2014/095) Prove the sum of any 10 consecutive terms of the Fibonacci sequence is multiple of 11?

proof:

this can be proved as below
$F_3 = F_2 + F_1$
$F_4 = F_3 + F_2=2F_2+F_1$
$F_5 = F_4 + F_3=3F_2+2F_1$
$F_6 = F_5 + F_4=5F_2+3F_1$
$F_7 = F_6 + F_5=8F_2+5F_1$
$F_8 = F_7 + F_6=13F_2+8F_1$
$F_9 = F_8 + F_7=21F_2+13F_1$
$F_{10} = F_9 + F_8=34F_2+21F_1$

Adding we get $F_1+F_2+F_3+F_4+F_5+F_6+F_7+F_8+F_9+F_{10}$
=$88F_2+55F_1= 11(8F_2+5F_1) =11F_7$

This can be proved further below as(more fun)

$F_1+F_2+F_3+F_4+F_5+F_6+F_7+F_8+F_9+F_{10}$
=$(F_1+F_2)+(F_3+F_4)+(F_5+F_6)+F_7+F_8+F_9+F_{10}$
= $F_3+F_5+F_7+F_7+F_8+F_9+(F_8+F_9)$
= $F_3+F_5+2F_7+2F_8+2F_9$
= $F_3+F_5+2F_7+2F_8+2(F_7 + F_8)$
= $F_3+F_5+4F_7+4F_8$
=  $F_3+F_5+4F_7+4(F_6+F_7)$
=  $F_3+F_5+8F_7+4F_6$
=  $F_3+(F_5+F_6) +8F_7+3F_6$
= $F_3+9F_7+3F_6$
= $F_3+F_6+ 9F_7+2F_6$
=$F_3+(F_4+F_5)+ 9F_7+2F_6$
= $(F_3+F_4)+F_5+ 9F_7+2F_6$ 
= $F_5+F_5+ 9F_7+2F_6$
= $2 F_5+ 9F_7+2F_6$
= $2(F_5+F_6)+ 9F_7$
= $2F_7 + 9F_7$
= $11F_7$
Done

2014/094) If $a+b+c=0$ then prove that $\dfrac{a^2}{a^2-bc} +\dfrac{b^2}{b^2-ca} +\dfrac{c^2}{c^2-ab} =2$

Solution

We are given $a + b+ c = 0\cdots(1)$

$(a+b+c)^2 = a^2 +b^2 + c^2 +2ab + 2bc + 2ca$

 so 

$a^2 +b^2 + c^2 = -2(ab +bc +ca)\cdots(2)$

 From (1)
$a(a+b+c) = 0$

$a^2 + ab + ca = 0$

or $a^2 - bc = -(ab + bc+ ca)\cdots(3)$

Similarly  

$b^2 - ca = -(ab + bc+ ca)\cdots(4)$

$c^2 - ab = -(ab + bc+ ca)\cdots(5)$

Thus, using (3), (4) & (5), we can rearrange the LHS as follows;

LHS = $\dfrac{a^2}{a^2-bc} +\dfrac{b^2}{b^2-ca} +\dfrac{c^2}{c^2-ab}$

 = $\dfrac{a^2 +b^2+ c^2}{-ab-bc-ca}$

= -$\dfrac{a^2 +b^2+ c^2}{ab+bc+ca}$

= -(- 2)$\dfrac{ab+bc+ca}{ab+bc+ca}$

= 2 

 

2014/093) Find the number of non zero integral solutions of $(1-i)^x=2^x$

We have

 $|(1-i)|=2$

and 

 $|2|=2$

 So no power other than zero shall meet modulus and x = 0 is the only solution as both sides are 1

Q2014/092) If a, b, c ϵ R, $x = a² - bc$, $y = b² - ca$, $z = c² - ab$, then prove that, $x³ + y³ + z³ - 3xyz$ is a perfect square

we have

$x^3 + y^3 + z^3 - 3xyz$
= $(x + y + z) (x^2 + y^2 + z^2 - xy - yz - zx)$
= $\dfrac{1}{2} (x + y + z) [(x - y)^2 + (y - z)^2 + (z - x)^2] \cdots(1)$... ( 1 )

