2011/111) Between 1000 and 2000 you can get all but one integers as the sum of positive consecutive integers. What is this number that you cannot get ?

It must have an odd factor it is proved in yahoo ans

first we know that each odd number can be expressed as sum of 2 consecutive numbers as

2n + 1 = n + (n+1)

Any even number is a power of 2 or can be expressed as power of 2 multiplied by an odd number >1

Say it is 2^p(2n+1) where p is highest power of 2 which is a factor.

Now 2n+1 can be written as n+ (n+1)

Now we have 2 cases 2^p <= n or 2^p > n

Case 1) 2^p < n

This can occur 2^p times but there shall be 2^p copies of n and 2^p copies on n + 1

In the kth copy subtract k-1 from n and add k-1 to n+ 1

So we get the values n-(2^p-1) to n+ 2^2p

The sum shall be 2^p ( n + 1 +n).

And all numbers are consecutive ( 1^st number >= 1) and positive

Case 2) 2^p > n

We take 2n+1 copies of 2^p

From the middle value at a distance of k we subtract k on from the left element and add k on the right element

So we have consecutive numbers 2^p-n to 2^p+ n

Sum = ½(2n+1)(2^p*2) = (2n+1) 2^p and as 2^p > n so all numbers are positive

Only power of 2 cannot be expressed as sum of consecutive number. Then the number is 1024 it has been proved in the link as above

2011/110) solve for a If a^n + a^(2n) = a^(3n) where n = 1+ log(base a)(cos(π/5)

a cannot be zero as base to a is taken

For a non trivial solution

a is not zero and dividing both sides by a^n we get

1+ a^n = a^2n

So a^2n = golden ration φ

n = 1+ log(base a)(cos(π/5)

so a^n = a * (cos(π/5)

= a * φ/2 = φ

Or a= 2

2011/109) How do you solve 1/x^2 + 1/y^2 = 1/z^2 in integers

1/x, 1/y . 1/z form numbers of (Pythagorean triplet assuming them to be integers and if they are not: which is generally true my multiplying by sutable nnumber ) so they are in the ratio

1/x : 1/y : 1/z = 2pq : p^2-q^2 : p^2 + q^2 (p and q are rational)

(knowing that (2pq)^2 + (p^2-q^2)^2 = (p^2+q^2) and you can find it in number theory book

or x: y : z : = 1/(2pq) : 1/ (p^2 - q^2) : 1/(p^2 + q^2)

or x: y: z = (p^2+q^2) (p^2-q^2) : 2pq(p^2+q^2) : 2pq(p^2-q^2)

putting p = m/r , q = n/r ( m,.n,.r are integers)

x:y:z = (m^2+n^2)(m^2-n^2) /r^4 : 2mn(m^2+n^2)/r^4 : 2mn(m^2-n^2)/r^ 4
or (m^2+n^2)(m^2-n^2) : 2mn(m^2+n^2) : 2mn(m^2-n^2)/


in a trivial case 1/x : 1/y: 1/z = 1/3: 1/4: 1/5 = 20:15: 12

2011/108) prove that sum of 2 sides of an euclidian triangle is greater than 3rd side

let the angles be A B C and opposite sides be a b c

we need to prove that a + b > c

we have A+B = 180-C

hence sin (A+B) = sin (180-C) = sin C

or sin A cos B + cos A sin B = sin C

as cos B < 1 and cos A < 1 we have (it is 1 at zero degree not possible)



sin A cos B + cos A sin B < sin A + sin B so sin C < sin A + sin B

as a/sin A = b/sin B = c/sin C we get c < a + b

proved

2011/107) Prove that... 1/sin(pi/7)=1/sin(2pi/7)+1/sin(3pi/7)

start from RHS as it is more complex

1/ (sin 2pi/ 7) + 1/(sin 3pi/7)
= 1/ ( sin 5pi/7) + 1/( sin 3pi/7) as sin 2pi/7 = sin (pi-2pi/7) = sin 5pi/7 and we can add with out a faction of the form pi/14)

