2015/010)If p is prime, then what is the remainder of (p - 2)! / p

say (p - 2)! ≡ x (mod p)
multiply by (p-1)

(p-1)! = x(p-1) mod p

so -1 = x (p-1) mod p as (p-1) ! = - 1 mod p

p-1 = -1 mod d

so -1 = x (-1) mod p = - x mod p

so -x mod p = - 1

or x mod p = 1

 so  remainder of (p - 2)! / p is 1

2015/009) find b(a+c) given they are roots of equation $\sqrt{2015}x^3-4031x^2+2=0$ and a > b > c

To avoid surds let $t=\sqrt{2015}$

so we get $tx^3 - (2t^2+1)x^2 + 2 = 0$
or $tx^3 - x^2 - 2t^2x^2+2=0$
or $x^2(xt-1) - 2(x^2t^2-1) = 0$
or $(x^2(xt-1) - 2 (xt-1)(xt+1) = 0$

so xt = 1 or $x^2-2xt -2 = 0$
2nd equation gives

$(x-t)^2 = t^2+ 2 = 2017$

so 3 roots are $x=\dfrac{1}{\sqrt{2015}}$ or $x= \sqrt{2015} \pm \sqrt{2017}$

clearly $ \sqrt{2015} +\sqrt{2017} \gt \dfrac{1}{\sqrt{2015}} \gt \sqrt{2015} - \sqrt{2017}$

so $a= \sqrt{2015} +\sqrt{2017}$
$b = \dfrac{1}{\sqrt{2015}}$
$c = \sqrt{2015} -\sqrt{2017}$

so b(a+c) = -2

Note:

I  have  taken the problem from  site http://mathhelpboards.com/challenge-questions-puzzles-28/find-b-c-14115.html#post66990.  I personally feel that my solution is simpler

2015/008) show that if $x,y,z$ are 3 sides of a triangle so are $\dfrac{1}{x+y},\dfrac{1}{x+z},\dfrac{1}{y+z}$

we need to show that for arbitrary sides of length x,y z

$\dfrac{1}{x+z} + \dfrac{1}{y+z}\gt \dfrac{1}{x+y}$

or $(y+z)(x+y) + (x+z)(x+y) \ge (x+z)(x+z)$
or simlifying
$x^2+y^2 + xy + yz + xz \ge z^2$


now
$x^2+y^2 + xy + yz + xz = z(x+y) + x^2+y^2 + xy$
$ \gt z(x+y) $
$\gt z^2$ as $x+y \gt z$

hence proved

I have taken the problem from http://mathhelpboards.com/challenge-questions-puzzles-28/lengths-three-segments-triangle-14063.html#post66760 where I provided the above solution

2015/007) $(x-1)^3$ is a factor of $P(x) = x^{10}+ax^2 + bx + c$ find a + 2b + 3c

$(x-1)^3$ is a factor of $P(x) = x^{10}+ax^2 + bx + c$

so $x^3$ is a factor of $P(x+1) = (x+1)^{10}+ a(x+1)^2+b(x+1) + c$

so in the above expression coefficient of $x^2$, x and constant term shall be zero

coefficient of $x^2$ = $\binom{10}{2}+ a = 0= 45 + a = 0 $

coefficient of $x$ = 10 + 2a + b = 0

constant term = 1 + a + b + c = 0

solving we get a = - 45, b = 80, c = -36

so a + 2b + 3c = 7


This I have solved at http://mathhelpboards.com/challenge-questions-puzzles-28/find-2b-3c-14031.html#post66646. There are 2 more solutions

2015/006) Find the maximum value of $P(x)=x^3 −3x$ on the set of all real x that satisfies the inequality $x^4 −13x^2 +36\le0 $.

we have $x^4-13x^2+ 36 \le 0$
or $(x^2-9)(x^2-4) \le 0$
or $4\le x^2\le 9$

so we get 2 ranges for x

1) $-3\le x\le -2 $
2)  $2\le x\le 3 $

we need to find the maximum of $x^3-3x$ in this ranges

$x^3-3x = x(x^2-3)$

in the 1st range x is negative  and x^2-3 is positive so in the range $x^3-3x$ is -ve

for $x\gt \sqrt{3}$ and hence for $x \ge 2$ both x and x^2- 3 are positive and increasing so the highest value is at highest x that is x = 3 so the value of $x^3-3x=18$

2015/005) if $l \ge m \ge n \gt 0$ show that

$\dfrac{l^2-m^2}{n}+ \dfrac{n^2-m^2}{l}+\dfrac{l^2-n^2}{m} \ge 4l - 3m +n$


as $l \ge m \ge n \gt 0$

we get $(l +m) \ge 2n$

or $\dfrac{l+m}{n} \ge 2$
or $\dfrac{l^2-m^2}{n} \ge 2(l-m) \cdots (1) $

further
 $(m+n ) \le  2l$
or $\dfrac{m+n }{l} \le 2$
or $\dfrac{m^2-n^2}{l} \le 2(m-n) $
or $\dfrac{n^2-m^2}{l} \ge 2(n-m) \cdots (2) $

also
$(l+n ) \ge  m$
or $\dfrac{l+n }{m} \ge 1$
or $\dfrac{l^2-n^2}{m} \ge l-n \cdots (3)$

adding (1), (2), (3) we get
$\dfrac{l^2-m^2}{n}+ \dfrac{n^2-m^2}{l}+\dfrac{l^2-n^2}{m} \ge 4l - 3m +n$

