Q2/131) Find 2^(1/2) in ratio of integer form(approximate taht is p/q where p and q are integers)

Solution      

We know 2^(1/2) is irrational and let us assume2^(1/2)  = x/y

So we get x^2 = 2y^2 or x^2-2y^2 = 0

This does not have integer solution.

But if we chose x^2-2y^2= 1 then it has  integer solution x= 3 , y=2 is one of them

for each of the equations we can generate solutions with higher x and y for solution of either one of the equation and as we get x and y higher we get x/y as close to 2^(1/2)

for that we need 2 solution ( or 1 solution can be used 2 times) ( assuming x/y and a/b)

x^2-2y^2= 1 ..1 => (x+ 2^(1/2)y)(x- 2^(1/2)y) = 1  …1

a^2-2b^2= 1 ..1 => (a+ 2^(1/2)y)(x- 2^(1/2)y) = 1  …2

multiply (1) by (2) to get (x+ 2^(1/2)y)( (a+ 2^(1/2)b) (x - 2^(1/2)y)(a -2^(1/2)b) = 1

or ( ax + 2by + 2^(1/2)(bx + ay))( ax + 2by- 2^(1/2)(bx + ay)) = 1 .. 3

so if (x,y) and (a,b) are solution then (ax+2by, bx+ay) is a solution

we can chose (x,y) and(a,b) as different of (x,y) = (a,b) also giving a solution

now taking (x,y) = (a,b) = (3,2) we get

(ax+2by, bx+ay) = (17,12) is one solution : note that 17^2 – 2 * 12^2 =1 confirms the same

Taking (x,y) = (3,2) and (a,b) = (17,12) we get

(ax+2by, bx+ay) = (99, 70) is next solution : note that 99^2 – 2* 70^2 =  1 confirms the same

We can proceed to get solutions as (3,2), (17,12), (99,70) and so on and 3/2(= 1.5 ) , 17/12 ( = 1.416 ) and 99/70 (=1.414285 = correct upto 4 decimal places) are approximately 2^(1/2) closer and closer . Few more iterations shall give the values much closer.

Q12/130) What is the sum of the squares of the roots of the equation x^2 - 7[x] + 5 = 0?

7[x]>  =  7x

hence any solution must satisfy the quadratic equation
x^2 - 7x + 5 <= 0
So x is between 0.8 and 6.2.(approximate)
It follows that the possible values of [x] are 0 to 6. So we need to check for of [x] from 0 to 6 and find x to be in proper range

There are altogether 4 distinct solutions.

They are sqrt(2), sqrt(23), sqrt(30), sqrt(37)

The sum of their squares is 2 + 23 + 30 + 37 = 32+60 = 92

Q2/129) For how many 2 digit numbers the sum of digits is greater than the product of digits?

Solution

(x+y) > xy
or  xy - x - y < 0
x(y-1) < y

  y = 0 for all x so all multiples of 10 is a set of solution 

x < y/(y-1)

or x < 1+ 1/(y-1)
so x has to be 1

so multiples of 10 or any one of digits should be 1

that 10, 20, 30, 40,50,60,70,.80,90, 11,12,13,14,15,16,17,18,19, 21,31,41,51,61,71,81,91

Q2/128) Solve in positive integers (1+1/x) (1+ 1/y))(1 +1/z) = 2

Without loss of generality we can choose x <=y <=z and answer  shall be a permutation of if

 

Now x < 4 as (5/4)^3 < 2 ( it is 125/64)

x cannot be 1 as (1+1/y)(1+ 1/z) = 1 has no solution

so x = 2 or 3

if x = 2 we get

3/2(y+1)(z+1) = 2yz

Or 3(y+1)(z+1) = 4 yz

Or yz – 3y – 3z = 3

(y-3)(z-3) = 12 by adding 9 on both sides

( y-3)(z-3) = 1 * 12 or 2 * 6 or 3 * 4

So (x,y,z ) = (2, 4, 15) , ( 2, 5, 9) , (2,6,7) are the 3 sets of solutions

if x = 3 we get

4/3(y+1)(z+1) = 2yz

Or 2(y+1)(z+1) = 3 yz

Or yz – 2y – 2z = 2

(y-2)(z-2) = 6

( y-2)(z-2) = 1 * 6 or 2 * 3

So (x,y,z ) = (3,3,8) , ( 3,4,5)

Q2/127) If a, b, c are different and the equations ax+a^2y+(a^3+1)=0 bx+b^2y+(b^3+1)=0 cx+c^2y+(c^3+1)=0 are consistent, prove that abc+1=0...

Let us have a function f(t) = xt+yt^2+(t^3+1)

We have f(a) = 0 , f(b) = 0, f(c) = 0

And f(t) = t^3 + yt^2 + xt + 1 …1

So a,b, c are zeroes of f(t) which is a cubic polynomial

So f(t) = m(t-a)(t-b)(t-c) …2

Comparing coefficient of t^3 from (1) and (2) m = 1 

Now constant term = -mabc = 1 or –abc = 1 or abc + 1 = 0

Proved

Q2/126) When does a/b repeat when it terminates

Reduce a/b to the lowest form p/q such that p and q are co-primes.

Now q is of the form 2^m5^nc

We can convert  power of 2 and 5 to power of 10 by multiplying properly so the denominator can be converted to 10^r c

If c is 1  then it terminates.

If c is not 1 then c is coprime to 10 and the remainder and hence the quotient repeats. 

Q12/125) Assume x and y are integers, such that (x^2+1)=2y. Now prove that y is the sum of squares of two integers?

x has to be odd so let x= 2m + 1

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

(x^2+1)/2 = y = 2m^2 + 2m + 1 = m^2 + (m+1)^2

proved

Q2/124) Show that in a sequences of 3 numbers one number is always divisible by 3

This can be proved by pigeon hole principle. As there are 3 numbers there are 3 remainders and the remainders can be 0 or 1 or 2 and no 2 remainder can be same if they where same then difference is divsible by 3 but difference cannot be be >2 so this is not possible.

So one of the remainders has to be zero.

Hence proved.

However this can be proved as below

Add 3 to all the terms starting from 1^st number and keep adding. Then we get all the terms to the right. Subtract 3 from 3^rd term and keep subtracting. We get all the numbers and none of them is divisible by 3. which is a contradiction. So one of the numbers has to be divisible by 3.

Q2/123) Find integer sided isosceles triangle whose area is integer

We have Pythagorean triplet that is integer length sides for a right angled triangle with hypotenuse c as

a= u^2-v^2
b= 2uv
and c = (u^2+v^2)

Now it may be noted that as b is even so ab/2 is integer.

