2017/036) Find the smallest multiple of 15 such that each digit of the multiple is either 0 or 8

Because it is multiple of 15 and 15 = 3 * 5 so it is multiple of 3 and 5. So the unit digit is zero. As 8 is not multiple of 3 there should be 3 8's for the number to be multiple of 3  and the digit cannot be the unit position. Hence the smallest number is 8880.

2017/035) a,b ,c are three distint real numbers and there are real numbers x,y such that $a^3+ax+y=0$, $b^3+bx+y=0$, $c^3+cx +y = 0$. Show that $a+b+c=0$

Let us consider the equation $f(p) = p^3 + px + y = 0$
The above equation is cubic in p and has 3 roots.
From the given condition a,b,c are three roots so the sum of roots = -  coefficient of $p^2$
hence a + b+ c = 0
Proved

2017/034) Let $p(x) = x^2 + bx +c $ where b and c are integers. if P(x) is a factor of both $x^4 + 6x^2 + 25$ and $3x^4+ 4x^2 + 28x + 5$, find the value of p(1)

Because P(x) devides $x^4 + 6x^2 + 25$ and  $3x^4+ 4x^2 + 28x + 5$ hence P(x) devides the GCD,
now $GCD(x^4 + 6x^2 + 25,3x^4+ 4x^2 + 28x + 5$
$= GCD(x^4 + 6x^2 + 25,3x^4+ 4x^2 + 28x + 5 - 3( x^4 + 6x^2 + 25))$
$= GCD(x^4 + 6x^2 + 25, -14x^2 + 28x -70))$
$= GCD(x^4 + 6x^2 + 25, x^2 -2x+5))$ deviding 2nd expression by -14
$= GCD(x^4 + 10x^2 + 25- 4x^2, x^2-2x + 5)$
$= GCD((x^2 + 5)^2- (2x)^2, x^2-2x + 5)$
$=GCD((x^2+2x+5)(x^2-2x+5),x^2-2x  + 5) $
$=x^2 - 2x + 5$
so as p(x) is of the form $x^2+bx+ c$ so $p(x) = x^2 - 2x+5$ and hence $p(1) = 4$

2017/033) If $a+b+c+d+e+f=0$ and $a^3+b^3+c^3+d^3+e^3+f^3=0$ and no 2 variables are additive inverse of each other then show that $(a+c)(a+d)(a+e)(a+f) = (b+c)(b+d)(b+e)(b+f)$

We have $(a+c+d) = - (b+e+f)\cdots(1)$
and $a^3+c^3+d^3= -(b^3+e^3+f^3)\cdots(2)$
cube both sides of (1) using $(x+y+z)^3 = x^3+y^3+z^3 + 3(x+y)(y+z)(z+x)$  to get
$a^3+c^3+ d^3 + 3(a+c)(a+d)(d+c) = -(b^3 + e^3 + f^3 + 3(b+e)(b+f)(e+f)$
or $(a+c)(a+d)(c+d) = - (b+e)(b+f)(e+f)\cdots(3)$ using (2)
similarly we have $(a+e)(a+f)(e+f) = - (b+c)(b+d)(c+d)\cdots(4)$
multiplying (3) and (4) we get the result
 

2017/032) If $a,b,c$ are in H.P then prove that $a^3b^3 + b^3c^3 + c^3 a^3 = (9ac-6b^2)a^2c^2$

$a,b,c$ are in HP
hence $\frac{1}{a} + \frac{1}{c} =  \frac{2}{b}$
cube both sides to get  $\frac{1}{a^3} + \frac{1}{c^3} + 3 \frac{1}{a}  \frac{1}{c}(\frac{1}{a} + \frac{1}{c}) =  \frac{8}{b^3}$
or   $\frac{1}{a^3} + \frac{1}{c^3} + 3 \frac{1}{a}  \frac{1}{c}\frac{2}{b}  =  \frac{8}{b^3}$
or $\frac{1}{a^3} + \frac{1}{c^3} + \frac{1}{b^3}  =  \frac{9}{b^3} -  3 \frac{1}{a}  \frac{1}{c}\frac{2}{b}$
multiply both sides by $a^3b^3c^3$ to get
$b^3c^3 + a^3b^3 + a^3c^3 = 9a^3c^3 -  6 a^2b^2c^2 = (9aac-6b^2) a^2c^2$

2017/031) Prove that A triangle ABC is equilateral if and only if $\tan\, A + \tan\, B + \tan\, C = 3\sqrt{3}$

First let us prove the if part
if ABC is equilateral we have $\tan\, A + \tan\, B + \tan\, C = \tan\, 60^{\circ}+\tan\, 60^{\circ} + \tan\, 60^{\circ} = 3\tan\, 60^{\circ} =\sqrt{3}$
Now for the other part
we have using AM GM inequality(for all 3 positive) 
$\frac{\tan\, A + \tan\, B + \tan\, C}{3} >=\sqrt[3]{\tan\, A\,  \tan\, B\, \tan\, C}$
or $\tan\, A + \tan\, B + \tan\, C >=3\sqrt[3]{\tan\, A\,  \tan\, B\, \tan\, C}$
and these are equal if $\tan\, A = \tan\, B = \tan\, C$ or $A=B=C$ and at this  we have the sum = $3\sqrt[3]3$

