ gives solution -91,- 5

2016/044) Sum to infinity $1+\frac{3}{1!}+\frac{5}{2!}+\frac{7}{3!}\cdots$

we have for n > 1$n^th$ term = $\frac{2n-1}{(n-1)!} = \frac{2n-2+1}{(n-1)!} = 2 \frac{1}{(n-2)!} +  \frac{1}{(n-1)!}$
so the sum = $1 + \sum_{n=2}^{+\infty} (2 \frac{1}{(n-2)!} +  \frac{1}{(n-1)!})$
$= 1 + \sum_{n=2}^{+\infty} (2 \frac{1}{(n-2)!} +  \frac{1}{(n-1)!})$
$= 1 + \sum_{n=2}^{+\infty} \frac{1}{(n-1)!} + 2 \sum_{n=2}^{+\infty} \frac{1}{(n-2)!}$
$= 1 + \sum_{n=1}^{+\infty} \frac{1}{n!} + 2 \sum_{n=0}^{+\infty} \frac{1}{n!}$
$= \sum_{n=0}^{+\infty} \frac{1}{n!} + 2 \sum_{n=0}^{+\infty} \frac{1}{n!}$
$= 3 \sum_{n=0}^{+\infty} \frac{1}{n!} =3e$

2016/043) Factor in real $(1-(ab+bc+ca))^2 + (a+b+c-abc)^2$

using $a^2+b^2 = (a+ib)(a-ib)$
we get  $(1-(ab+bc+ca))^2 + (a+b+c-abc)^2$
$ = ( 1 +(a+b+c) i - (ab+bc+ca) - abci))  ( 1 - (a+b+c) i - (ab+bc+ca) + abci)$
$= ( 1 +(a+b+c) i + (ab+bc+ca)i^2  + abci^3 ))$
 $ ( 1 - (a+b+c) i + (ab+bc+ca)^2 - abci^3)$
$= ( 1 +ai)(1+bi)(1+ci)(1- ai)(1-bi)(1-ci)$
$= ( 1 +ai)(1-ai)(a+bi)(1- bi)(1+ci)(1-ci)$
$= ( 1 +a^2)(1+b^2)(1+c^2)$

2016/042) If $(m_r,\frac{1}{m_r})$ for r from 1 to 4 are 4 points that lie on a circle show that $m_1m_2m_3m_4= 1$

we have general equation of a circle
$x^2+y^2+ 2gx + 2fy + c = 0$
let a point $p,\frac{1}{p}$ be on the circle
then we get
$p^2 + \frac{1}{p^2} + 2gp + \frac{2f}{p} + c = 0$
or $p^4 + 1 + 2gp^3 + 2fp + cp^2=0$
or $p^4 + 2gp^3 + cp^2 + 2fp + 1=0$
the 4 roots are $m_1,m_2,m_3,m_4$ and hence product of roots = $m_1m_2m_3m_4=1$ (the constant term)

2016/041) If $ax + y + 1= 0, x +by+1=0,$ $x+y+c=0$ are concurrent then prove that $\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$

proof
We have
$a = - \frac{y+1}{x}$
or $1- a = \frac{x+y+1}{x}$
or $\frac{1}{1-a} = \frac{x}{x+y + 1} \cdots 1$
Similarly
or $x +by+1=0$
or  $x+1 = - by$
or $b = -\frac{x+1}{y}$
or $1-b= \frac{x+y+1}{y}$
or $\frac{1}{1-b} = \frac{y}{x+y+1}\cdots (2)$

and $x+y+c=0$
$=>c = x + y$
$=>1-c = x+y+1$
or $\frac{1}{1-c} = \frac{1}{x+y+1}\cdots(3)$

adding all 3 we get the
$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = \frac{x+ y + 1}{x+y+1} =1$

Proved

2016/040) Derive the identity $u_{n+3} = 3u_{n+1} - u_{n-1}$ where $u_n$ is the $n^{th}$ fibonacci number and $n >= 2$

$u_{n+3} = u_{n+2} + u_{n+1}$
$= u_{n+1} + u_{n} + u_{n+1}$
$= 2u_{n+1} + u_{n}$
$= 2u_{n+1} + u_{n+1} - u_{n-1}$
$= 3 u_{n+1} -  u_{n-1}$

2016/039) if $x=log_abc, y= log_bca, z= log_cab$ then show that $\frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} = 1$

we have $x=log_abc$ hence $x + 1 = log_abc + log_aa = log_aabc$ or $\frac{1}{x+1} = log_{abc}a$
 similarly $\frac{1}{y+1} = log_{abc}b$ and $\frac{1}{z+1} = log_{abc}c$
 Hence $\frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} = log_{abc}a + log_{abc}b = log_{abc}c = log_{abc}abc =1$