Now if we double any of the base and hypotenuse as the two other sides we have area is integer

So the 3 sides are (u^2+v^2, u^2+v^2, 4uv) or (u^2+v^2, u^2+v^2, 2u^2-2v^2)

Q2/122) : find a , b, c,d such that 1/a = 1/b + 1/c + 1/d

We know that

1/n = 1/(n+1) + 1/(n(n+1))  .. (1)

Puting n = a we get

1/a = 1/(a+1) + 1/(a(a+1))

Now putting n = a(a+1)

We get

1/(a(a+1)) = 1/(a(a+1) + 1) + 1/((a(a+1))(a(a+1) + 1) =

= 1/(a^2 + a + 1) + 1/(a(a+1)(a^2+a+1))

Hence b= a+ 1

c= a^2 + a + 1 and d = abc is the solution

for example if a= 2 we have

1/2 = 1/3 + 1/6 ... using 1

1/6 = 1/7 + 1/42 using 1

hence 1/2 = 1/3 + 1/7+ 1/42 ( a= 2 , b= a+1 = 3, c= a^2 + a + 1 = 7 and d= abc = 42)

we can proceed indefinitely to any number of reciprocals. and for any starting reciprocal as well

Q2/121) The set of all integers n for which sqrt(n^2 + n) is an integer is.

a) the set {0,-1};
b) a finite set with at least three elements;
c) an infinite set;
d) none of these sets;

Proof:
sqrt(n^2 + n) = sqrt(n(n+1)) is inetger

n and n+1 are coprimes so either n= 0 or n = - 1 or n and n+1 both squares

n = x^2 and n+1 = y^2 => 1 = (x+y)(y-x) => x+y = 1 and y-x = 1 => x = 0 y =1 => n = 0

or x+y = -1 and y-x = -1 => x = 0 y =-1 => n = 0

so only solution 0 and -1 so ans is a)

Q12/120) Let a,b and c be the sides of a right angled triangle. Let theta be the smallest angle of this triangle.?

f 1/a, 1/b, 1/c are also the sides of a right angled triangle then show that Sin(theta) = (sqrt(5) - 1)/2;

Proof:
Let a < b < c


And theta (say t opposite to smaller side)

Sin t = a/c and cos t = b/c

Now 1/a > 1/b > 1/c and as it is right angled triangle we have

(1/a)^2 = (1/b)^2 + (1/c)^2

Or

(c/a)^2 = (c/b)^2 +1

(1/sin ^2 t) = (1/ cos^2 t ) +1

Or cos^2 t = sin ^2 t + sin ^2 t cos^2 t

Or 1- sin ^2t = sin ^2 t + sin ^2 t (1- sin ^2 t)
= 2 sin ^2 t – sin ^4 t
Or sin ^4 t – 3 sin ^2 t + 1 = 0

Sin^2 t = (3 +/- sqrt(9-4))/2 = (3 – sqrt(5))/2 , + cannot be taken as it shall be >1 not possible


So sin t = sqrt((3 – sqrt(5))/2) say sqrt(x) – sqrt(y) as there is sqrt(5) in square

So squaring we get

x+y – 2sqrt(ab) = (3 –sqrt(5))/2

so x+y = 3/2 and sqrt(xy) = sqrt(5)/4) or xy = 5/16

we can solve it as x = 5/4 and y= 1/4 or y = 1/4 and x = 5/4

so sin t = +/-(sqrt(5/4) - sqrt(1/4))

positive value to be taken as sint > 0 so sin t = sqrt(5/4) – sqrt(1/4) = (sqrt(5)-1)/2
 

Q12/119) Let a < b < c < d be four real numbers, such that all six pairwise sums ...? are distinct. The values of the smallest four pairwise sums are 1, 2, 3, and 4 respectively. What are the possible values of d?

Then the six pairwise sums

a+b, a+c, a+d, b+c, b+d, c+d

are all distinct.
there are 2 possibilities
case 1:
a+d < b+ c

Now, a+b = 1, a+c = 2, a+d = 3, b+c = 4.

∴ (a+c) - (a+b) = 2-1

∴ c-b = 1

∴ (b+c) + (c-b) = 4+1 ∴ 2c = 5 ∴ c = 5/2

∴ a+c = 2 gives a = 2-c = 2-(5/2) = -1/2

∴ a+d = 3 gives d = 3-a = 3-(-1/2) = 7/2

∴ d = 7/2. ........ Ans for case 1.
 case2 : b+ c< a + d
then we get a+b = 1, a+c = 2, b+c =3 a +d = 4 => 2a + b + c = 3 => a = 0 b = 1 c = 2 and d = 4

so d =4 is another solution for case 2

so d can be 7/2 or 4

Q12/118) Given 4 positive integers a,b,c and d such that a^5=b^4, c^3=d^2 and c−a=19, what is d−b?

as c^3 = d^2 so c will be a square let c = x^2

as a^5 = b^4 so a = y^4 for some y

now
c-a = 19
=> x^2 - y^4 = 19
=> (x-y^2)(x+y^2) = 19
hence x - y^2 =1 and x+y^2 = 19 as 19 is a prime
so x = 10 and y = 3

so a = y^4 or a^5 = y^20 = b^ 4 or b= y^5 = 243

c= x^2 => c= 100 and hence d^2 = 10^6 and so d = 1000

d-b = 1000 - 243 = 757

Q12/117 )100 positive integers are written in a row. The average of the first and second numbers is 1. The average of the second and third numbers is 2. The average of the third and fourth numbers is 3. This pattern continues, and the average of the 99th and 100th number is 99. What is the 100th number?

Average if 1st and 2nd number is 1 so sum is 2 so both positive integers have to be 1

so a1= 1
a2 = 1

now average of 2nd and 3rd is 2 so sum = 4 so a3= 3

average of 3rd and 4th is 3 so sum = 6 so a4= 3

average of 4th and 5th is 4 so sum = 6 so a5= 5

so we have nth term is n when n is odd and n-1 when n is even

this can be proved as below

for n odd 2 numbers are n and n and average = n
for n even the number is n-1 and next number is n+1 and average is n

so 100th number = 99

The integer 2006 is the product of three prime numbers p, q, and r. If p + q = r^2 (p + r), what is the value of q?

