2010/034) Factorise x^4+ x^2+1

x^3 and x are missing so we can try to do it by difference of squares adding and subtracting x^2 that is middle term

= (x^4+2x^2+1)- x^2
= (x^2+1)^2 - x^2
= (x^2+x+1)(x^2-x+1)

2010/033) Let S be the sum of all the real coefficients of the expansion of (1 + ix)^2009. what is log_2(S)?

Let S be the sum of all the real coefficients of the expansion of (1 + ix)^2009. what is log_2(S)?

the sum of all the real coefficients of the expansion of (1 + ix)^2009 is real part of (1+i)^2009 by putting x = 1

now (1+i) = sqrt(2) cis(pi/4) where cis(x) = cos x + i sin x

so (1+i)^2009 = sqrt(2)^2009(cis (2009pi/4))

so real part = sqrt(2)^2009 cos(2009pi/4) = sqrt(2)^2009 * cos(pi/4) = sqrt(2)^2008 = 2^1004

hence log_2(s)= 1004

2010/032) If p and q are distinct primes, show that p^q + q^p ≅ (p + q) mod pq

If p and q are distinct primes, show that p^q + q^p ≅ (p + q) mod pq.
as

q is prime

so p^(q-1) mod q = 1 (as per fermats little theorem)


so p^(q-1) = mq + 1

multiply by p on both sides p^q = mpq + p = p mod pq

similarly as q is prime q^p = p mod pq

adding we get (p^q + q^p) mod pq = (p+q)

proved

2010/031) 5-digit number whose half is a perfect cube and one-third is a perfect square?

5-digit number whose half is a perfect cube and one-third is a perfect square?

solution
we have
2n^3 = 3p^2
so n has to be multiple of 3 say 3x and p multiple of 2 say 2a
2(3x)^3 = 3(2a)^2
or 54x^3 = 12a^2
or 9x^3 = 2a^2
so a has to be divisible by 3 (say a = 3y) (sqrt of 9) and x by 2 say x = 2b
so n= 6y and p = 6b

and 9(2y)^3 = 2(3b)^2

or 4y^3 = b^2

so b has to be divisible by 2 say b = 2c that is p = 12c

then 4y^3 = 4c^2 or y^3 = c^2 so y has to be a prfect square and c a perfect cube that it

y= m^2 and c= m^3

so n = 6m^2 and p = 12m^3

now it shall not be difficult to find 2n^3 or 2(6m^2)^3 which is a 5 digit number

by checkin we find that it is m = 2 so n = 24 and ans = 2 * 24^3 or 27648

2010/030) factor a^2(b+c) + b^2(a+c) + c^2(a+b) + 2abc

a^2(b+c) + b^2(a+c) + c^2(a+b) + 2abc
= a^2(b+c) + (b^2a + b^2 c + ac^2 + bc^2 + 2abc)
= a^2(b+c) + a(b^2+c^2 + 2ab) + (b^2 c + bc^2)
= a^2(b+c) + a(b+c)^2 + bc(b+c)
= (b+c)(a^2+ ab + ac + bc)
= (b+c)(a+b)(a+c)

2010/029) Find the maximum value of, (√p + √q + √r ). (plz see details. . . )?

Let α, β, γ > 0 and (α + β + γ) = π/2. If p = (tan α tan β) + 5, q = (tan β tan γ ) + 5
and r = (tan γ tan α) + 5 ,
find the maximum value of (√p + √q + √r ).

tan γ = tan [ (π/2) - (α+ß) ] = cot(α+ß) = 1 / tan(α+ß)

tan γ = (1 - tan α tan ß) / ( tan α + tan ß )

tan γ tan α + tan ß tan γ = 1 - tan α tan ß

(r-5) + (q-5) = 1 - (p-5)

p + q + r = 16 ................. (1)

Let p = a^, q= b^2 and r = c^2

We are given p+q+r = 16 => a^2+b^2+c^2 = 16

We need to maximize a + b + c

We know
(a+b+c)^2 = a^2+b^2 +c^2 + 2ab + 2bc + 2ca
(a-b)^2 = a^2 + b^2 – 2ab
(b-c)^2 = b^2 + c^2 – 2bc
(c-a)^2 = c^2 + a^2 – 2ac

Adding we get (a+b+c)^2 +(a-b)^2 + (b-c)^2 + (c-a)^2 = 2 (a^2+b^2 + c^2) = 32
Clearly a+b+c is maximum when a=b=c because as (a-b)^2 + (b-c)^2 + (c-a)^2 can not be lower than zero and is zero when a= b= c

So 2 (3a^2) = 32 or a^2 = p = 16/3

so maximum value of (√p + √q + √r ) = 4 √3

2010/028) If x + y + z = xyz

Prove that x(1-y^2)(1-z^2) + y(1-z^2)(1-x^2) + z(1-x^2)(1-y^2) = 4xyz?

proof

If x + y + z = xyz,

putting x = tan A, y = tan B, z = tan C,

tan A + tan B + tan C = tan A tan B tan C, i.e., S₁ = S₃

∴ tan ( A+B+C ) = ( S₁ - S₃ ) / ( 1 - S₂ ) = 0

∴ A+B+C = nπ for integral n.

2A + 2B +2C = 2nπ

∴ tan ( 2A+ 2B+ 2C ) = tan 2nπ = 0

∴ tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C

∴ 2x / (1-x²) + 2y / (1-y²) + 2z / (1-z²) = [ 2x / (1-x²) ]•[ 2y / (1-y²) ]•[ 2z / (1-z²) ]

∴ x(1-y²)(1-z²) + y(1-z²)(1-x²) + z(1-x²)(1-y²) = 4xyz ............

Hence proved