2016/038) Solve the equation $3x-7y \equiv 11 \pmod {13}$

to solve this we solve 2 equations below and combine the two
$3x - 7y = 11\cdots(1)$
$3x - 7y = 13\cdots(2)$
let us solve $3x - 7y = 11$ first
we need to find solution of $3x-7y =1$. this can be done by extended euler algorithm but
we see that $(-2,-1)$ satisfy $3x-7y = 1$ so (-22,-11) satisfies $3x-7y = 11$
$(-22 + 7t, -11 + 3t)$  is a solution. choose t to be 4 to get (6,1) to be a soution.
for solving $3x - 7y = 13$ we have a solution $(-26,-13)$ so also $( -26 + 7t, - 13+ 3t)$ and
 putting t = 4 we get $(2,-1)$ a solution.
So solution for $3x-7y \equiv 11 \pmod {13}$ is $(6 + 2t, 1 - t)$. because the value need to be computed in mod 13 so the ans is $(6 + 2t \pmod {13} , (1 - t) \pmod {13})$.  

2016/037) Solve the Diophantine below $12x + 25 y = 331$

Taking mod 25 we get
$12x \equiv 6 \pmod {25}$
we need to find inverse of $12\pmod {25}$
It can be done many ways , as we see that $12 * 2 = 24$ so multiply both sides by 2 to get
$24 x \equiv 12 \pmod {25}$
or $ x \equiv -12 \pmod {25}$ or  $ x \equiv 13 \pmod {25}$
so $x = 13$ is one of the values and putting $x = 13$ in given equation we get $y = 6$
so one solution $ x= 13,y = 6$ and as $12 * (-25) + 25 * 12 = 0$ we get generic solution
$ x = 13 - 25t, y= 6 + 12t$

2016/036) Prove that $\arg[(a+bi)(c+di)]$

$=\arg(a+bi)+\arg(c+di) \pmod \pi $
Proof:
we have $(a+bi)(c+di) = (ac - db) + (bc + ad)i$
hence $\arg[(a+bi)(c+di)] = \arg (ac - db) + (bc + ad)i$
or  $\arg[(a+bi)(c+di)] = \arctan (\frac{bc + ad}{ac-db})$
$= \arctan (\frac{\frac{b}{a}+ \frac{d}{c}}{1 - \frac{b}{a}\frac{d}{c}})$
$= \arctan (\tan ( \arctan(\frac{b}{a}) + \arctan(\frac{d}{c}))$
$= \arctan(\frac{b}{a}) + \arctan(\frac{d}{c})\pmod \pi $
$= \arg(a+bi)+\arg(c+di) \pmod \pi $

2016/035) If one root of $4x^2+2x-1= 0 $ be $\alpha$ then show that other root is $4\alpha^3-3\alpha$

We have $4\alpha^2+2\alpha-1= 0\cdots(1) $
We need to prove
$4(4\alpha^3-3\alpha)^2+2(4\alpha^3-3\alpha)-1= 0 $
We have $4\alpha^2+2\alpha-1= 0$
Hence $4\alpha^3+2\alpha^2-\alpha= 0$
So $4\alpha^3 = -2\alpha^2 + \alpha$
or $4\alpha^3 -3\alpha = - 2\alpha^2 - 2 \alpha=  - \frac{1}{2}( 4\alpha^2 + 4\alpha)$
$= - \frac{1}{2}( 1 - 2\alpha + 4\alpha)= - \frac{1}{2}( 1 + 2\alpha) \dots(2)$
Hence $4(4\alpha^3-3\alpha)^2+2(4\alpha^3-3\alpha)-1= 0 $
 = $4(\frac{1}{2}( 1 + 2\alpha))^2- 2 (\frac{1}{2}( 1 + 2\alpha)-1$ using (2)
 = $( 1 + 2\alpha))^2- ( 1 + 2\alpha)-1$
 = $1 + 4 \alpha + 4\alpha ^2 - 1 -2 \alpha -1$
 = $4\alpha ^2 + 2 \alpha -1= 0$ from (1)

2016/034) Show that $x^2-xy+y^2 - 4x - 4y + 16=0$ has only one solution in real x,y. find it

 we have multiplying by 2
$2x^2-2xy+2y^2 - 8x - 8y + 32=0$
or $(x^2 - 2xy + y^2) + (x^2-8x + 16) + (y^2 - 8y + 16) = 0$
or $(x-y)^2 +(x-4)^2 + (y-4)^2=0$
hence we have each of the squares zero giving $x=y=4$ the only solution

2016/033) If $a,b,c$ are in AP and $a^2,b^2,c^2$ are in HP then prove that either $a=b=c$ or $a,b,\frac{-c}{2}$ are in GP

we have $2b = a + c\cdots(1)$
$\frac{2}{b^2} = \frac{1}{a^2} + \frac{1}{c^2}$
or  $2a^2c^2 = b^2c^2 + b^2 a^2 = b^2(a^2+c^2) = b^2((a+c)^2 - 2ac) = b^2(4b^2 - 2ac)$
or $a^2c^2 = 2b^4 - b^2ac$
or $2b^4 - b^2ac - a^2c^2 = (2b^2+ac)(b^2-ac) = 0$
$b^2= ac =>  4b^2 = 4ac = (a+c)^2 => (a-c)^2 = 0 => a = c = b$
or $2b^2+ ac = 0 => a,b ,\frac{-c}{2}$ are in GP