2006 = 2•17•59 (prime factorization)
so p q rare permutions of 2,17,59
now r^2 divides p+q so 17 and 59 are too large for r so r= 2
so p+q = 4 p + 8
or 3p = q - 8
so p < q-8 => so p=17 and 3p = 51 and q = 59 satisfying the condition
hence q = 59

What is the nth term that goes 1,2,2,3,3,3,4,4,4,4, etc

as n occurs n times

the last position of n = n(n+1)/2

last position of (n-1) = n(n-1)/2

so the position n(n-1)/2 + 1 to n(n+1)/2 contains n

kth position contains n and

n(n-1)/2 < k <= n(n+1)/2

or n(n-1) < 2k <= n(n+1)

n^2 - n < 2k <= n^2 + n

multiply by 4

4n^2 - 4n < 8k <= 4n^2 + 4n
(2n-1) ^2 - 1 < 8k <= (2n +1)^2 - 1
(2n-1) ^2 < 8k +1 <= (2n +1)^2

so n = (ceil(sqrt(8k+1) -1)/2

find last 3 digit of sum (9^k) k = 1 to 400

the sum is S = [1<= n <= 400] 9^n

we have 1000 = 8 * 125 so to compute S mod 8 and S Mod 125

now 9 = 1 mod 8 so 9^n = 1 mod 8

there are 400 numbers each is 1 mod 8 so sum = 0 mod 8

sum = 9/8(9^400-1)

as 9 and 125 are coprime as per http://www.cut-the-knot.org/blue/Euler.s…

9^φ(125) mod 125 = 1

now 125 = 5^3 so φ(125) = 125(1-1/5) = 100

so 9^(100) = 1 mod 125
or 9^400 = 1 mod 125
so 9/8(9^400-1) = 0 mod 125

now S = 0 mod 8 and S = 0 mod 125 and

hence S = 0 mod 1000 so last 3 digits are zero

Prove that average of the numbers : 2 sin2° , 4 sin4° , 6 sin6° ..... , 180 sin 180° = cot 1°

we have cos (n-1) - cos (n+1) = 2 sin n sin 1 ,,,1

now 180 sin 180 = 180 sin 0
178 sin 178 = 180 sin 2

so total sum = 180 ( sin 0 + sin 2 + sin 4 + ... + sin 88) + 90 sin 90
= 90 ( 2 sin 0 + 2 sin 2 + 2 sin 4 + 2 sin 88) + 90 sin 90

as there are 90 numbers average
= 2 sin 2 + 2 sin 4 + 2 sin 88 + sin 90 as
= ( cos 1 - cos 3)/ sin 1 + (cos 3 - cos 5)/ sin 1 + ( cos 87- cos 89 / sin 1 + 1
= ( cos 1 - cos 89)/ sin 1 + 1
= cot 1 - sin 1/sin 1 + 1 as cos 89 = sin 1
= cot 1 - 1 + 1
= cot 1

Given log(base a)x*log(base b)x + log(base b)x*log(base c)x +log(base c)x*log(base a)x = log(base a)x*log(base b)x*log(base c)x. prove that x = abc

devide both sides by

log(base a)x*log(base b)x*log(base c)x.
to get 1/ log(base c)x + 1/log(base a)x + log(base b)x = 1

now 1/ log(base c)x = log (base x) c

so we get log (base x) c + log (base x) a + log (base x) c = 1

or aply the product rule to get

log (base x) abc = 1

so abc = x^1 = x

proved

solve 2sin(10º)sin(20º + θ) = sin(θ)

we have
2 sin(10º)sin(20º + θ) = 2 sin(10º)( sin20º cos θ+ cos 20º sin θ ) = sin θ

or 2 cos θ sin 10º sin20º
=  sin θ( 1- 2 sin 10º cos 20º)
= 2 sin θ( sin30º - sin 10º cos 20º)
= 2  sin θ(sin(20º + 10º) -  sin 10º cos 20º)
= 2  sin θ(sin 20º cos 10º)
= >  tan θ = tan 10º or θ = 10º

prove that 2^n > n^3 for n > 10

Basic Step:
When n = 10, the inequality 2^n > n^3 is true as 2^10 = 1024 > 1000

Inductive Step:
Assume n = k is true, where k ≥ 10, then
2^k > k^3 ................ (1)

we have for k >= 10

k^3 > 10k^2 > 4k^2 or  3k^2 + k^2 > 3k^2 + 3k + 1

So, from (1)
2(2^k)> 2(k^3)
=>2(2^k) > k^3 + k^3
=>2^(k+1) > k^3 + 3k^2 + 3k + 1
=>2^(k+1) > (k + 1)^3

So n = k + 1 is also true

so induction step is proved and hence proved

 

Starting at the origin, a beam of light hits a mirror (in the form of a straight line) at point A = (4, 8) and is reflected to point B = (8, 12). Compute the exact slope of the mirror.

Slope of the incident ray = (8-0)/(4-0) = 2 ( angle A)

Slope of the reflected ray = (12-8)/(8-4) = 1( angle B)

Now let us find the line that bisects the 2 rays

Tan (A+B) = (tan A + tan B)/(1- Tan A tan B) = (1+2)/(1-2) = - 3

Let C be the angle bisector between A and B

A + B = 2C say m is slope of the bisector

So we get 2m/(1-m^2) = - 3

Or 2m = -3 + 3m^2
Or 3m^2 – 2m – 3 = 0
=> m = (1/6) [2 + √(4 + 36)] ... [m positive taken to the correct slope
=> m = (1/3) [1 + √(10)].

this is the slope of the mirror

The reason for the slope to be positive is that from 4,8 one line goes to (8,12) that is line is pointing towards 1st quadrant and (0,0) is to the direction of 3rd quadrant. so the tangent should have a slope between 1 and 2 and not the normal

Resolve the following expression into partial fraction: X^2/(x+1)^3

This type of problem can be done as

Let x^2/(x + 1)^3 = A/(x + 1) + B/(x + 1)^2 + C/(x + 1)^3
=> x^2 = A(x + 1)^2 + B(x + 1) + C
x = - 1 => C = 1

Comparing coefficients of x^2 and x,
A = 1 and 2A + B = 0 => B = - 2A = - 2
=> x^2/(x + 1)^3 = 1/(x + 1) - 2/(x + 1)^2 + 1/(x + 1)^3.

Which is correct and nothing is wrong in it. This is conventional approach.

However seeing that (x+1)^3 in the denominator it can be done by putting

x+ 1 = t or x= (t-1)
we get x^2/(x+1)^3 = (t-1)^2/t^3 = (t^2 - 2t + 1)/t^3 = 1/t - 2/t^2 + 1/t^3 

             = 1/(x+1) - 2/(x+1)^2 + 1/(x+1)^3

which reduces the number of steps.  

find the smallest positive integer value of n for which {(1+i)^n}/{(1-i)^(n-2)} is a real number ?

