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
 $a^2+b^2+c^2  \ge a+b+c$

2014/088) For how many different integral values of b are both roots of $x^2+bx-16=0$ integers?

16 = 1* 16
= 2 * 8
= 4 * 4
= (-1) * (-16)
= (-2) * (-8)
= (-4) * (-4)

there are 6 sets of b as it can be factored in 6 ways

the values being 1+16 = 17, 2+8 = 10, 4 + 4 = 8, -1 - 16 = - 17, -2 -8 = -10 and -4 -4 = =8

2014/087) Solve $\sqrt{x-2}- \sqrt{x-7}=1$

Solution

we are given


$\sqrt{x-2}- \sqrt{x-7}=1\cdots (1)$

we have $(x-2) - (x-7) = 5\cdots(2)$

dividing we get

$\sqrt{x-2}+ \sqrt{x-7}=5\cdots (3)$
adding (1) and (3)

$2\sqrt{x-2}=6$
or  $\sqrt{x-2}=3$
or x= 11

2014/086) if $a\sin\,x=b\sin(x+\frac{2\pi}{3})=c\sin(x+\frac{4\pi}{3})$

prove that $ab+bc+ca=0$ and $\dfrac{1}{a} +\dfrac{1}{b}+\dfrac{1}{c}=ab+bc+ca$

proof
say
$a\sin\,x=b\sin(x+\dfrac{2\pi}{3})=c\sin(x+\dfrac{4\pi}{3})=k$
  then
$\sin\, x= \dfrac{k}{a}$
 $\sin(x+\dfrac{2\pi}{3})=\dfrac{k}{b}$
 $\sin(x+\dfrac{4\pi}{3})=\dfrac{k}{c}$
  

so $k(\dfrac{1}{a} + \dfrac{1}{b} +  \dfrac{1}{c}) = ( \sin\,x +\sin(x+\frac{2\pi}{3})+ \sin(x+\frac{4\pi}{3}))$
=$\sin\, x + 2 \cos \dfrac{2\pi}{3} \sin\, x$
=  $\sin\, x + 2 \cdot \dfrac{-1}{2} \sin\, x$
=  0

as k is not zero

$\dfrac{1}{a} + \dfrac{1}{b} +  \dfrac{1}{c}= 0$  and multiply both sides by abc to get

bc + ca + ab = 0

as both are zero they are same

2014/085) prove that $121^n – 25^n + 1900^n – (-4)^n$ is divisible by 2000

proof:

We have 2000 = 125 * 16
So it should be divisible by 125 * 16 or 125 and 16 as these are co-primes

$a^n-b^n$  is divisible by $a-b$
hence
$121^n – 25^n$ is divisible by $121- 25 = 96= 16 * 6$ hence 16
$1900^n – (-4)^n$ is divisible by $1900- (-4) = 1904 = 16 * 1 19$ hence 16
So $121^n – 25^n + 1900^n – (-4)^n$ is divisible by 16

$121^n – (-4)^n$ is divisible by $121 + 4 = 125$
$– 25^n + 1900^n$ is divisible by $1900 – 25 = 1875 = 125 * 15$ hence 125
So $121^n – 25^n + 1900^n – (-4)^n$ is divisible by 125

As the number is divisible by 125 and 16 hence the product that is 2000

2014/084) 2 urns balls reducing problem

there are 2 urns with some balls in each of them. Either one can remove same number of balls from each or one can double the number of balls in one of them. How to empty the 2 urns.

Solution

Say urn containing larger number of balls is A and number of balls m and the other urn is B and number of balls n. There are 2 method and 1^st method is straight forward and 2^nd method is more elegant.


Method 1)

we have to reduce both and if n =1 then double the number of balls and double again.

If n = 2 then double the number of balls.

if n >2 remove n-2 balls from both then balls in A >2 (as m >n) and in B 2 and double the number of balls..

As long as number of balls in A >4 keep doubling the number of balls in B and remove 2 from each.

As the number of balls in A keeps decreasing by 2 it shall be reduced to 3 or 4(depending on odd or even after removing m-2). In case it is 4 then after doubing the number of balls in B it becomes 4 then remove 4 from both.

In case it is 3 and in b 2 (do not double it) remove 1 from both so number of balls left is 2 in A and 1 in B and then double the the number of balls in b to 2 and remove 2 balls from both the urns..

Method 2)

Let k be the highest number such that m > 2**k( that is if m is a power or 2 then take m/2) .

Keep doubling the number of balls in B so that N(the number after doubling) N >2**k

If both numbers are same then remove them and both the urns are empty

m > 2**K and n > 2**K then remove 2**K then m <= 2**K and n < 2**K.

Interchange the roles of m and n if m < n.

So each time the value is larger urn becomes ½ or less of the previous. And the lower one becomes not more than the previous larger Keep repeating the process until both are same as they are reducing the 2 numbers will converge to be same.

To illustrate as an example let m = 69 and n = 10

Keep doubling n, i.e 20 , 40, 80

Remove 64 that is m = 5 n = 16( > 8)

Then keep doubling m = 10

Remove 8 so m =2 and n = 8

Keep doubling m till it is 8 and remove 8 from both