2016/032) if $a,b,c$ are positive real numbers show that $(1+a)^7(1+b)^7(1+c)^7 > 7^7 a^4b^4c^4$

we have
$(1+a)(1+b)(1+c) = 1 + a + b + c + ab + bc + ca + abc $
$>     a + b+ c+ ab+bc+ca + abc\cdots(1)$
By am gm inequality we have $\frac{a + b+ c+ ab+bc+ca + abc}{7} >= \sqrt[7]{a^4b^4 c^4}$
or  $a + b+ c+ ab+bc+ca + abc >= 7 \sqrt[7]{a^4b^4 c^4}$
or $ 1  + a + b+ c+ ab+bc+ca + abc > 7 \sqrt[7]{a^4b^4 c^4}$
or $ (1  + a)( 1+ b)(1+ c) > 7 \sqrt[7]{a^4b^4 c^4}$ from (1)
hence  $(1  + a)^7( 1+ b)^7(1+ c)^7 > 7^7 a^4b^4 c^4$

2016/031) Find 3 digit number which is same as sum of factorials

Let the number be $100 a + 10b+c$
none of $a,b, c$ can be $> 5$ as $6! = 720$ and $7! = 5040 > 1000$
one of them that is b or c= 5 ( a cannot be 5 as $5! = 120$ and $5! + 4! + 3! < 200$)
so $a = 1, b= 5, c = ?$ or $a = 1, b = ? , c = 5$ ( it has to be $< 5$)
if  $a = 1 , b = 5$ we get $1 + 120 + c ! > 150$ and $< 160$
$39 >c! > 29$ and so there is no c
if $a = 1, c = 5$ we get $1 + 120 + b! = 105 + 10 b$
so b = 4 and the number is 145

2016/030) Let $R= (5\sqrt{5} + 11)^{2n+1}$ and $f= R- \lfloor R \rfloor $ where $\lfloor R \rfloor$ is greatest integer functon. Prove that $Rf = 4^{2n+1}$

we have $(5\sqrt{5} - 11) = \frac{(5\sqrt{5} - 11)(5\sqrt{5} - 11)}{5\sqrt{5} + 11}$
$= \frac{125-121}{5\sqrt{5} + 11} = = \frac{4}{5\sqrt{5} + 11} < 1$
$R-r =   (5\sqrt{5} + 11)^{2n+1} -  (5\sqrt{5} - 11)^{2n+1}$
$= \sum \limits_{i=0}^{2n+1}{2n+1\choose i} (5\sqrt{5})^i 11^{2n+1-i} -\sum \limits_{i=0}^{2n+1}{2n+1\choose i} (5\sqrt{5})^i  (-11)^{2n+1-i} $
$= 2\sum \limits_{i=0}^n{2n+1\choose i} (5\sqrt{5})^{2i} 11^{2n+1-1}$
$= 2\sum \limits_{i=0}^n{2n+1\choose i} 125^i 11^{2n+1-1}$ an integer
so $r = R -\lfloor R \rfloor$
So $r = f = (5\sqrt{5} - 11)^{2n+1}$
Hence
$Rf = (5\sqrt{5} +  11)^{2n+1}(5\sqrt{5} - 11)^{2n+1}$
$= ((5\sqrt{5} + 11)(5\sqrt{5} - 11))^{2n+1} = (125-121)^{2n+1} = 4^{2n+1}$

2016/029) Solve $2\log_x a + \log_{ax} a + 3\log_b a =0$ where $a > 0$ and $b=a^2 x$

We have $\frac{2}{\log_a x}  + \frac{1}{\log_a ax} + \frac{3}{\log_a a^2 }x= 0$
$=> \frac{2}{\log_a x} + \frac{1}{\log_a x + 1} + \frac{3}{\log_a x + 2}=0$
$=>  \frac{2}{p } + \frac{1}{1+p } + \frac{3}{2 + p}= 0$ where $p = \log_a x$
$=>2(1+p)(2+p) + p(2+p) + 3p(1+p) = 0$
$=> 4 + 6p + 2 p^2 + 2p + p^2 + 3p + 3p^2 = 6p^2 + 11p + 4$
$ = 6p^2 + 8p + 3p+4 = (3p+4) (2p+1)=0$
$=> p = \frac{-4}{3}$ or $\frac{-1}{2}$
Hence $x= a^{-\frac{4}{3}}$ or $a^{\frac{-1}{2}}$

2016/028) Let $a,b$ be roots of equation

Let $a,b$ be roots of equation $x^2-10cx+11d=0$ and $c,d$ be roots of equation
$x^2-10ax-11b=0$ then find the value of $a+b+c+d$ when a,b,c,d are all distinct.

