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