((1+i)/(1-i)) = (1+i)^2/2

so ((1+i)/(1-i))^n = (1+i)^(2n)/2^n)

so {(1+i)^n}/{(1-i)^(n-2)} = (1+i)^(2n)/(2^n) (1-i)^2
= ((1+i)^n(1-i))^2 /(2^n)
= (1+i)^(n-1)/2^(n-1)

so (1+i)^(n-1) must be real or n-1 = 0 (by inspection) or n = 1 is the lowest n

if one needs to solve for all n the 1+ i makes 45 degrees with x axis and n = 4t is the solution ( t integer)

solve x^2 - y^2 = 2764 for integer x and y > 0

x^2 - y^2 = (x + y)(x - y) = 2764
 so both (x + y) and (x - y) are factors of 2764.

x+ y and x- y both are even or odd ( as one even and on odd shall  give fractions)

 as product is even so x + y and x-y both are even

Now factorize 2764 into product of primes: 2764 = 4*691 = 2*2*691

so x+ y= 1382 and x- y =2 => x = 692 and y =690 is the only solution

show that x = cos(π/10) is a root of the equation 16x^4 - 20x^2 + 5 = 0.

proof:

let y =  π/10

we have

 cos5y = 16cos^5 y - 20cos^3 y + 5cosy

or 0 =  16cos^5 y - 20cos^3 y + 5cosy

= (16cos^4 y - 20cos^2 y + 5)cos y

as cos  y is not zero 

 so (16cos^4 y - 20cos^2 y + 5) = 0

hence x =  cos(π/10) is a root

Prove that (sinA+sinB)(sinB+sinC)(sinC+sin A) > sin A sinB sinC in a triangle ABC

we have in a triangle a + b > c

so sin A + sin B > sin C ..1

as a/ sin A = b/ sin B = c/ sin C

similarly

( sin B + sin C) > sin A ..2

sin C + sin A > sin B ...3

by multiplying (1) (2) and (3) and knowing that both sides are positive ( sin is positive for any angle of a triangle)

we get the result

(sinA+sinB)(sinB+sinC)(sinC+sinA)>sinA sinB sinC

proved

solve 3 * 4^x - 6^x = 2 * 9^x

we have 4^x = 2 ^ 2x
6^x = 2^x 3^ x
9^x = 3^2x

let 2^x = a and 3^x = b ( we have a and b are not independent of each other) ..1

3a^2 - ab = 2b^2

or 3a^2 - ab - 2b^2 = 0

or (3a + 2b)(a-b) = 0

a = b or (a/b) =1 = (2/3)^x ( from 1) so x = 0

or 3a = - 2b which is not possible as a and b both >0 from (1)

so x = 0

show that for any real a,b both not zero a^2+ab+ b^2 > 0

proof:
      a^2+ab+b^2 = (a^3-b^3)/(a-b)

without loss of generality let a <= b ( as the expression is symetric)

if a < b

then a^3 < b^3 and both a^3-b^3 and a-b are -ve and so ratio is positive

if a = b then a^2 + ab + b^2 = 3a^2 and is zero only if a = 0 that is a=b = 0

proved

solve arcsin(x)+arctan(x) = 0

we note that for x > 0 arc sin x and arctan x both are > 0
for x < 0 arcsin x and arctan x both are < 0

and  at x = 0 both are zero so x = 0 is the solution

Suppose two integers, m and n differ by 1. Suppose also that x, y, and a are integers such that x + a = m^2 and y + a = n^2. Show that xy + a is a perfect square.

without loss of generality as it is symmetric let m = n+ 1 
x=(n+1)^2-a ;
y=n^2 -a

xy + a = ( (n+1)^2 - a)( n^2 - a) +a
= (n^2)(n+1)^2 - (2) (n)(n+1) (a) + a^2
= ((n)(n+1) - a)^2

therefore xy+a is a perfect square.

What is the smallest symmetrical number greater than 56,789 which is exactly divisible by 7?

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

How many 5-digit numbers divisible by 11 are there containing each of the digits 2,3,4,5,6?

The sum of digits = 2 + 3+ 4 + 5 + 6 = 20

now for the number to be divisible by 11 the sum of digits at odd place should be same as sum of digits at even place or difference multiple of 11

say o is sum od digits at odd place and e sum of digits at even place.

o + e = 20
o- e = 11n and |o - e| < 20 as o+ e = 20

o-e should be even hence should be 0

so e = o = 10
sum of 2 digits ( even placed) and 4 + 6 (only combination)

so 4 and 6 can go to even place

2,3,5 should go to odd place

even position digits can be permuted( 2 ways)  among them selves and odd placed as well ( 6 ways)

so number of numbers = 2 * 6 = 12

If (x+ 1/x)^2 = 3 then the value of x^206 + x^200 + x^90 + x^84 + x^18 + x^12 + x^6 + 1

x^206 + x^200 + x^90 + x^84 + x^18 + x^12 + x^6 + 1

= x^200(x^6+1) + x^84(x^6+1) + x^12(x^6+1) + (x^6+1)

= (x^6+1)(x^200 + x^ 84+ x^12 + 1)

x^2 + 1/x^2 + 2 = 3
so x^2 + 1/x^2 = 1
or x^4 + 1 = x^2
or x^4 - x^2 + 1 = 0
or x ^6 + 1 = 0 ( multiplying by x^2 + 1)

hence given sum = 0          

prove that (3)*5^(2n+1)+2^(3n+1) is divisible by 17

= 3 * 5 * 5^(2n) + 2 * (2^3)^n mod 17
= 15 * 25^n + 2 * 8^n mod 17
= 1 5 * 8^n + 2 * 8^n mod 17 as 25 = 8 mod 17
= 17 * 8^n

hence divisible by 17

calculate 2008 ! mod 2011

we see that 2011 is a prime number

so as per wilson theorem http://en.wikipedia.org/wiki/Wilson%27s_…

2010! mod 2011 = - 1

as 2010 mod 2011 = - 1 so 2009! mod 2011 = 1

so 2008 ! mod 2011 = inverse(2009) mod 2011 = - (inverse 2 mod 2011)

you can compute inverse using extended euclid algorithm http://en.wikipedia.org/wiki/Extended_Eu…

but there is a short cut.

as 2 * 1006 = 1 mod 2011 so inverse 2 is 1006
so 2008 ! mod 2011 = - (inverse 2 mod 2011) = - 1006 mod 2011 = 1005

What is the remainder of (x^100 - 4x^98 + 5x + 6) divided by (x³ - 2x² - x + 2)?

The remainder shall be quadratic say f(x) = ax^2 + bx + c

Now (x³ - 2x² - x + 2) = (x-2)(x-1)(x+1)
Let g(x) = (x^100 - 4x^98 + 5x + 6)- (ax^2 + bx + c)
g(1) , g(2), g(-1) all are 0
g(1) = 1- 4 + 5 + 6 – (a + b + c) = 0 or a + b+ c = 8 ..1
g(-1) = 1 – 4 – 5 + 6 – (a-b+ c) = 0 or a-b+c = -2 ..2
g(2) = 2^100- 2^100 + 10 + 6 – (4a + 2b+c) = 0 or 4a + 2b + c = 16 ..3

subtract (2) from (1) 2b = 10 or b = 5
so a + c = 3 and 4a + c = 6=> a = 1 and c = 2

remainder = x^2 + 5x + 2

show that p^4+p^2−2 is a multiple of 72 for prime p > 3

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

if p is a prime > 3 then 3 does not devide p so p+1 or p-1 is divisible by 3 and so p^2-1 is divisible by 3
as p^2 + 2 = p^2-1 + 3 so p^2 + 2 is divisible by 3

so (p^2 - 1)(p^2 + 2) is divisible by 9

now as p + 1 and p-1 are 2 consecutive even numbers one of them is divsible by 4 and other by 2 so product is diviisble by 8

so the number is divisible by 72

Is the sum of positive numbers < 10^6 close to even square more or less than sum of numbers close to odd square ?

