Friday, April 29, 2016

2016/040) Derive the identity $u_{n+3} = 3u_{n+1} - u_{n-1}$ where $u_n$ is the $n^{th}$ fibonacci number and $n >= 2$

$u_{n+3} = u_{n+2} + u_{n+1}$
$= u_{n+1} + u_{n} + u_{n+1}$
$= 2u_{n+1} + u_{n}$
$= 2u_{n+1} + u_{n+1} - u_{n-1}$
$= 3 u_{n+1} -  u_{n-1}$

2016/039) if $x=log_abc, y= log_bca, z= log_cab$ then show that $\frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} = 1$

we have $x=log_abc$ hence $x + 1 = log_abc + log_aa = log_aabc$ or $\frac{1}{x+1} = log_{abc}a$
 similarly $\frac{1}{y+1} = log_{abc}b$ and $\frac{1}{z+1} = log_{abc}c$
 Hence $\frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} = log_{abc}a + log_{abc}b = log_{abc}c = log_{abc}abc =1$

2016/038) Solve the equation $3x-7y \equiv 11 \pmod {13}$

to solve this we solve 2 equations below and combine the two
$3x - 7y = 11\cdots(1)$
$3x - 7y = 13\cdots(2)$
let us solve $3x - 7y = 11$ first
we need to find solution of $3x-7y =1$. this can be done by extended euler algorithm but
we see that $(-2,-1)$ satisfy $3x-7y = 1$ so (-22,-11) satisfies $3x-7y = 11$
$(-22 + 7t, -11 + 3t)$  is a solution. choose t to be 4 to get (6,1) to be a soution.
for solving $3x - 7y = 13$ we have a solution $(-26,-13)$ so also $( -26 + 7t, - 13+ 3t)$ and
 putting t = 4 we get $(2,-1)$ a solution.
So solution for $3x-7y \equiv 11 \pmod {13}$ is $(6 + 2t, 1 - t)$. because the value need to be computed in mod 13 so the ans is $(6 + 2t \pmod {13} , (1 - t) \pmod {13})$.  

Wednesday, April 27, 2016

2016/037) Solve the Diophantine below $12x + 25 y = 331$

Taking mod 25 we get
$12x \equiv 6 \pmod {25}$
we need to find inverse of $12\pmod {25}$
It can be done many ways , as we see that $12 * 2 = 24$ so multiply both sides by 2 to get
$24 x \equiv 12 \pmod {25}$
or $ x \equiv -12 \pmod {25}$ or  $ x \equiv 13 \pmod {25}$
so $x = 13$ is one of the values and putting $x = 13$ in given equation we get $y = 6$
so one solution $ x= 13,y = 6$ and as $12 * (-25) + 25 * 12 = 0$ we get generic solution
$ x = 13 - 25t, y= 6 + 12t$

Tuesday, April 26, 2016

2016/036) Prove that $\arg[(a+bi)(c+di)]$

$=\arg(a+bi)+\arg(c+di) \pmod \pi $
Proof:
we have $(a+bi)(c+di) = (ac - db) + (bc + ad)i$
hence $\arg[(a+bi)(c+di)] = \arg (ac - db) + (bc + ad)i$
or  $\arg[(a+bi)(c+di)] = \arctan (\frac{bc + ad}{ac-db})$
$= \arctan (\frac{\frac{b}{a}+ \frac{d}{c}}{1 - \frac{b}{a}\frac{d}{c}})$
$= \arctan (\tan ( \arctan(\frac{b}{a}) + \arctan(\frac{d}{c}))$
$= \arctan(\frac{b}{a}) + \arctan(\frac{d}{c})\pmod \pi $
$= \arg(a+bi)+\arg(c+di) \pmod \pi $

2016/035) If one root of $4x^2+2x-1= 0 $ be $\alpha$ then show that other root is $4\alpha^3-3\alpha$