we have $a,b$ are roots of $x^2-10cx + 11d = 0$
Hence $a+b = 10c\cdots(1)$
$ab= 11d\cdots(2)$
similarly
$c+d= 10a\cdots(3)$
$cd = 11b\cdots(4)$
from (1) and  (3)
$a+b+c+d = 10(c+a)$ or $b+d = 9(a+c)\cdots(5)$
from (2) and (4)
$abcd = 121bd$ or $ac = 121\cdots(6)$
now because a and b satisfy $x^2-10cx + 11d = 0$
we have
$a^2-10ca + 11d = 0$
similarly
$c^2 10ca + 11b = 0$
add to get $(a+c)^2 -2ac - 20ac +11(b+d) = 0$
or $(a+c)^2 - 22 * 121 + 99(a+c) = 0$
So $a + c = - 22\, or\,  121$
$a + c = -22$ means $a = c = -11$  wich is inadmissible
so $a + c = 121$
so $a + b+ c+ d = 10(a+ c) = 1210$

2016/027) Solve for integers $a,b,c$ given $a^2+b^2+c^2 + a + b+ c = 1$

Multiplying by 4 and adding 3 on both sides we get
$(4a^2+4a+1) + (4b^2 + 4 b+ 1) + (4c^2 + 4c+1) = 7$
or $(2a+1)^2 + (2b+1) ^2 + (2c+1)^2   = 7$
as for n odd $n^2 = 1 \, mod \, 8$
we have $LHS = 3\, mod \, 8$ and $RHS = 7 \,mod \, 8$ so no solution

2016/026) find the highest 3 digit prime factor of ${2000 \choose 1000}$

we have
${2000 \choose 1000} = \frac{2000!}{1000!1000!}$
so the prime p occurs $ \lfloor \frac{2000}{p} \rfloor - 2\lfloor \frac{1000}{p} \rfloor $ times
now if p is 3 digit $> \frac{2000}{3}$ or $>666$ then
$ \lfloor \frac{2000}{p} \rfloor = 2 $
$ \lfloor \frac{1000}{p} \rfloor = 1 $
so $ \lfloor \frac{2000}{p} \rfloor - 2\lfloor \frac{1000}{p} \rfloor =0 $
if  is $< 666$
$ \lfloor \frac{2000}{p} \rfloor - 2\lfloor \frac{1000}{p} \rfloor >= 1 $
so largest p is largest prime $< 666$ and it is $661$

2016/025) find the least possible value of a + b where a and b are positive and 11 divides $a+ 13b$ and 13 divides $a + 11 b$

$11$ divides $a + 2b$ and hence $11$ divides $6a + 12b$ or $11$ divides $6a + b$
$13$ divides $a - 2b$ and hence $13$ divides $6a - 12b$ or $13$ divides $6a + b$
so $6a + b$ is divisible by $11$ and $13$ and hence $143$
say $6a +b = 143 t\cdots(1)$
$6a + 6b = 143t + 5b = 144 t + 6b - ((t+b)$
so $t + b$ is divisible by $6$ and hence $t + b > 6 \cdots(2)$
$6(a+b) = 143t + 5b = 138 t + 5(t+b) >=168$
hence $a + b >= 28$
From (2) putting $t = 1$ we get $b= 5$ and from (1) we get $a = 23$ so $a=23$ and $b=5$ satisfies
the condition so $a+b$ lowest value is 28

2016/024)integrate $\int_{x = 0}^{\pi}\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} dx$

as we have $\int_{x = a}^{b} f(a) = \int_{x = a}^{b} f(b)$
$\int_{x = 0}^{\pi}\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} dx$
$=\int_{x = 0}^{\pi}\frac{e^{\cos(\pi-x) }}{e^{\cos (\pi-x) }+e^{-\cos (\pi- x) }} dx$
$=\int_{x = 0}^{\pi}\frac{e^{^-\cos x}}{e^{- \cos (x) }+e^{\cos  x}} dx$
$=\int_{x = 0}^{\pi}\frac{e^{^-\cos x}}{e^{\cos (x) }+e^{-\cos  x}} dx$
hence
$\int_{x = 0}^{\pi}\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} dx= \int_{x = 0}^{\pi}\frac{e^{^-\cos x}}{e^{\cos (x) }+e^{-\cos  x}} dx$
$=\frac{1}{2}\int_{x = 0}^{\pi}(\frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} + \frac{e^{^-\cos x}}{e^{\cos (x) }+e^{-\cos  x}})dx$
$=\frac{1}{2}\int_{x = 0}^{\pi} 1 dx = \frac{\pi}{2}$