A number is near even square if it is closer to even square than odd square else if is close to odd square. 

Let us take the numbers from (2n)^2 upto (2n+2)^2-1 that is from even square to the number 1 less than next even squares  that is from (2n)^2 upto (2n+2)^2-1.

That is from 4n^2 to 4n^4 + 8n + 3

Average of numbers = (8n^2 + 8n+ 3)/2

Total number of numbers = (8n+4)

Sum of numbers = (4n+2)(8n^2+8n + 3) ..1

Nuw numbers near odd square (2n+1)^2 (or 4n^2 + 4n + 1) are from 4n^2+ 2n+1 to 4n^2 + 6n + 2

Average of numbers = 2

Number of numbers = 4n+ 2

So sum of numbers near odd square = (2n+1)( (8n^2+ 8n + 3) ..2

Which is ½ of sum of numbers

So sum of numbers near odd square = sum of numbers near even square

taking the ranges from 0 to 2^2-1, 2^2 to 4^2- 1 so on upto 998^2 to 1000^2-1 we can each odd sum is same as even sum and hence sum of close to odd square is same as sum of close to even square.

2arctan(sqrtx)-arcsin((x-1)/(x+1)) is equal to pi/2?

it is defined for x >= 0

now let y = arctan(sqrtx)

cos 2arctan(sqrtx)
= cos (2y)
= (cos^2 y - sin ^2 y)/(cos^2 y + sin ^2 y)
= (cos^2 y - sin ^2 y)/(cos^2 y + sin ^2 y) as denominator is 1
= ( 1- tan ^2 y)/(1+ tan ^2 y)
= (1-tan ^2 y)/(1+tan ^2 y)
= (1-x)/(x+1)

so 2arctan(sqrtx)-arcsin((x-1)/(x+1)
= arc cos((1-x)/(x+1)) - arcsin((x-1)/(x+1))
= arc cos((1-x)/(x+1)) + arcsin((1-x)/(x+1)) as arcsin t = - arcsin - t
= pi/2 as arccos t + arcsin t = pi/2

(if t >= 0 both in 1st quadrant and sum is between 0 and pi)
if ( t < 0 then arc cos is between pi/2 and pi and arc sin < 0) so sum between 0 and pi) this is mentioned for fixing the range

α & β are roots of ax^2+bx+c=0 a,b,c real constants & λ is a constant find equation with roots α+ λ and β+ λ?

α & β are roots of f(x) = ax^2+bx+c=0

so α+ λ and β+ λ are root of f(x - λ)

the reason is x = α+ λ => x - λ = α and x = β + λ => x - λ = β

f(x) = ax^2+bx+c=0
=> f(x - λ) -= a (x - λ)^2 + b(x - λ) + c = 0

or ax^2 + x(b- 2a λ) + (c- bλ+ aλ^2) = 0

find x such that 3^x-x^2 is divisible by 5?

as 3 and 5 are coprimes

so 3^4 = 1 mod 5 as per Fermats little theorem

now 3^(4k+1) = 3 mod 5
3^(4k+2) = 4 mod 5
3^(4k+3) = 2 mod 5
3^(4k) = 1 mod 5

now (5k+m)^2 mod 5 = 0 if m= 0, 1 if m= 1 or 4, 4 if m = 2 or 3 mod 5

now 3^x= x^2 mod 5 if

x =  0 mod 4 and 1 mod 5 ( in both cases remainder 1)
or x = 0 mod 4 and 1 mod 5( in both cases remainder 1)
or x = 2 mod 4 and 2 or 3 mod 5

x = 0 mod 4 and 1 mod 5 => x= 16 mod 20 (checking 0,4,8,12,16)
or x = 0 mod 4 mod 5 => x = 4 mod 20 (checking 0,4,8,12,16)
x =2 mod 5 and 2 mod 4 => x = 2 mod 20 as remainders are 2
or x = 2 mod 4 and 3 mod 5 => x= 18 mod 20 (checking 2,6,10,14,18)

 so solutions x = 2 mod 20, 4 mod 20, 18 mod 20, 16 mod 20

This could be solved by using Chinese remainder theorem but bing smapll number I checked for direct

There exists positive integers x,y such that both the expressions (3x+2y) and (4x-3y) are exactly divisibe by ?

Options a) 11 b) 7 c) 23 d) 17

solution
Let us eliminate y from a combination of the two

3(3x+2y) + 2( 4x -3y) = 17x

as sum of combination is divisible by 17 so ans is (d) 17 provided one of them is divisible by 17

this can be shown if we have x and y such that 3x+ 2y is divisible by 17

x= 3 , y = 4 satisfies it so (d) is solution

if tanh (x/2) = tan(x/2) then show that cosh x cosx = 1.

cos (x) = (1- tan ^2 (x/2))/(1+ tan ^2 (x/2))
= (1- tanh ^2(x/2))/( 1+ tanh^2 (x/2)) as tan (x/2) = tanh (x2)
= ( cosh^(x/2) - sinh^2 (x/2))/( cosh^(x/2) + sinh^2 (x/2))
= 1/ cosh x

so cosh x cosx = 1.

proved

A^2 - bc = 3 b^2 - ca = 4 c^2 - ab = 5 then value of a+b+c = ?

options : 0 1/2 1 -1?

solution
a^2 - bc = 3 ..1
b^2 - ca = 4 ...2
c^2 - ab = 5 ..3

subtract (2) from (1)

a^2-b^2 +c (a-b) = -1

or (a+b+c)(a-b) = - 1

subtract (3) from (2)

(a+b+c)(b-c) = - 1

so a- b = b-c or a,b,c are in AP

now let a= b- t and c= b+ t
from 2nd equation

b^2 - (b^2 - t^2) = 4

or t = +/- 2

so a-b = 2 or -2

so (a+b+c) = 1/2 or - 1/2

as 1/2 is a choice and - 1/2 is not a choice so ans is 1/2

this is subject to the condition that equations are consistent.

else we need to evaluate a,b,c and then the sum

Show that all numbers of the form 12008, 120308, 1203308, 12033308, … are divisible by 19

Let the nth number be

100x + 8

the (n+1)st number be 1000x + 308

the difference is 900x + 300 = 300(3x + 1)

if 100x + 8 = 0 mod 19

then 5x + 8 = 0 mod 19
20x + 32 = 0 mod 19
or x + 13 = 0 mod 19
or 3x + 39 = 0 mod 19
or 3x +1 = 0 mod 19
so difference is divisible by 19 if nth number is divisible by 19

1st number is divisible by 19 and hence the difference and and so the second one

using the logic above we see that next one and so so
I have proved by mathematical induction

Solve polynomial equation when sum of 2 roots is equal to sum of the other 2 roots?

x^4 - 8x^3 - 14x^2 + 8x -15

let 4 roots be a,b,c,d

sum of 4 roots = 8

sum of 2 roots say ( a+ b) = (c+d) so (a+b)= (c+d) = 4

(x-a)(x-b)(x-c)(x-d) = 0

(x^2 - (a+b) x + ab)(x^2 - (c+d)x + cd) = 0

or (x^2 - 4x + ab)(x^2- 4x+ cd) = 0

say ab = m and cd= n

we get (x^2-4x + m)(x^2 - 4x + n)
= x^4 - 8x^3 + x^2(m + n + 16) - 4x(m+n) + mn

so mn = - 15, m+ n = - 2 , m+n+16 = - 14 (from coefficient)

so equations are consistent

solving we get m = - 5, n = 3

we get (x^4- 4x - 5)(x^2 - 4x + 3)

= (x-5)(x+1) (x-3)(x-1)

What are the two natural no. whose difference is 66 and the least common multiple is 360?

