2014/083) Show that no member of the infinite series: 10001, 100010001, 1000100010001, 10001000100010001, … is prime.

that is between each sucessive 1 there are 3 zeroes.

first we note that the smallest number 10001 = 73 * 137 and not a prime

then from the 2nd term onwards (now n >=2)
Write the series as
$\displaystyle\sum_{i=0}^{n}10^{4i}= \dfrac{10^{4n+4}-1}{10^4-1}$
=  $\dfrac{(10^{2n+2}+1)(10^{2n+2}-1)}{10^4-1}$
as $n \ge 3$ we have 2 factors $\ge 10^6-1$ and denominator 9999 so we shall have 2 distinct factors and hence composite
 

2014/082) solve in integer

Alan and Bob have a whole number of dollars. Alan says to Bob, “If you give me  3 dollars, I will have n times as much as you”. Bob saya to Alan, “If you give me n dollars, I will have 3 times as much as you”.

 Let alan have a dollars and Bob b dollars
so we get
$a+3=n*(b-3)$

 $b+n=3*(a-n)$

from 1st we get $a = (nb-3n-3)$
put in 2nd to get
$b+ n = 3(nb-4n-3) = 3nb -12n -9$
or $n=\dfrac{b+9}{3*b-13}$
as $n\gt 0$ , $a\gt 0$,$b\gt0$ we get  $3b\gt13\  or\  b > 4$

$b+ 9\gt3b - 13$ or $22\gt\ 2b$ or $b\lt 11$
now $b+ 9$ should be divisible by $3b - 13$
or $3b + 27$ should be divisible by $3b - 13$


or 40 should be divsible by 3b - 13 or 3b -13 should be( 2 mod 3) so we get factors of 40 (2 mod 3)  2 or 5 or 8 or 20 giving b = 5 or 6 or 7 or 11

putting those 4 values of b we find

a=11, b=5, n=7
a=6, b=6, n=3
a=5, b=7, n=2
a=5,b= 11,n=1 

2014/081) factor $(x+1) (x+3) (x+5) (x+7)+16 $

$(x+1) (x+3) (x+5) (x+7)+16$
= $((x+1)(x+7))( (x+3)(x+5)) + 16$
= $(x^2 + 8x + 7) (x^2+ 8x+ 15)+ 16 $
=$ ((x^2 + 8x + 11) - 4)(x^2 + 8x+ 11) + 4) + 16$
= $(x^2+ 8x+ 11)^2 – 16 + 16$
= $(x^2+ 8x + 11)^2 $

2014/080) solve $n^2+20n+11=m^2$ for integer n,m

 adding 89 on both sides we get

We have $n^2 + 20n + 100 = m^2 + 89$

$(n+10)^2 = m^2 + 89$ or taking $n+ 10 = p$ we get
$p^2 – m^2 = 89$ or $(p+m)(p-m) = 89$
As 89 is prime there are 4 cases
1)$p+m = 89,p-m = 1 => p = 45, m = 44\ or\ n = 35, m = 44$
2)$p+m = – 89,p-m = – 1 => p =- 45, m = – 44\ or \ n = – 55 , m = -44$
3)$ p +m = -1, p-m = – 89$
4) $p+m = – 89, p-m = 1$
one can solve (3) and (4) to given (35,-44) and (-55,44)

2014/079) Suppose that a, b, c are sides of of a triangle. Prove that:

$a^2(b + c - a) + b^2(c + a - b) + c^2(a + b - c) \le 3abc$

proof:

For the angles of any triangle always
$\cos\ A + \cos\ B + \cos\ C ≤ 1.5$

further

$c² = a² + b² - 2ab\ cos\ C$
hence $c³ = a²c + b²c - 2abc\cos\ C$

$a²c + b²c - c³ = 2abc\cos\ C$
similarly
$ab² + ac² - a³ = 2abc\cos\ A$
$bc² + a²b - b³ = 2abc\cos\ B$

adding we get


$a²c + b²c - c³ + ab² + ac² - a³ + bc² + a²b - b³ = 2abc (\cos\ A + \cos\ B + \cos\ C)$

as

$\cos\ A + \cos\ B + \cos\ C ≤ 1.5$

$a²c + b²c - c³ + ab² + ac² - a³ + bc² + a²b - b³ ≤ 3abc$
or $(a²b + a²c - a³) + (b²c + ab² - b³) + (ac² + bc² - c³) ≤ 3abc$

or $a²(b + c - a) + b²(c + a - b) + c²(a + b - c) ≤ 3abc$
hence proved

2014/078) given: $(1+\sqrt{1+x}\tan\ x=1+\sqrt{1-x}$ then prove that $\sin\ 4x=x$

Let $x = \sin\ 2y$

Then $\sqrt{1+x} = \sin\ y + \cos\ y$

and $\sqrt{1-x} = \sin\ y - \cos\ y$

hence

$\tan\ x = \dfrac{1+ \sin\ y - \cos\ y}{1+ \sin\ y + \cos\ y}$

using if $\dfrac{a}{b} =\dfrac{c}{d}$ => $\dfrac{a+b}{a-b} =\dfrac{c+d}{c-d}$

$\dfrac{\tan\ x + 1}{\tan\ x -1} = \dfrac{ 1+ \sin\ y}{-\cos\ y}$

$\dfrac{\sin\ x + \cos\ x }{\sin\ x -\cos\ x} = \dfrac{ 1+ \sin\ y}{-\cos\ y}$

$\dfrac{(\sin\ x + \cos\ x) ^2}{\sin ^2 x -\cos ^2 x} = \dfrac{1+ \sin\ y}{ \cos\ y}$

$\dfrac{1+ \sin\ 2x}{-\cos\ 2x} = \dfrac{ 1+\ sin\ y}{ \cos\ y}$

From the above $y = 2x$ so $x = \sin\ 4x$ proved

2014/077) Solve for x

$(5 + 2\sqrt{6})^{x^2-3}+ (5 - 2\sqrt{6})^{x^2-3}= 10$

let $y =5 + 2\sqrt{6}\cdots (1)$
so we have $\dfrac{1}{y} =5 - 2\sqrt{6}\cdots (2)$

now let $z= (5 + 2\sqrt{6})^{x^2-3}$

so $\dfrac{1}{z}= (5 - 2\sqrt{6})^{x^2-3}$

hence $z+\dfrac{1}{z}= 10$

or $z^2-10z+z= 0$
hence $z= 5 + 2\sqrt{6}$ or $z= 5 - 2\sqrt{6}$

 $z= 5 + 2\sqrt{6}$
=> $x^2-3 =1$ or $x=\pm 2$

$z= 5 - 2\sqrt{6}$
=> $x^2-3 =-1$ or $x=\pm \sqrt{2}$

2014/076) Solve for largest x

 $\dfrac{6}{x-6}+ \dfrac{8}{x-8}+\dfrac{20}{x-20}+\dfrac{22}{x-22}=x^2-14x-4$


add 1 to each term onLHS and so 4 to RHS to get

$\dfrac{x}{x-6} + \dfrac{x}{x-8} + \dfrac{x}{x-20} + \dfrac{x}{x-22} = x^2-14x$

so one solution is x = 0 and further we deviding by x we get
$\dfrac{1}{x-6} + \dfrac{1}{x-8} + \dfrac{1}{x-20} + \dfrac{1}{x-22} = x-14$

put y = x - 14 to get


$\dfrac{1}{y+8} + \dfrac{1}{y+6} + \dfrac{1}{y-6} + \dfrac{1}{y-8} = y$

or 

$\dfrac{1}{y+8} + \dfrac{1}{y-8} + \dfrac{1}{y+6} + \dfrac{1}{y-6} = y$

or

$\dfrac{2y}{y^2-64} + \dfrac{2y}{y^2-36} = y$

so y = 0

or 
$\dfrac{2}{y^2-64} + \dfrac{2}{y^2-36} = 1$

or
$2((y^2-36) + y^2-64))= (y^2-36)(y^2-64)$
or $2((2y^2-100))= (y^2-36)(y^2-64)= y^4-100y^2+ 36 *64$
or $y^4- 104y^2+48^2+200=0$
or $(y^2-52)^2 = 200$
$y^2 = 52 \pm 10\sqrt{2}$
we should take the higher of the 2 and add 14 to get the largest x or x = 14 + $\sqrt{52+10\sqrt{2}}$