2016/023) Show that $\lim_{n\to\infty} \frac{1}{n+1} + \frac{1}{n+2}\frac{1}{n+3} +\cdots + \frac{1}{6n} = ln 6$

we have $\lim_{n\to\infty} \frac{1}{n+1} + \frac{1}{n+2}\frac{1}{n+3} +\cdots + \frac{1}{6n}$
$= \lim_{n\to\infty} \sum_{r=1}^{5n} \frac{1}{n+r}$
$= \lim_{n\to\infty}\frac{1}{n} \sum_{r=1}^{5n} \frac{1}{1+\frac{r}{n}}$
$=\int_{x = 0}^{5}\frac{1}{1+x}dx =  \bigl[ln (1+x)\bigr]_1^5 = \ln 6 - \ln 1 = \ln 6 $

2016/022) Show that $a^b+b^a > 1$ for a , b positive

if a or b are above 1  we are done,.
so let us assume both $< 1$
by Bernoulli Inequality  we have
$(1+x)^n \>= 1+nx$
so $(1+\frac{a}{b})^{\frac{1}{a}}  \ge 1+\frac{1}{b}$
so $(1+\frac{a}{b})^{\frac{1}{a}} > \frac{1}{b}$
or $1+\frac{a}{b} > (\frac{1}{b})^a$
or $\frac{a+b}{b} > (\frac{1}{b})^a$
or $\frac{b}{a+b} < b^a$
similarly
$\frac{a}{a+b} < a^b$
Adding, we get $1 < a^b + b^a$
or $a^b+b^a > 1$

2016/021) w,x,y,z are positive real numbers that satisfy :

$xyz+xy+xz+yz+x+y+z=1\cdots(1)$
$wxy+wx+wy+xy+w+x+y=9\cdots(2)$
$wxz +wx+wz+xz+w+x+z=9\cdots(3)$
$wyz +wy+wz+yz+w+y+z=5\cdots(4)$

Solution

from (1)

$xyz+xy+xz+yz+x+y+z+1=2$
or $(x+1)(y+1)(z+1)=2\cdots(5)$
similarly from (2) (3) and (4)
$(y + 1)(x + 1)(w + 1) = 10\cdots(6)$
$(z + 1)(x + 1)(w + 1) = 10\cdots(7)$
$(y+1)(y+1)(w+1) = 5\cdots(8)$
From (5) (6) (7) and (8) we get
$(x+1)(y+1)(z+1)(w+1) = 10\cdots(9)$
dividing (9) by (5), (6), (7), (8) we get
$(w+1) = 5, y+1 = z+ 1 = 1, x + 1 = 2 => x = 1, w = 4, y= z= 0$

2016/020)Solve $\cos^{-1}\frac{x^2-1}{x^2+1} +\frac{1}{2} \tan^{-1}\frac{-2x}{1-x^2}=\frac{2\pi}{3}$

we have $\tan (2y) = \frac{2\tan y }{1-\tan^2 y}$
hence   $\tan (- 2y) = \frac{-2\tan y }{1-\tan^2 y}$
hence if $x = \tan y$
 $-2y = \tan ^-1 \frac{-2x}{1-x^2}$
or  $\tan ^{-1} \frac{-2x}{1-x^2}= -2 \tan ^{-1} x\cdots(1)$
further $\cos(2y) = \sin ^2 y - \cos^2 y = \frac{\sin ^2 y - \cos^2 y}{\sin ^2 y + \cos^2 y}$
or $\cos(2y) = \frac{\tan  ^2 y - 1}{\tan  ^2 y + 1}$
or $\cos(\pi - 2y) = \frac{1 - \tan  ^2 y}{\tan  ^2 y + 1}$
or $\pi - 2y = \cos ^{-1} \frac{1 - \tan  ^2 y}{\tan  ^2 y + 1}$
putting $\tan y = x$ we get  $\pi - 2 \tan ^{-1} x = \cos ^{-1} \frac{1 - x^2}{1 + x^2}\cdots(2)$
putting values from (1) and (2) in given equation we get
$\pi - 2\tan^{-1} x - (1/2)*2\tan^{-1} x = \frac{2\pi}{3}$
or  $3\tan^{-1}(x) = \frac{\pi}{3}$
or  $\tan^{-1} (x) = \frac{\pi}{9}$
Thus $x = \tan\frac{\pi}{9}$

2016/019) If $\theta = \frac{\pi}{2^{n+1}}$ then show that $2^n cos\theta \ cos 2 \theta \cos 2^2\theta \cdots cos 2^n \theta = 1$

$2^n cos\theta \cos 2 \theta \cos 2^2\theta \cdots cos 2^n \theta$
$= (2\cos\theta) (2\cos 2 \theta)(2 \cos 2^2\theta) \cdots (2cos 2^n \theta)  = 1$
$= (\frac{\sin 2\theta}{\sin \theta}) (\frac{\sin 4\theta}{\sin 2\theta})
          \cdots (\frac{\sin 2^n\theta}{\sin 2^{n-1}\theta})  = 1$
$=\frac{\sin 2^n \theta }{\sin \theta}$
$==\frac{\sin 2^n\frac{\pi}{2^n+1})}{\sin \frac{\pi}{2^n+1}}$
$==\frac{\sin (\frac{\pi}{2^n}-\frac{\pi}{2^n+1}) }{\sin \frac{\pi}{2^n+1}}$
$==\frac{\sin (\frac{\pi}{2^n+1}) }{\sin \frac{\pi}{2^n+1}}=1 $

2016/018) Show that function $f(x) = | x+2 | $ is continuous at $x = - 2$ but not differentiable at x = - 2.