Note: 
This problem I have taken from http://mathhelpboards.com/challenge-questions-puzzles-28/inequality-challenge-x-12071.html#post65756  where I was the only person who answered the same

2015/004) Find x,y when $ x \cos \dfrac{\pi}{y} = \sqrt{8 + \sqrt{32 + \sqrt{768}}}$

we have

$768 = 1024 * \dfrac{3}{4}= (32 * \dfrac{\sqrt{3}}{2})^2 = (32 *\cos(\dfrac{\pi}{6}))^2$

so $\sqrt{768} = 32 *\cos(\dfrac{\pi}{6})$
so
 $32 + \sqrt{768} = 32 *(1+\cos(\dfrac{\pi}{6})$
= $64 *\cos^2(\dfrac{\pi}{12}$ using $1+\cos 2x = 2 cos^2 x$

so
 $\sqrt{32 + \sqrt{768}} = 8 \cos(\dfrac{\pi}{12})$
so
 $8 + \sqrt{32 + \sqrt{768}} = 8 ( 1 + \cos(\dfrac{\pi}{12})) = 16 \cos^2 \dfrac{\pi}{24}$

so $\sqrt{8 + \sqrt{32 + \sqrt{768}}} = 4 \cos \dfrac{\pi}{24}$

so x = 4, y = 24 or (x,y) = (4,24)

Note:
I have solved this as a problem of the week at http://mathhelpboards.com/potw-secondary-school-high-school-students-35/problem-week-145-january-5th-2015-a-13891.html

2015/003) If $X$ and $Y$ are real root for the equations $x^3+3x^2+6x+20=0$ and $y^3+6y^2+15y-2=0$, find the sum of $X$ and $Y$

Now let
$P(x) = x^3+3x^2+6x+20$ and
$G(y) = y^3+6y^2+15y-2=0$

now let us check for p(x)

to eliminate the $x^2$ term I put x -1 to get
$P(x-1) = (x-1)^3 + 3(x-1)^2 + 6(x-1) + 20$
$= x^3-3x^2+ 3x - 1 + 3x^2 - 6x + 3 + 6x - 6 + 20$
$= x^3 + 3x + 16$

now let us check for g(y)
to eliminate the $y^2$ term I put y -2 to get
$G(y-2) =(y-2)^3+6(y-2)^2+15(y-2)-2$
$=y^3 - 6y^2 + 12y - 8 + 6y^ - 24y + 24 + 15y - 30 - 2$
$=y^3 + 3y - 16$

now if we put $R(x) = x^3 + 3x - 16$ then we get
P(x-1) = R(x) 
G(y-2) = - R(-y)

or P(x) = R(x+1) and G(y) = - R(y + 2)

so both are same and odd functions so
for the zeros of P and G the 2 values shall be -ve of each other hence

X + 1 = - (Y +2) or  X + Y  = - 3

I had solved incorrectly at http://mathhelpboards.com/challenge-questions-puzzles-28/find-x-y-13744.html#post65445.

2015/002) show that if $3x^2+x=4y^2+y$ then each of $x-y$ , $3x+3y+1$ and $4x+4y+1$ is a perfect square

we have

$3x^2+x = 4y^2 + y$
or $3x^2- 3y^2 + x - y = y^2$
or $3(x+y)(x-y) + (x-y) = y^2$
or $(3x+3y+1)(x-y) = y^2\cdots\, 1$
also
$4x^2-4y^2 + x-y=x^2$
or $4(x-y)(x+y) + (x-y) = x^2$
or $(x-y)(4x+4y+1) = x^2\cdots\,2 $
multiply (1) and (2) to get

$(x-y)^2(3x+3y+1)(4x+4y+1)=x^2y^2$
or$(3x+3y+1)(4x+4y+1)=(\dfrac{xy}{x-y})^2$

so $(3x+3y+1)(4x+4y+1)$ is a perfect square

as $4(3x+3y+1) - 3(4x+4y+1) = 1$ so $(3x+3y+1)$ and $(4x+4y+1)$ both are perfect squares and the from (1) or (2) $(x-y)$ is a perfect square

2015/001) Find all solutions to $z^2 + 4z^* + 4 = 0$ where z is a complex number and $z^*$ is conjugate

Let z = a + ib

you get $(a+ib)^2 + 4(a-ib) + 4 = 0$

expand

$a^2 - b^2 + 2aib + 4a - 4bi + 4 = 0$
or
$(a^2 - b^2 + 4a + 4) + 2abi- 4bi = 0$

equate imaginary and real parts to zero to get

$(a^2 - b^2 + 4a + 4) = 0 \cdots 1$
and $2abi - 4bi = 0 => b = 0$ or a = 2

solve 1 using b = 0 to get a= - 2
sollve 1 using a= 2 to get b = $\pm4$

so (-2,0) is a solution so z = - 2
(2, 4i) and (2,-4i) are 2 other solutions so $z = 2 \pm 4i$ 

refer to https://in.answers.yahoo.com/question/index;_ylt=Aj51ZIERhFIw57.9V.6ln8.QHQx.;_ylv=3?qid=20120424152238AA6VkyD