2009/004) Prove the identity sin^2Acos^2B-cos^2Asin^2B = sin^2A-sin^2B

sin^(2)Acos^(2)B-cos^(2)Asin^(2)B
= sin ^2 A(1- sin ^2 B) - cos^(2)Asin^(2)B
= sin ^2 A - sin ^2 A sin ^2 B - cos^(2)Asin^(2)B
= sin ^2 A - sin ^2 B(sin ^2 A + cos ^2 A)
= sin ^2 A - sin ^2 B

2009/003) factor (x-y)^5+(y-z)^5+(z-x)^5

if a + b+ c = 0

then (a+b)^5 = -c ^5

so a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a^4 b + b^5 = - c^5

so a^5 + b^5 +c^5 = - (5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a^4 b)
= -5 ab(a^3 + 2 a^2b + 2 a b^2 + b^3)
= - 5 ab((a+b)^3 - (a^2 b + ab^2)
= - 5ab((a+b)3 - ab(a+b))
= - 5ab(a+b)((a+b)^2 - ab)
= 5abc(c^2-ab) as a+b = - c

as (x-y)+ (y-z) + (z-x) = 0

we get 5(x-y)(y-z)(z-x)((z-x)^2 - (x-y)(y-z))
= 5(x-y)(y-z)(z-x)(z^2 + x^2- 2xz -xy + xz +-y^2 -xy)
= 5(x-y)(y-z)(z-x)(x^2+y^2+z^2 - xy - yz - zx)

2009/002) show: arcsin[(4/√41)]+arcsin[(1/√82)=pi/4?

the things become easy in case we convert arc sin to arc tan

let x = arcsin 4/√41

so sin x = 4/√41

cos x = sqrt(1- sin ^2 x) = sqrt(1-16/41) = 5/√41

so tan x = 4/5

now let sin y = 1/√82

so cos y =√(1-1/82) = 9/√(82)

so tan y = 1/9

we heed to find tan (x+y) when tan x = 4/5 and tan y = 1/9

tan (x+y) = (tan x+ tan y)/(1-tan x tan y) = (4/5+1/9)(1-4/5*1/9) = (41/45)/(41/45) = 1

so x + y = tan ^-1 1 = pi/4 ( this is so because x < pi/4 and y < pi/4 so sum <= pi/2

2009/001) Find the smallest positive integer x for which 7x^25 - 10 is completely divisible by 83.

We are given
7x^25 = 10 mod 83We should make the coefficient of x^25 as 7
to get rid of 7 multiply by inverse of 7we have GCD(7,83) = 1using extended eulers algorithm 6 = 83- 7 *11 1 = 7- 6 = 7*12 – 83
so inverse of 7 is 12multiply by 12 on both sides knowing 7*12 = 1 mod 83 we get x^25 = 10 * 12 mod 83 = 37
now we need to raise a power so that x^82 = 1 mod 1now we need to find reciprocal of 25 mod 82again using extended eulers algorithm82 = 3*25 + 7
7 = 82 - 3* 2525 = 7*3 + 4 or4= (25-7*3) = (25-3*(82-3*25) = 10 * 25 - 3*82now knowing 1 = 2*4 - 7 = 2(10*25-3*82) - (82-3*25)= 23*25 - 7*82
so reciprocal of 25 is 23so raise the number to the power 23x= 37^23 mod 83= 37* (37*2)^11 mod 83= 37 * (1369)^11 mod 83= 37 * 41^11 mod 83= (37*41)* 41 ^10 mod 83= 23 *41 ^10 mod 83= 23 *(41^2)^5 mod 83= 23 *21^5 mod 83= 23*21 * 21^4 mod 83= 483 * 21^4 mod 83= 68 * 21^4 mod 83= 68 * 441 * 441 mod 83= 69

2008/014) Solve the equation 4xABCD=DCBA

A has to be < 3 because 3* 4 = 12 so RHS is a 5 digit number
A cannot be 1 as from RHS A has to be even.
So A has to be 2.
now B can be either 1 or 3 or 5 or 7 but B < 5 because 4*25 = 100 that is 5 digit
So AB = 21 or 23
if AB = 23 DC >= 92do D = 9 which is not possible as 4*8 is 2 ending but 4*9 is notAB = 21
So D = 8
so the number = 4*(2108+10C) = 8032+100C
or 8432+40C = 8012+ 100C
or 60C = 420
so C =7
so number = 2178*4 = 8712

2008/013) 2 is the only prime sum of 2 positive cubes

We know a^3 + b^3 = (a+b)(a^2-ab+b^2)
if a= b then a^3+b^3 = 2 a^3 which not a prime unless it is 2
with out loss of generality we can assume a> b
now a+b >=2
a^2+b^2-ab = a(a-b) + b^2 > b^2 so
a^2 + b^2-ab > 1
as it has 2 factors and both are >2 a^3+b^3 cannot be prime or in other words a prime number > 2 cannot be sum of 2 positive cubes

2008/012) x+y+z=6 , x^2+y^2+z^2=306 ,find maximum and minumum of XYZ

x+y+z = 6
x^2+y^2+ z^2 = 306
find minumum of maximum of xyz
the three unknowns need not be positive.

x+y + z = 6 ..1

x^2+y^2+z^2=306.. 2

square (1)
x^2+y^2 + z^2 + 2xy + 2yz + 2xz = 36

so xy+xz+yz = (36-306)/2 = -135

now let us take

p(t) = (t-x)(t-y)(t-z)
= t^3 - t^2(x+y+z) +t(xy+yz+xz) - xyz

= t^3 - 6t^2 + 135t - k
k = t^3 - 6t^2 - 135t
k is maximum or minimum when t^3 - 6t^2 - 135t
is

now you can differentiate wrt t and equate to zero

3 t^2 - 12 t - 135 = 0

or t^2 - 4t - 45 = 0

(t-9) ( t + 5 ) = 0

t =9 gives t^3 - 6t^2 - 135t = 729- 486- 1215 = -972 minimum
t =-5 gives -125-150+ 675 = 400 maximum