We have $f(x) = | x+2 | $
or $f(x) = x + 2$ for $x >=-2$ and $f(x) = -x-2$ for $x < -2$
at x = -2 the right hand limit is 0 and the left hand limit is 0
so it is continuous at x = -2
differentiating from left we get $f'(x) = -1$ and differentiating from right $f'(x) = 1$
as left hand derivative and right hand derivative are not same of it is
not differenctiable

2016/017) In a quadrilateral ABCD ab is the smallest and CD is the largest side. Prove that angles -- (1) $A > C$ and $B > D$

Join AC. In triangle ABC $AB < BC$ so $\angle BAC > \angle BCA$
in triangle ADC $CD > AD$ so $\angle CAD > \angle DCA$
adding above 2 we get the result. Similarly the 2nd part

2016/016) Show that $4 * (29!) + 5! \equiv 0 (\,mod\, 31) $

because 31 is prime we have as per wilson's theorem
$30!  \equiv -1 (\,mod\, 31)\cdots(1) $
and also $30 * (-1) = -30  \equiv 1 (\,mod\, 31) =>30^{-1} = \equiv 1 (\,mod\, 31)\cdots(2)$
from (1) and (2)
$29!  \equiv 1 (\,mod\, 31)$
or $ 4 * 29!  \equiv 4 (\,mod\, 31) $
or $ 4 * 29! + 5!   \equiv 4 + 120 (\,mod\, 31) \equiv  124 (\,mod\, 31) \equiv 0 (\,mod\, 31)$

2016/015) Solve for x $(5+2\sqrt 6)^{x^2-3} + (5-2\sqrt 6)^{x^2-3} = 10$

We have $(5+ 2 \sqrt6)(5-2\sqrt6) = 25 - 24 =1$
So if $t= 5 + 2 \sqrt 6$ then $\dfrac{1}{t} = 5 - 2\sqrt 6$
So we get
$t^{x^2-3} + \dfrac{1}{t^{x^2-3}} = 10$
let $t^{x^3-3} = p\cdots(1) $
so we get
$ p + \dfrac{1}{p} = 10$
or $p^2 - 10p +1 = 0$
or $p = 5 + 2 \sqrt 6 $ or $p= 5 - 2\sqrt 6= (5+ 2 \sqrt 6)^{-1}$
Hence from (1) and above $x^2-3 =1  => x = \pm 2$
or $x^2 -3 = -1 => x = \pm \sqrt 2$

2016/014) if $x,y,z$ are in H.P then show that $log(x+z)+ log (x+z-2y) = 2log(x-z)$

We have $x,y,z$ are in HP
so $\dfrac{1}{x} + \dfrac{1}{z} = 2\dfrac{1}{y}$
or $ y(x+z) = 2xz$
Now $(x+z)(x+z-2y) = (x+z) ( x + z - \dfrac{4xz}{x+z})$
$= (x+z)^2 - 4xz$
$= x^2 + z^2 + 2xz -4xz$
$= x^2 + z^2 -2xz = (x-z)^2$
taking log of both sides we get the result

2016/013) Prove that $\cos(\tan^{-1}(\sin(\cot^{-1} x)))= \sqrt{\frac{x^2+1}{x^2+2}}$

Let $\cot^{-1} x = y$
so $\cot y = x$
so $\csc^2 y = x^2+1$
so $\sin  y = \frac{1}{\sqrt{x^2+1}}$
so $\sin\cot^{-1} x = \frac{1}{\sqrt{x^2+1}}$
similarly  $\cos \tan ^{-1}x = \sqrt{\frac{1}{x^2+1}}$
Hence $\cos(\tan^{-1}(\sin(\cot^{-1} x)))$
$=\cos(\tan^{-1}\frac{1}{\sqrt{x^2+1}})$
$=\frac{1}{1+\frac{1}{x^2+1}}$
$=\frac{1}{\sqrt{1+\frac{1}{x^2+1}}}$
$=\frac{1}{\sqrt{\frac{x^2+2}{x^2+1}}}$
$=\sqrt{\frac{x^2+1}{x^2+2}}$

2016/012) Suppose that $f(x+3)=3x^2+7x+4\cdots(1) $ and $f(x)=ax^2+bx+c\cdots(2)$ What is $a+b+c$

we have from 2nd equation $f(1) = a + b + c\cdots(3)$
Now we need to evaluate f(1)
as $f(x+3) = 3x^2 + 7x + 4$
so $f(x) = 3(x-3)^2 + 7(x-3)  + 4\cdots(4)$
putting x = 1 we get $f(1) = 3 * (-2)^2 + 7 * (-2) + 4 = 2$
so $a + b+ c = 2$