Let the GCD be x

then 2 numbers are mx,nx: with out loss of generality let m > n

now LCM = mnx ( as m and n are copimes)

so difference = (m-n) x = 66

mnx = 360

now 11 does not devide x so 11 devides (m-n)

m-n = 11 => x = 6 so mn = 60

m-n = 11
mn = 60

can be solved to give m = 15, n= 4 so numbers are 90 and 24


if m-n = 22 then x= 3 and

so mn = 120 does not have integer solution

m-n = 33 => x = 2 so mn = 180 does not have integer solution

m-n = 66 => x = 1 so mn = 360 does not have integer solution

so only solution (90,24)

How many numbers have exactly 15 divisors till 10000

a number of the form p1^a1 * p2^a2 * p3^a3 ... (p1 , p2, p3 etc are distinct primes ) has number of factors

(a1+1)(a2+1)(a3+1) ....

now we need to factor 15 in as many ways we can ( 15 * 1 or 3 * 5)

15 *1 -> a1 = 14

but 2^14 is 16384 > 10000 so no solution

now case 3 * 5

so a1 = 2, a2 = 4

so number n is of the form = p1^4 p2^2 where p1 and p2 are primes and p1 is not p2 ( it is convenient to keep power in decreasing order so to have lesser computations

p1^2 p2^4 < = 10^ 4

p1 = 2 => p2^2 <= 625 or p2 < 25 that is p2 = 3, 5 , 7, 11,13, 17,19, 23

p1 =3 => p2^2 < 123 or p2 <= 11 that is p2 =2, 5,7, 11
p1 = 5 => p2^2 < 16 or p2 = 2 or 3

p1 = 7=> p^2 <= 4 or p2 = 2

p1 = 11 => p1^4 > 10^4 not possible

so the numbers are

2^4q^2 where q in 3, 5 , 7, 11,13, 17,19, 23 ( 8 numbers)
or 3^4 q^2 where q in 2. 5,7, 11( 4 numbers)
5^4 * 2^2 or 5^2 * 3^2 or 7^4 *2^2 ( 3 numbers) that is 15 numbers

Let a1,a2....,an be a random permutation of [1,2,...n] where n is an odd natural number.

Show that the product

P = (a₁-1)*(a₂-2)*...*(a_n-n)

is an even integer.

proof:

we know (a1+ ..+ an) = ( 1+ 2...+n) as LHS is a pemuation of 1 to n

so (a1-1)+ (a2- 2) + ... + ( an-n)= 0

now there are odd number of terms on the left and all the terms on left cannot be odd as sum is 0 that is even

so one the terms has to be even

hence product is even. Done

taken from puzzle-a-day.blogspot.in/2010/10/even-product.html

If the function ax^2 + bx + c has a minimum value of -5 when x = -1 , and 0 when x = -2, find the values of a, b and c.

It has minimum at x = -1

so it is of the form

f(x) = m(x+1)^2 + n


x = - 1 gives n = - 5


so f(x) = m(x+1)^2 - 5

at x = -2 it is zero so 0 = m- 5 => m= 5

so f(x) = 5 (x+1)^2 - 5 = 5x^2 + 10x

so a= 5 and b= 10 and c = 0

Prove the identity (1+cosA-sinA)/(1+cosA+sinA )= sec A - tan A

we have
(1- sin A)/cos A = (1- sin A)(1+ sin A)/(cos A (1+sin A)) = cos^2 A/(cos A(1+ sin A))
= cos A/(1+ sin A)

so
(1- sin A)/cos A = cos A/(1+ sin A)

now if a/b = c/d then a/b = (a+c)/(b+d)

so (1- sin A)/cos A = cos A/(1+ sin A) = (1- sin A + cos A)/(1+ sin A + cos A)

or (1- sin A + cos A)/(1+ sin A + cos A) = (1- sin A)/ cos A = sec A - tan A

Find the smallest number that is made up of each of the digits 1 through 9 exactly once and is divisible by 99.

To determine if a number is divisible by 99 it needs to be divisible by 9 and 11, both of which can be done easily.

Sum of digits is 1+2+3+4+5+6+7+8+9 = 45 which is divisible by 9.

So any permutation of the digits which give a new number shall be divisible by 9.

We need to construct a permutation of 123456789 which is divisible by 11 and it should give the smallest number.

Let sum of digits in odd position be O and sum of the digits in even position be E

O + E = 45

O- E should be odd and multiple of 11

O-E = 33 => E =6 and sum of 4 digits cannot be 6 (it is minimum 10)

O-E = 11 => E = 17

O - E = - 11 => E = 28

O-E = - 33 => E = 39 is not possible as it is maximum 6+7+8+9 = 30

So let us consider 2 cases

E= 17 and E= 28

E= 17 consider 1^st

It can be seen as sum of 4 numbers (< 10 that is digits) and one can be even and 3 odd. Or 3 even and 1 odd.

If the digit is even then this does not get swapped with a digit in odd position and if is odd then it gets swapped. So to keep the number as small as possible the odd digit should be highest and even digits lowest hence one odd digit

So we get the 4 digits 17= 2 + 4 + 6 + 5 => this shall cause odd digit 5 to be swapped with 8

(If we chose 2 + 4 + 3 + 8 then digit 3 gets swapped with 6 and we get a larger number)

So even position digits 2,4,5,6 (sort it to get lower number)

Odd position digits 1,3,7,8,9

And hence number 123475869

If we consider 28 the digits shall be (4789) and 2 goes to odd position and and we get a larger number compared to 123475869 as a larger digit comes there

So solution is number 123475869

This question I picked from http://trickofmind.com/?p=154#comments where you can find some discussions

3sinA + 5cosA = 5 then 3cosA - 5sinA = ?