We have $4\alpha^2+2\alpha-1= 0\cdots(1) $
We need to prove
$4(4\alpha^3-3\alpha)^2+2(4\alpha^3-3\alpha)-1= 0 $
We have $4\alpha^2+2\alpha-1= 0$
Hence $4\alpha^3+2\alpha^2-\alpha= 0$
So $4\alpha^3 = -2\alpha^2 + \alpha$
or $4\alpha^3 -3\alpha = - 2\alpha^2 - 2 \alpha=  - \frac{1}{2}( 4\alpha^2 + 4\alpha)$
$= - \frac{1}{2}( 1 - 2\alpha + 4\alpha)= - \frac{1}{2}( 1 + 2\alpha) \dots(2)$
Hence $4(4\alpha^3-3\alpha)^2+2(4\alpha^3-3\alpha)-1= 0 $
 = $4(\frac{1}{2}( 1 + 2\alpha))^2- 2 (\frac{1}{2}( 1 + 2\alpha)-1$ using (2)
 = $( 1 + 2\alpha))^2- ( 1 + 2\alpha)-1$
 = $1 + 4 \alpha + 4\alpha ^2 - 1 -2 \alpha -1$
 = $4\alpha ^2 + 2 \alpha -1= 0$ from (1)

Sunday, April 24, 2016

2016/034) Show that $x^2-xy+y^2 - 4x - 4y + 16=0$ has only one solution in real x,y. find it

 we have multiplying by 2
$2x^2-2xy+2y^2 - 8x - 8y + 32=0$
or $(x^2 - 2xy + y^2) + (x^2-8x + 16) + (y^2 - 8y + 16) = 0$
or $(x-y)^2 +(x-4)^2 + (y-4)^2=0$
hence we have each of the squares zero giving $x=y=4$ the only solution

2016/033) If $a,b,c$ are in AP and $a^2,b^2,c^2$ are in HP then prove that either $a=b=c$ or $a,b,\frac{-c}{2}$ are in GP

we have $2b = a + c\cdots(1)$
$\frac{2}{b^2} = \frac{1}{a^2} + \frac{1}{c^2}$
or  $2a^2c^2 = b^2c^2 + b^2 a^2 = b^2(a^2+c^2) = b^2((a+c)^2 - 2ac) = b^2(4b^2 - 2ac)$
or $a^2c^2 = 2b^4 - b^2ac$
or $2b^4 - b^2ac - a^2c^2 = (2b^2+ac)(b^2-ac) = 0$
$b^2= ac =>  4b^2 = 4ac = (a+c)^2 => (a-c)^2 = 0 => a = c = b$
or $2b^2+ ac = 0 => a,b ,\frac{-c}{2}$ are in GP

Saturday, April 23, 2016

2016/032) if $a,b,c$ are positive real numbers show that $(1+a)^7(1+b)^7(1+c)^7 > 7^7 a^4b^4c^4$

we have
$(1+a)(1+b)(1+c) = 1 + a + b + c + ab + bc + ca + abc $
$>     a + b+ c+ ab+bc+ca + abc\cdots(1)$
By am gm inequality we have $\frac{a + b+ c+ ab+bc+ca + abc}{7} >= \sqrt[7]{a^4b^4 c^4}$
or  $a + b+ c+ ab+bc+ca + abc >= 7 \sqrt[7]{a^4b^4 c^4}$
or $ 1  + a + b+ c+ ab+bc+ca + abc > 7 \sqrt[7]{a^4b^4 c^4}$
or $ (1  + a)( 1+ b)(1+ c) > 7 \sqrt[7]{a^4b^4 c^4}$ from (1)
hence  $(1  + a)^7( 1+ b)^7(1+ c)^7 > 7^7 a^4b^4 c^4$

2016/031) Find 3 digit number which is same as sum of factorials

Let the number be $100 a + 10b+c$
none of $a,b, c$ can be $> 5$ as $6! = 720$ and $7! = 5040 > 1000$
one of them that is b or c= 5 ( a cannot be 5 as $5! = 120$ and $5! + 4! + 3! < 200$)
so $a = 1, b= 5, c = ?$ or $a = 1, b = ? , c = 5$ ( it has to be $< 5$)
if  $a = 1 , b = 5$ we get $1 + 120 + c ! > 150$ and $< 160$
$39 >c! > 29$ and so there is no c
if $a = 1, c = 5$ we get $1 + 120 + b! = 105 + 10 b$
so b = 4 and the number is 145