2014/075) if $ax^2+bx+c = 0 $ has no real roots and $a+b+c \lt 0$ then show that $c \lt 0$

let
$f(x) = ax^2 + bx +c = 0$
if f(x) has no real root then f(x) is positive for all x or it is -ve for all x

$f(1) = a + b+ c \lt 0$

hence $f(0) = c \lt 0$

2014/074) integrate $\dfrac{\sin\ x}{a \cos \ x + b\sin\ x} $

Back ground information
$∫ \dfrac{\sin\ x}{\sin \ x + \ cos\ x} dx $
= $∫\dfrac{1}{2} \dfrac{2\sin\ x}{\sin \ x + \ cos\ x} dx $
= $∫\dfrac{1}{2} \dfrac{(\sin\ x+\cos \ x) +(\sin\ x-\cos \ x) }{\sin \ x + \ cos\ x} dx $
= $∫\dfrac{1}{2} (1+ \dfrac{\sin\ x-\cos \ x }{\sin \ x + \ cos\ x} dx $
=  $∫\dfrac{1}{2}dx + \dfrac{1}{2} \dfrac{\sin\ x-\cos \ x }{\sin \ x + \ cos\ x} dx $
= $\dfrac{1}{2}x - \dfrac{1}{2} ln \mid (\sin \ x + \ cos\ x)\mid $

as $\dfrac{\cos\ x}{\sin \ x + \ cos\ x}  $
=  $1-  \dfrac{\sin\ x}{\sin \ x + \ cos\ x} $

so $∫ \dfrac{\cos \ x}{\sin \ x + \ cos\ x} dx $
=  $\dfrac{1}{2}x + \dfrac{1}{2} ln \mid (\sin \ x + \ cos\ x)\mid $
Now integral

now we need to inetgrate
$\dfrac{\sin\ x}{a \cos \ x + b\ sin\  x} $

as derivate of  $a \cos \ x + b\sin \  x $ is $- a \sin  \ x + b\cos \   x $

we can put  $\sin\ x= m(a \cos \ x + b\ sin\  x) + n (- a \sin  \ x + b\cos \   x) $

comparing coefficients we get 1 = ma + nb and 0=mb-na

solving these 2 we get

$m=\dfrac{b}{b^2+a^2}$ and $n=  \dfrac{- a}{b^2+a^2}$

so  $\sin\ x=\dfrac{b}{b^2+a^2}(a \cos \ x + b\ sin\  x) - \dfrac{a}{b^2+a^2} (- a \sin  \ x + b\cos \   x) $

so $\dfrac{\sin\ x}{a \ cos \ x + b\sin \ x }= \dfrac{b}{b^2+a^2} - \dfrac{a}{b^2+a^2} \dfrac{- a \sin  \ x + b\cos \   x} {a \cos \ x + b\sin \ x } $

so integral of above = $ \dfrac{b}{b^2+a^2} x - \dfrac{a}{b^2+a^2} ln \mid a \cos \ x + b\sin \ x \mid $

2014/073) factor $bca^2 + bcd^2 + adb^2 +adc^2$

this looks non trivial but we can treat is a quadratic in a
bca^2 + a(db^2 + dc^2) + bcd^2

now product of bc and $bcd^2$ is $b^2c^2d^2$ which is product of $db^2$ and $b^2$

so we get $(bca^2 + adb^2) + (adc^2+bcd^2)$
=$ab(ac+bd) + cd(ac + bd)$
= $(ac+bd)(ab+cd)$