2016/011) Simplify $a^2(b+c)+b^2(c+a)+c^2(a+b)$ , if a,b,c are in Arithmetic Progression

$a^2(b+c)+b^2(c+a)+c^2(a+b)$
$= a^2(b+c+a) + b^2(c+a+b) + c^2(a+b+c) - (a^3+b^3+c^3)$
$= (a+b+c) (a^2+b^2+c^2) - ((a+b+c)(a^2+b^2 + c^2-ab-bc-ca) + 3abc)$
(because $a^3+b^3+c^3 - 3abc = (a+b+c) ( a^2+ b^2+ c^2 - ab- bc-ca))$
$= (a+b+c)( ab + bc + ca) - 3abc$
a b c are in ap so $a+c = 2b$ so $a+b+c = 3b$
now $ab+bc+ ca = b(a+c) + ca = 2b^2 + ca$
so sum = $3b(2b^2+ ca) - 3abc = 6b^3$

2016/010) Solve $|e^{it} - 1| = 2$ for $-\pi<\theta<=\pi$

$e^ {it} = \cos t + i \sin t$
so $e^{it} - 1 = (\cos t-1) + i \sin t$
take mod and square
$(\cos t-1)^2 + \sin ^2 t = 4 $
or $\cos^2 t - 2 \cos t + 1 + \sin ^2 t = 2 $
or $2 - 2\cos t = 4$ or $cos t = - 1$ and hence $t = \pi$

2016/009) What is the value of c such that a straight line exists which intersects $f(x)=x^4+9x^3+cx2+9x+4$ at 4 points

$f(x)=x^4+9x^3+cx^2+9x+4$
for a straight line to intersect at 4 distinct points
$f^{''}(x)$ must have 2 roots
The reason
if it has no root $f'(x)$ is either positive or -ve and so f(x) is monotonically
increasing or decreasing. as it is cubic polynomial with leading coefficient positive
it is increasing. hence no line can intersect at more than 2 points
if it has one zero then it does not have any point of inflection
now $f^{''}(x)=12x^2+54x+2c=12(x+\frac{9}{4})^2+2(c-\frac{243}{8})$
it has a double root when $c<\frac{243}{8}$
hence $c<\frac{243}{8}$

2016/008) Simplify $\frac{x^4}{(x-y)(x-z)}+\frac{y^4}{(y-z)(y-x)}+\frac{z^4}{(z-x)(z-y)}$

$\dfrac{x^4}{(x-y)(x-z)}+\dfrac{y^4}{(y-z)(y-x)}+\dfrac{z^4}{(z-x)(z-y)}$
=  - ($\dfrac{x^4}{(x-y)(z-x)}+\dfrac{y^4}{(y-z)(x-y)}+\dfrac{z^4}{(z-x)(y-z)})$
= - $(\dfrac{x^4(y-z) + y^4(z-x) + z^4(x-y)}{(x-y)(y-z)(z-x)})$
now
$x^4(y-z) + y^4(z-x) + z^4(x-y)$
= $x^4(y-z) + yz(y^3-z^3) - x (y^4-z^4)$
= $x^4(y-z) + yz(y-z)(y^2+yz+z^2) - x(y-z)(y^3 + y^2z + yz^2 + z^3)$
= $(y-z)(x^4 + yz(y^2 +yz+z^2) - xy(y^2 + yz + z^2) - xz^3)$
= $(y-z)(x^4 + (y^2+yz+z^2)(yz-xy) - xz^3)$
= $(y-z)(x(x^3-z^3) + y(z-x)(y^2 + yz + z^2)$
=$(y-z)(z-x)(y(y^2 + yz + z^2) - x(x^2 + zx + z^2)$
= $(y-z)(z-x)(y^3 + y (yz+ z^2) - x^3 - x(zx + z^2)$
= $(y-z)(z-x)(y^3-x^3 + (y^2z + yz^2 - zx^2 - z^2 x)$
= $(y-z)(z-x)((y-x) (x^2 + xy + y^2) + (z(y^2 - x^2) +z^2(y-x))$
= $(y-z)(z-x)((y-x)(x^2 + xy + y^2 + z(y+x) + z^2)$
= $(-(x-y)(y-z)(z-x)(x^2 + y^2 + z^2 + xy+yz+zx)$
So the given expression
= $x^2 + y^2 +z^2 + xy + yz+ xz$
hence $\dfrac{x^4}{(x-y)(x-z)}+\dfrac{y^4}{(y-z)(y-x)}+\dfrac{z^4}{(z-x)(z-y)}= \dfrac{1}{2}((x+y)^2 + (y+z)^2 + (x+z)^2$

2016/007) How many integer values of x and y are there such that $4x+7y=3$ while $ |x| < 500$ and $|y| < 500$