2017/030)In a triangle if $\cot\,A\,\cot\,B\,\cot\,C$ are in AP then $a^2,b^2,c^2$ are in $\cdots$ progression (IIT1985-5 marks)

We have $\cot\,A\,\cot\,B\,\cot\,C$  are in AP
so
$\cot\, A + \cot\, C = 2\cot\, B$
or $\frac{\cos\, A }{\sin\, A }  +\frac{\cos\, C }{\sin\, C } = \frac{\cos\, B }{\sin\, B }$
or $\frac{\cos\, A \sin\, C + \sin\, A  \cos\, C}{\sin\, A \sin\, C } = \frac{\cos\, B }{\sin\, B} $
or $\frac{\sin ( A + C) }{\sin\, A \sin\, C }  = \frac{\cos\, B }{\sin\, B} $
or  $\frac{\sin B}{\sin\, A \sin\, C }  = \frac{\cos\, B }{\sin\, B} $ as in a triangle $\sin (A+B) = \sin\,c$
or  $\sin^2 B = 2\sin\, A \sin\, C\cos \,B $
using law of sin an cos we get
$b^2 = 2ac\frac{a^2+c^2-b^2}{ac}$
or $b^2 = a^2+c^2-b^2$
or $2b^2 = a^2 +c ^2$
hence $a^2,b^2,c^2$ are in AP

2017/029) Find the exact value for the real root of the equation $x^3+3x-2=0$

Say $f(x)=x3+3x-2=0$
$f(x)$  changes sign once so there is one positive root.
$f(-x)=-x^3-3x-2$ changes sign zero times so no -ve root
let $x=2sinh\, t$
so $ x^3+3x=2=>8\sinh\,3t+6sinh\,t=2$
or $4sinh\,3t+3sinh\,t=1$
or $sinh\,3t=1$ or $3t=sinh^{-1}1$
hence $x=2sinh(\frac{sinh^{-1}}{3})x$
this is the only real root

2017/028) Find 3 consecutive numbers that can be expressed as sum of 2 non-zero squares

We have $(2n)^2+ (2n)^2 = 8n^2$
and $(2n+1)^2 + (2n-1)^2 = 8n^2 + 2$
If we can find $8n^2 + 1  = a^2+ b^2$ for some a, b  then we are through.
we haave
$(2n - a)^2   +   (2n + a - 1)^2   =   8n^2   +   1   -   (4n - (2a^2 - 2a))$
If we can have  n   and   a   such that
$4n   -   (2a^2 - 2a)   =   0$ then we are through
or $2n   =   a^2 - a   =   a(a - 1)$
or $n = \frac{a(a-1)}{2}$

as there are infinite number of a's we find infinite nuber of solutions

a = 2, gives n= 1 2n-a = 0 so we put
a=3 , gives n= 3 we get

$6^2+6^2 = 72$
$3^2 + 8^2 = 73$
$5^2+7^2 = 74$
similarly we can find more solutions by putting any value of a.

2017/027) Find all integers n such that $(n^2-n-1)^{n+2}= 1$

There are 3 cases when
$x^y= 0$ when y = 0 or x =1 or x = -1 and y even

hence  $(n^2-n-1)^{n+2}= 1$

when
1) n+2 = 0 or n = -2
or

2) $n^2 -n -1 = 1$
or $n^2 - n -2 = 0$ or $(n-2)(n+1) = 0$ giving n = 2 or -1

or
3) $n^2 -n -1 = -1 $   and n+2  is even
in this case
$x = -1 =>n^2 -n -1 = -1$ $n^2-n=0$ so n= 0 or n =1. but n+2 need to be even so n= 0

so solution set n= 0 or 2 or -2 or -1

2017/026) if $x=\frac{4ab}{a+b}$ find the value of $\frac{x+2a}{x-2a}+ \frac{x+2b}{x-2b}$

we have
$x=\dfrac{4ab}{a+b}$
hence $\dfrac{x}{2a} = \dfrac{2b}{a+b}$
using componendo dividendo we get
 $\dfrac{x+ 2a}{x-  2a} = \dfrac{2b + (a +b)  }{2b - (a+b) } =  \dfrac{3b + a }{b- a}\cdots(1)$
similarly   $\dfrac{x+ 2b}{x-  2b} = \dfrac{2a + (a +b)  }{2a - (a+b) } =  \dfrac{3a + b}{a-b}\cdots(2)$
Adding (1) and (2) we get
 $\dfrac{x+2a}{x-2a}+ \dfrac{x+2b}{x-2b} =  \dfrac{3b + a}{b-  a}  +  \dfrac{3a + b}{a- b} $
$= \dfrac{3b + a}{b-  a}  -  \dfrac{3a + b}{b-a}$
$= \dfrac{3b + a-3a -b}{b-  a} =  \dfrac{2b-2a}{b-a} = \dfrac{2(b-a)}{b-a} = 2$