( a sin A + b cos A )^2 + ( b sin A - a cos A) ^2

= a^2 sin ^2 A + b^2 cos^2 A + 2ab sin A cos A + b^2 sin ^2 A + a^2 cos ^2 A - 2ab cos A sin A
= a^2 + b^2

so
(3sinA + 5cosA)^2 + (5 sin A - 3 cos A)^2 = 3^2 + 5^2

so 5^2 + (5 sin A - 3 cos A)^2 = 3^2 + 5^2

so (5 sin A - 3 cos A) = +/- 3

or 3 cos A - 5 sin A = +/- 3

Let x and y be positive real numbers. Prove that x^7 + y^7 >= x^4*y^3 + x^3*y^4

x^7 + y^7 - (x^4*y^3 + x^3*y^4)
= x^4(x^3-y^3) - y^4(x^3 - y^3)
= (x^4-y^4)(x^3-y^3)
= (x^2-y^2)(x^2+y^2) (x-y)(x^2 + xy + y^2)
= (x-y) (x+ y)(x^2+y^2) (x-y)(x^2 + xy + y^2)
= (x-y)^2 (x+ y)(x^2+y^2) (x^2 + xy + y^2)

as each term is non negative so
x^7 + y^7 - (x^4*y^3 + x^3*y^4) >= 0 ( > if x and y are not same and = if same)

or x^7 + y^7 >= x^4*y^3 + x^3*y^4

Prove that 1^k+2^k+3^k+...+n^k is devisible by n(n+1)/2 where k is any odd positive integer & n>1 or n=1?

as k is odd we have (a^k+b^k) is divisible by (a+b)

now

n^k + 1 ^k is divisible by (n+1)
(n-1)^k + 2^k is divisible by (n+1)
if n is odd
((n+1)/2)^k is divisible (n+1)/2 so sum is divisible by (n+1)/2

if n is even then there are n/2 terms each is divisible by (n+1)

we consider 2 cases

case 1 n odd :
----------------------
the sum is divisible by (n+1)/2

using the same analogy 1^k+2^k+3^k+...+ (n-1)^k is divisible by n(as n-1 is even ) and hence 1^k+2^k+3^k+...+ (n-1)^k + n^k is divisible by n

so it is divisible by n(n+1)/2 as they are co-primes.

case 2 n be even
------------------------------
the sum is divisible by (n+1)

using the same analogy 1^k+2^k+3^k+...+ (n-1)^k is divsible by n/2 (as n-1 is odd) and hence 1^k+2^k+3^k+...+ (n-1)^k + n^k is divisible by n/2

so it is divisible by n(n+1)/2

so in both cases is is divisible by n(n+1)/2

I am sure some one like will come with a better solution

if n is even we have the sum divisible by n+ 1


if n is odd the middle term is ((n+1)/2)^k

so sum is divisible by (n+1)/2 both cases

prove that the value of (sin x*cos 3x)/(cos x*sin 3x) cannot lie between 1/3 and 3.

it is tan x/ tan 3x

= tan x/ (tan x [ 3 - tan^2(x) ]/ [1 - 3tan^2(x) ])
= (1 -3 tan ^2 x)/(3 - tan^2 x )

say tan x = y which can take any value and

t = (1-3y^2)/(3-y^2)

or 3t - ty^2 = 1- 3y^2

or (3-t)y^2 + 3t - 1 = 0

or (3-t)^2y^2 = (3t-1)(t-3)

as LHS >=0 so RHS >=0

(3t-1)(t-3) >= 0 => t >= 1/3 and t >=3 => t >= 3

or t <= 1/3 and t <=3 => t <= 1/3

so cannot lie between 1/3 and 3.

If a and b are the roots of x^2+x+1=0, what is the equation with roots a^19 and b^7?

a^2 = - (a + 1)

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

so a^19 = a

similiary

b^7 = b

so roots are a^19 and b^7=> a and b are roots so eqution is x^2+x+1= 0

find x^4 + y^4 + z^4 when

x+y+z=3
x^2+y^2+z^2=5
x^3+y^3+z^3=7

solution

because there are 3 variables let them be solution of

f(t) = t^3 + at^2 + bt + c = 0

sum of roots = 3 so a = - 3

b = (xy + yz + zx) = 1/2((x+y+z)^2 - (x^2+y^2+z^2)) = 2

t^3 - 3t^2 + 2t +c = 0

f(x) = x^3 - 3x^2 + 2x + c = 0
f(y) = y^3 - 3y^2 + 2y + c = 0
f(z) = z^3 - 3z^2 + 2 z + c = 0

so add to get (x^3 + y^3 + z^3) - 3(x^2 + y^2+ z^2) + 2 (x+y+z) + 3c = 0

or 7 - 3 * 5 + 2 * 3 + 3c = 0

or 3c = 2

so f(t) = t^3 - 3t^2 + 2t + 2/3 = 0

multiply by t to get

t^4 - 3 t^3 + 2t^2 + 2/3 t = 0

x,y ,z satisfy this and adding putting x, y, z for t and adding we get

x^4 + y^4 + z^4 = 3(x^3+y^3+z^3) - 2 (x^2 + y^2 + z^2 ) - 2/3(x+y+z)
= 3 * 7 - 2 * 5 - 2/3 * 3 = 21-10 - 2 = 9

(x^2+y^2)/2 where x and y have the same parity. will it result a sum of 2 perfect squares

in other words

x^2 + y^2 =2 (a^2 +b^2) and x^2 + y^2 is even

the solution becomes obvious if you realize that

(a+b)^2 + (a-b)^2 = 2(a^2+b^2)

so x = a+b and y = a- b is a solution and both a and b should be even or odd

Does sqrt(-2i) = 1-i or i-1

both 1-i or i-1 when squared give - 2i

so both are square roots

but when we need to compute sqrt(-2i) this is the principal sqrt and and as per wiki the principal sqrt is defined to be

if z = r e^it then z^(1/2) = r^(1/2)e^(it/2)

-2i = 2 e^i(- pi/2) (note that that angle should be between (-pi to pi] that is pi is inculded and not -pi

so sqrt(-2i) = sqrt(2) e^(-pi/4)i = sqrt(2)( cos (-pi/4) +i sin (-pi/4)) = 1 -i

The no. of roots of equation x^2+x+3+2sinx=0,x belongs to (-180 to +180) is?

x^2+x+3+2sinx
= ( x^2+ x + 1/4)+ 11/4 + 2 sin x
= (x+ 1/2)^2 + 11/4 + 2 sin x

the lowest value of (x+ 1/2)^2 = 0
lowest value of 2 sin x = - 2

so lowest value = 11/4 -2 = 3/4

so it can never be zero

so ans is zero

simplify Sin π/9 sin 2π/9 sin π/3 sin 4π/9

we know sin sin π/3 = √(3)/2

2 Sin π/9 sin 2π/9 sin π/3 sin 4π/9 = Sin π/9 sin π/3 ( 2 sin 2π/9 sin 4π/9)

= Sin π/9 sin π/3 ( cos 2π/9 – 2 cos 2π/ 3) using 2 sin A sin B = cos |(A-B)| - cos (A+B)

= Sin π/9 in π/3 ( Sin π/9 cos 2π/9 + 1/ 2 )

So 4 Sin π/9 sin 2π/9 sin π/3 sin 4π/9 = = Sin π/9 in π/3 ( 2 Sin π/9 cos 2π/9 + 1)

= sin π/3 ( 2Sin π/9 ( 1- 2 sin ^2 π/9) +1 Sin π/9)

= sin π/3 ( 3 Sin π/9 - 4 sin ^3 π/9)
= sin π/3 sin π/3 using sin 3x = 3 sin x – 4 sin ^ 3 x
= sin^2 π/3 = 3/4

Or Sin π/9 sin 2π/9 sin π/3 sin 4π/9 = 3/16

Prove that the expression x^2 - (a+b+c)x+a^2+b^2+c^2+2bc -ca-ab can never be negative if x,a,b,c, are real.