1st let us find one solution
this can be found by any method but as we see that $7- 4 = 3$
so $(-1,1)$ is a solution
as coeffcient of y is larger so we need to restrict x between $- 500$ to $500$
general solution is $x = -1+ 7t$ and $y = 1 + 4t$
Now $- 500 < x < 500$ or -$ 500 < 1 + 7t < 500$ or
$-501 < 7t < 499$ or $-71 <= t <= 71$ so 143 values

2016/006) show that 13 is the largest number that divides 2 numbers of the form $(n^2+3),(n+1)^2+3$

 let us find the GCD

$GCD(n^2+3,(n+1)^3+3) = GCD(n^2+3, n^2 + 2n + 4)$
$=GCD(n^2+3, n ^2+2n+4-(n^2+3))= GCD(n^2+3,2n+1)$
$=GCD(2*(n^2+3), 2n+ 1)$ we can multiply 1st one by 2 as 2nd one is odd
$=GCD(2n^2+6,2n++1)$
$=GCD(2n^2+6-n(2n+1),2n+1)$
$=GCD(6-n, 2n+ 1)$
$= GCD(12-2n,2n+1)$ we can multiply 1st one by 2 as 2nd one is odd
$= GCD(13,2n+1)$
So 13 is the largest number that divides both and when 2n+1 = 13 mod 13 or n= 6 mod 13

2016/005) Find sum $\sin^3x + \sin^3 (x+\frac{2\pi}{3}) + \sin^3 (x+\frac{4\pi}{3})$

$\sin^3x + \sin^3 (x+\frac{2\pi}{3}) + \sin^3 (x+\frac{4\pi}{3})$
$=\sin^3x + \sin^3 (x-\frac{\pi}{3}) + \sin^3 (x+\frac{\pi}{3})$
$= \frac{1}{4}(3\sin x - \sin 3x) + \frac{1}{4} (3\sin(x - \frac{\pi}{3}) - \sin(3x - \pi)) +
  \frac{1}{4}(\sin(x + \frac{\pi}{3}) - \sin(3x + \pi))$
$= \frac{1}{4}(3\sin x - \sin 3x + 3\sin(x - \frac{\pi}{3} ) + \sin3x - 3\sin(x + \frac{\pi}{3}) + \sin 3x)$
$= (3/4) [\sin x - \sin 3x + \sin(x - \frac{\pi}{3}) + \sin(x + \frac{\pi}{3}]$
$=\frac{3}{4}(\sin x - \sin 3x - 2 \sin x \sin\frac{\pi}{6})$
$= \frac{3}{4}(\sin x - \sin3x - \sin x)$
$= - \frac{3}{4}\sin 3x$
The following has been used in proving the above identity

$1) \sin^3 x = \frac{3\sin x - \sin3x}{4}$
$2) \sin A + \sin B = 2\sin \frac{A+B}{2} \sin \frac{A-B}{2}$

2016/004) $(1+ni)^2$ is purely imaginary find n.

We have $(1+ni)^2 = 1 +2ni- n^2$
for it to be imaginary we need to have real part zero or $1-n^2 = 0$ or $n = \pm 1$

2016/003) In an AP sum of p terms is same as sum of q terms. Show that $(p+q)^{th}$ term is zero

let $1^{st}$ term be $a$ and difference $d$
let $k^{th}$ term be $t_k$ and sum upto k term be $S_k$
$t_{p}= a + (p - 1) d$
So $S_p= \frac{a + (p-1) d}{2} * p$
sum of q terms $S_q= \frac{a + (q- 1)d}{2}* q = \frac{a + (p-1) d}{2}* p$ (as both are same)
So $(a-d) q + dq ^2 = (a-d) p + dp^2$
or $(a-d)(q-p) = d(p^2 - q^2)$
so $(a-d) = -d (p+ q)$
or $a + (p+q-1) d = 0$ and this is the $(p+q)^{th}$ term and it is zero

2016/002) 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
$n^4 + 4^n= n^4 + 2n^2*2^n + 4^n - 2n^2*2^n$
$=(n^2 + 2^n)^2 - 2n^2*2^n$
$=(n^2 + 2^n)^2 - n^2*2^{n+1}$
$= (n^2 + 2^n - n * 2^{\frac{n+1}{2}})(n^2 + 2^n + n * 2^{\frac{n+1}{2}})$
Therefore, if n is odd it has 2 factors so composite

So it is composite for any n

2016/001) How to find the nth term of a sequence whose consecutive terms difference' is in A.P?

for example $1,3,6,10,15\cdots$

Because difference is AP let us find the differences  $2,3,4,5$
$t_{n+1} - t_{n} = n$
so $t_n$ should be order 2 polynomial
say $t_n = an^2 + bn + c$
hence $t_{n+1}-t_n=a((n+1)^2-n^2) + b=2an +a +b = n $
so $2a = 1$ and $a+b = 1 => a = b = \frac{1}{2}$
putting $n = 1$ we get $c = 0$
so $t_n = \frac{1}{2}(n^2 + n) = \frac{1}{2}n(n+1)$