2015/020) If $(x+yi)^3 = u + vi$ Prove that $\dfrac{u}{x} + \dfrac{v}{y} = 4(x^2 - y^2)$

$(x+yi)^3 = u + vi$
expand LHS

$x^3 + 3ix^2 y - 3 xy^2 - iy^3 = (x^3 - 3y^2) + i(3x^2y - y^3)$
= $x(x^2- 3y^2) + iy(3x^2 - y^2)$

equate real part on both sides and imaginary parts

to get $u = x(x^2-3y^2)$
$v = y(3x^2-y^2)$

so $\dfrac{u}{x} + \dfrac{v}{y}= (x^2-3y^2) + (3x^2-y^2) = 4x^2 - 4y^2$

2015/019) Prove that there exists non negative integers a and b such that n=4a+5b for n >=12

we can specify from 12 to 15 as 4a+ 5b with a and b non -ve as below


12 = 3 * 4 + 0 * 5
13 = 2 * 4 + 1 * 5
14 = 1 * 4 + 2 * 5
15 = 0 * 4 + 3 * 5

Now for numbers > 15 if the number is
  4a then taken12 and add 4a-12
if 4a + 1 then 13 + 4(a-3)
if 4a + 2 then 14 + 4(a-3)
if 4a + 3 then 15 + 4(a-3)
as a >= 3 we have the result

2015/018) 9 balls problem

Of nine similar balls, Eight of them weigh the same. How can you find the ball that is different by using a balance and only three weighing?

Solution

Take 2 sets of 3 balls each. If they are of same weight then the 3^rd set of balls say (1),(2),(3) contain the different ball.  Now weigh (1) and (2). If they are same then 3 is the different ball. If they are different say (1) is heavier then weigh (1) and (3), if they are different the (1) is different ball.  If they are same then (2) is the different ball.

Now say 1^st set is heavier.

We compare the 1^st set and 3^rd set. If 1^st one is heavier then this contains the heavy ball. Compare 2 balls of 1^st set and if they are of same weight then 3^rd ball is different and if not same then heavier ball is different.  If the 2 are same. Then 2^nd set contains the lighter ball. Compare 2 balls of second set and if they are of same weight then 3^rd ball is different and if not same then lighter ball is different

 

2015/017) If the (m+1)th,(n+1)th,& (r+1)the terms of an A.P are in G.P & m,n,r are in H.P .Show that the ratio of the common difference to the first term in the A.P is -2/n

Let 1^st term be a and difference be d

so (m+1)st term a + md

(n+1)st term a + nd

(r+1) st term a + rd

these are in GP

so $(a+md)(a+rd) = (a+nd)^2$

or $(m+r) ad + rmd^2 = 2nad + n^2d^2$

or $ad(m+r – 2n) = d^2(n^2 – rm)$

or $\dfrac{d}{a} = \dfrac{m+ r – 2n}{n^2-rm}$

as m,n,r are in HP so $\dfrac{1}{m} + \dfrac{1}{r} = \dfrac{2}{n}$

or $r + m = \dfrac{2rm}{n}$

so from (1)

$\dfrac{d}{a} = \dfrac{\frac{2rm}{n} – 2n}{n^2-rm} =\dfrac{-2}{n}$

2015/016) What is the smallest symmetrical number greater than 56,789 which is exactly divisible by 7

Solution
the number has to be form

10000x+1000y+100z+10y+z
= 10001x+ 1010y + 100z

10001x + 1010y+100z mod 7 = 5x + 2y + 2z mod 7

x cannot be < 5. so let x = 5 and let us look for solution y minimum 6

so 25+2y+2z mod 7 = 0 or 2y+2z = 3 mod 7 or y + z = 5 mod 7

y+z = 5 no solution
so y + z = 12 if y = 6 z = 6 not possible

so y = 7 and z = 5 possible

number = 57575

it is 7 * 8225
If we did not find a solution with x = 5 then we should have tried at x = 6 refer to https://in.answers.yahoo.com/question/index?qid=20101001195414AAyI1cu

 

2015/015) The roots of the cubic x^3-9x^2+mx-24=0 are in AP. Find m and the roots