$x + y + z$
= $a^2 - bc + b^2 - ca + c^2 - ab$
= $\dfrac{1}{2}[(a - b)^2 + (b -c)^2 + (c - a)^2] \cdots (2)$

further

$x - y$
= $(a^2 - bc) - (b^2 - ca)$
= $a^2 - b^2 + ca - bc$
= $(a - b) (a + b) + c (a - b)$
= (a + b + c) (a - b)

Similarly,
$y - z = (a + b + c) (b - c)$ and
$z - x = (a + b + c) (c - a)$

=> $(x - y)^2 + (y - z)^2 + (z - x)^2$
= $(a + b+ c)^2 ((a - b)^2 + (b - c)^2 + (c - a)^2)\cdots ( 3 )$

Plugging results ( 2 ) and ( 3 ) into ( 1),

$x^3 + y^3 + z^3 - 3xyz$
= $\dfrac{1}{2} * \dfrac{1}{2}((a - b)^2 + (b -c)^2 + (c - a)^2) *$

            $(a + b+ c)^2 ((a - b)^2 + (b - c)^2 + (c - a)^2)$
= $(\dfrac{1}{2} (a + b + c) ((a - b)^2 + (b - c)^2 + (c - a)^2))^2$

which is a perfect square.

2014/091) The LCM and GCD of two composite numbers total 111. What are the numbers?

Let the gcd be t
the 2 numbers are at and bt with a < b where a and b are coprimes or a  =1

Therefore, the LCM = abt

 so abt + t = 111 or 111 = t(1+ab) = 3 * 37

then we have following cases
t = 1 , ab = 110 then a = 1, b= 110 giving 2 numbers 1 and 110
                                 a = 2, b= 55 giving 2 numbers 2 and 55
                         or a = 5 , b = 22 giving 2 numbers 5 and 22
                         or a = 10, b = 11 giving 2 numbers 10 and 11
t=3, ab = 36 then a = 1, b= 36 giving 2 numbers 3 and 108
                             a = 4, b= 9 giving 2 numbers 12 and 27
t = 37, ab =2 then a = 1, b=2 giving 37 and 74

2014/090) Factor (x+1)(x+2)(x+3)(x+6)-3x^2

we can do by expanding the same but can use the simple method as below

1 * 6 = 2 * 3

so $(x+1)(x+2)(x+3)(x+6) - 3x^2$
= $(x+1)(x+6)(x+2)(x+6)- 3x^2$
= $(x^2 + 7x + 6)(x^2 + 5x + 6) - 3x^2$
= $((x^2 + 6x+ 6) + x)((x^2 + 6x + 6) -x) - 3x^2$
= $(x^2+ 6x + 6)^2 - x^2 - 3x^2$
= $(x^2 + 6x+6)^2 - 4x^2$
= $(x^2 + 6x + 6)- (2x)^2$
= $(x^2 + 8x + 6) (x^2 + 4x+ 6)$

above 2 cannot be factored further

2014/089) show that if $abc = 1$ then $a^2+b^2+c^2\ge a + b + c$

we have by AM GM inaquality

$a^2+1 \ge\ 2a \cdots (1) $
$b^2+1 \ge\ 2b \cdots (2) $
$c^2+1 \ge\ 2c \cdots (3) $

further

$\dfrac{a+b+c}{3} \ge \sqrt[3]{ abc}$
or  $\dfrac{a+b+c}{3} \ge 1$
or $a+b+c \ge 3$
or $(a+b+c-3 )\ge 0\cdots(4) $

adding  (1) (2) and (3) we get

$a^2+b^2+c^2 + 3 \ge 2(a+b+c)$
or $a^2+b^2+c^2  \ge a+b+c + (a+b+c-3)$
using (4) we get
 $a^2+b^2+c^2  \ge a+b+c$