= ( sin 3pi/7 + sin 5pi/7) / ( sin 5pi/7 sin 3pi/7)
= (2 sin 4pi/7 cos pi/7)/ ( ( sin 5pi/7 sin 3pi/7)
as sin 4pi/7 = sin 3pi/7 and one is in numerator and another in denominator
= (2 sin 3pi/7 cos pi/7)/( sin 5pi/7 sin 3pi/7)
= (2 cos pi/7)/ ( sin 5pi/7)
= (2 cos pi/7)/( sin 2pi/7) as sin 5pi/7 = sin 2pi/7
= ( 2 cospi/7 )/ ( 2 sin pi/7 cos pi/7)
= 1/ sin pi/7 = LHS

2011/106) A natural number n is chosen strictly between two consecutive perfect squares.

The smaller of these two squares is obtained by subtracting k from n an the larger is obtained by adding l to n. Prove that n-kl is a perfect square.

Proof:

Let the number n be between a^2 and (a+1)^2

As per given condition

n- a^2 = k …1

(a+1)^2 –n = l ….2

Adding (1) and (2)

k + l = 2a + 1

or l = 2a + 1 - k

now

n- kl = (a^2+k) – k(2a+1-k)

= a^2 + k – 2ak –k + k^2 = a^2 – 2ak + k^2 = (a-k)^2

Hence proved

that n-kl is a perfect square.

2011/105) Prove a^4 + b^4 + c^4 >= abc(a+b+c)

this can be done in 2 steps

we know by AM GM inequality

a^4+b^4 > = 2 a^2b^2
b^4+c^4 >= 2 b^2 c^2
c^4 + a^4 >= 2 a^2^2

adding the 3 above and deviding by 2 we get

a^4+b^4+c ^4 >= a^2b^2+b^2c^2 + c^2 a^2 ...1

now we repeat the process of a^2b^2 , b^2 c^2 and c^2 a^2 to get as below

a^2 b^2 + b^2 c^2 > = 2 b^2ac
b^2c^2 + c^2 a^2 >= 2 c^2ab
c^2a^2 + a^2 b^2 >= 2 a^2bc

adding the above and deviding by 2 we get

a^2b^2 + b^2 c^2 + c^2 a^2 >= (b^2ac+c^2ab+a^2bc) or abc(b+c+a) ,.2

from (1) and (2) it follows

a^4 + b^4 + c^4 >= abc(a+b+c)

2011/104) If a=(root 2+1)^ 1/3 - (root 2 -1 )^1/3 then value of a ^ 3 +3a-2 is zero. explain

a = -(root 2+1)^ 1/3-(root 2 -1)^ 1/3

so a + (-(root 2+1)^ 1/3) +(root 2 -1)^ 1/3 = 0

using x+y+z= 0 => x^3 +y^3 + z^3 = 3xyz

we get

a^3 + (-(root 2+1)) + ((root 2-1)) = 3a((-(root 2+1))((root 2-1)

or a^3 - 2 = - 3a and hence a^3+3a - 2 = 0

2011/103) Find sufficient condition on a, b, c that the roots of x³ + ax² + bx + c = 0 are in arithmetic progression.

Let the roots be m-d, m and m+d

then as per vieta's formula http://en.wikipedia.org/wiki/Vi%C3%A8te%…
1)
sum of roots = m-d + m + m+d = - a or 3m = - a
2)
b= m(m-d) + m(m+d) + (m-d)(m+d) = 3m^3 - d^2 or d^2 = 3m^2 - b or (a^2)/3- b
3)
product of roots = m(m-d)(m+d) = m(m^2 - d^2) = - c

or (-a/3)((a/3)^2 - (a^2/3 -b) = -c

or(- a/3)(2a^2/9 - b) = - c

or (a/3) (b - 2a^2/9) = c

or (a/27) (9b - 2a^2) = c
above condition is necessary and sufficient