The roots are in AP so they are , a-b, a, , a +b and b > 0

sum of roots = 3a = 9 hence a = 3

now product of roots = a(a-b)(a+b) = 24 or 3(3-b)(3+b) = 24 or $9-b^2 = 8$ and hence b = 1 

so roots are p = 2, q = 3, r = 4 

m = pq+ qr + rp = 26 = 2* 3 + 3 * 4 + 4 * 2 = 26 

hence roots are 2,3, 4 and m = 26

2015/014) if $a^3-3a^2b = 2005$ and $b^3-3b^2a = 2004$ then find

$(\dfrac{b_3-a_3}{b_3})(\dfrac{b_2-a_2}{b_2})(\dfrac{b_1-a_1}{b_1})$ 

We have

$a^3-3a^2b = 2005\cdots(1)$
and


$b^3-3b^2a = 2004\cdots(2)$

now subtract 2nd equation from 1st

$a^3-3a^2b+3ab^2-b^3= 1$
or $(a-b)^3=1$

$(a_1-b_1)\,(a_2-b_3)\,(a_3-b_3)$ are roots of equation $x^3-1=0$

so $(a_1-b_1)(a_2-b_3)(a_3-b_3) = 1\cdots(3)$

further $a=b+1$ and putting in (2) we get
$b^3-3b^2(b+1)=2004$
or   $2b^3 +3b^2= -2004$
so product of roots = $b_1b_2b_3=- 1002\cdots(4)$

from (3) and (4)

$(\dfrac{b_3-a_3}{b_3})(\dfrac{b_2-a_2}{b_2})(\dfrac{b_1-a_1}{b_1})= \dfrac{1}{1002}$

Note:

I have taken the problem in the site at  http://mathhelpboards.com/challenge-questions-puzzles-28/evaluate-value-product-14240.html#post67603

Where you can find some other different correct solutions.

Unforunately the problem stated in the link was wrong and I corrected the same

Q2015/013) solve for x, y, z when

$x+y + z = - 3 \cdots 1$


and $x^3+y^3 + z^3- 20(x+3)(y+3)(z+3)= 2013\cdots(2)$

We have

$x+y + z = - 3 \cdots 1$


and $x^3+y^3 + z^3- 20(x+3)(y+3)(z+3)= 2013\cdots(2)$

from (1)

so $x+y = -(3+z)\cdots (3)$
$y + z = -(3+x)\cdots (4)$
$z+x = -(3+y)\cdots(5) $

$(x+y+z)^3 = x^3+y^3+z^+3(x+y)(y+z)(z+x)$

or $-27 = 2013 + 20(x+3)(y+3)(z+3) - 3(z+3)(x+3)(y+3)$ (LHS from (1) and RHS from (2), (3),(4),(5)

so $ 17(x+3)(y+3)(z+3) = -(2013+27)=-2040$
or $(x+3)(y+3)(z+3)= - 120$

further x + 3 + y +3 + z + 3 = 6

so we need 3 integers product is -120 and sum 6 and  the numbers are 10,2,-6

so z = 7, y = -1, x = - 9

3x + y + 2z = - 18- 1 + 14 = - 5

Note:

I  have  taken the problem from the site http://mathhelpboards.com/challenge-questions-puzzles-28/find-3x-y-2z-14136.html#post67082

where till this date I was the only solution provider

2015/012) find $12x^4-2x^3-25x^2+ 9x + 2017$ given $x= \dfrac{\sqrt{5}+1}{4}$

we have $4x-1 = \sqrt5$

squaring and reordering we get
$16x^2-8x -4-0$
or $4x^2-2x-1 = 0 \cdots (1)$

now deviding $12x^4-2x^3-25x^2+ 9x + 2017$ by $(4x^2-2x-1)$ we find that

$12x^4-2x^3-25x^2+9x+2017$
$=(4x^2-2x-1)(3x^2+x+5) + 2012= 2012$ using (1)

2015/011) prove that if $a_1$,$a_2$ and $a_3$ are altitudes of a triangle and r is

  
radius of inscribed triangle then

$\dfrac{1}{a_1} + \dfrac{1}{a_2} + \dfrac{1}{a_3}= \dfrac{1}{r}$

let x,y,z be the sides of triangle with corresponding altitudes $a_1$.$a_2$ and $a_3$ and area be A

so $xa_1 = ya_2 = za_3= 2A$

so
$\dfrac{1}{a_1} + \dfrac{1}{a_2} + \dfrac{1}{a_3}$

= $\dfrac{x}{2A} + \dfrac{y}{2A} + \dfrac{z}{2A}$

=  $\dfrac{x+y+z}{2A}$

= $\dfrac{2s}{2A}$

= $\dfrac{s}{A}\cdots(1)$

now because

$r^2= \dfrac{(s - x)*(s - y)*(s - z)}{s}$

= $\dfrac{s(s - x)*(s - y)*(s - z)}{s^2}$
=  $\dfrac{A^2}{s^2}$
so $r = \dfrac{A}{s}$

Putting the above in (1) we get the result

Note:

I  have  taken the problem from the http://mathhelpboards.com/potw-secondary-school-high-school-students-35/problem-week-148-january-26th-2015-a-14133.html where this appeared as problem of the week. Other solution provided there are more elegant