There are two proofs to it

1) let us multiply by 4 to make it denominator free

4x^2 - 4(a+b+c)x+4(a^2+b^2+c^2+2bc -ca-ab)

= (2x - (a+b+c))^2 + 4(a^2+b^2+c^2+2bc -ca-ab) - (a+b+c)^2

= (2x - (a+b+c))^2 + 4(a^2+b^2+c^2+2bc -ca-ab) - (a^2+b^2 + c^2 + 2ab + 2bc + 2ca)

= (2x - (a+b+c))^2 + (3a^2+3b^2+3c^2+6bc -6ca-6ab)

= (2x - (a+b+c))^2 + 3(a^2+b^2+c^2+2bc -2ca-2ab)

= (2x - (a+b+c))^2 + 3(a^2+(b+c)^2 -2a(b+c))
= (2x - (a+b+c))^2 + 3(a-(b+c))^2

as it it sum of squares it cannot be -ve

Method 2

Using the quadratic formula gives:

x = (-(-(a+b+c) ± √((-(a+b+c))² - 4(a²+b²+c²+2bc-ca-ab)))/2

So if ((-(a+b+c))² - 4(a²+b²+c²+2bc-ca-ab)) < 0 it will only have complex roots and so wont cross the x axis. Since the coefficient of x² is positive this means it would only be positive. Therefore one need to prove ((-(a+b+c))² - 4(a²+b²+c²+2bc-ca-ab)) < 0

o r

-3a² -3b² - 3c² + 6ab + 6ac - 6bc < 0

factoring gives
-3(a-b-c)² < 0

which is true unless a-b-c = 0 it will have complex roots and so only be positive.

If a-b-c = 0 that means it has equal roots and so will only touch the x axis without crossing it and so will only be positive or 0.

Prove that n^5 - n is divisible by 30

proof:
we have n^5 - n = n(n^4-1) = n(n^2-1)(n^2+1) = n(n+1)(n-1) (n^2+1)

as n(n+1)(n-1) is product of 3 consecutive numbers it is divisible by 6

now under mod 5
n^2 + 1 = n^2- 4 = (n+2)(n-2)
so n^5 - n = n(n+1)(n-1)(n+2)(n-2)

as it is product of 5 consecutive numbers it is divisible by 5

as it is divisible by 5 and 6 hence 30

proved

solve (x+2)(x+3)(x+8)(x+12)=4x^2?

we see 2 * 12 = 3 * 8 = 24

so group as follow

(x+2)(x+12)(x+3)(x+8)=4x^2

=> (x^2 + 14 x + 24) (x^2 + 11x + 24) = 4x^2
let x^2 + 11x +24 be y
(y+ 3x) y = 4x^2
y^2 + 3xy - 4x^2 = 0
(y- x)(y+ 4x) = 0

y =x => x^2 + 11x + 24 = x => x^2 + 10x + 24 = 0 => (x+4)(x+6) = 0 => x = -4 or - 6

y+ 4x = 0 => x^2 + 15x + 24 = 0

this can be solved for complex solutions

x = -15/2 - SQRT(129)/2 or x = -15/2 + SQRT(129)/2

Show 1 + cisx + cis2x = (1-cis3x)/(1-cisx)?

we know
cis x = cos x + i sin x

(cis x)^n = cos ^n x + i sin ^n x = cis nx as per http://en.wikipedia.org/wiki/De_Moivre%2…


so LHS = 1 + cisx + cis2x = 1 + cis x + cis ^2 2x

let cis x = t (not required but makes understanding easier)

LHS = 1 + t + t^2 = (1-t^3)/(1- t) = (1- cis ^3 x)/(1- cis x) =
= ( 1- cis 3x )/(1- cis x) = RHS

Let a, b, and c be the roots of the polynomial p(x)=x^3+2x^2+3x+4.

Let g(x) be the monic polynomial?

whose roots are a+b, a+c, and b+c (each with multiplicity 1). Determine g(x).

sum of the roots = - coefficient of x^2

hence

a + b + c = -2

So a+b = -2-c

a +c = - 2-b

b+c = - 2- a

so we need to find equation whose roots are 2-a, 2-b 2-c

p(x)=x^3+2x^2+3x+4 has roots a,b,c

so p(-x) = -x^3 + 2x^2 – 3x + 4

or f(x) = x^3 – 2x + 3x – 4 (dividing p(-x) by – 1) has roots –a , -b ,- c

so f(x+2) = (x+2)^3 – 2(x+2)^2 + 3(x+2) – 4 has roots -2 – a , -2 –b , -2 - c

so g(x) = f(x+ 2) = x(x+2)^2 + 3 x+ 6- 4

= x(x^2 + 4x+ 4)+ 3x + 6 -2

= x^3 + 4x^2 + 7x + 4 is the required result.

If x=(√(a+2b)+√(a-2b))/(√(a+2b)-√(a-2b)) … then prove that bx2-ax+b=0

x =(√(a+2b)+√(a-2b))/(√(a+2b)-√(a-2b))

use Componendo and dividendo

(x+1)/(x-1) = √(a+2b)/√(a-2b)

square both sides

(x^2 + 2 x + 1)/(x^2-2x + 1) = (a+2b)/(a-2b)

again use Componendo and dividendo

(x^2 + 1)/(2x) = (a)/(2b)

or bx^2 +b = ax
or bx^2 - ax + b = 0

Note: other method can be used but it is simpler when we have form (a+b)/(a-b)

Source(s):

http://en.wikipedia.org/wiki/Componendo_…

Let x = ln(sec(y) + tan(y). Show that sec(y) = cosh (x).?

Remember that cosh(x) = (1/2) * (e^(x) + e^(-x)) ..1

and as sec^2 y- tan ^2 y = 1 so 1/(sec y + tan y) = sec y - tan y ..2

x = ln(sec(y) + tan(y))

so e^x = sec(y) + tan(y) ...3
so e^-x = 1/( sec y + tan y) = sec y - tan y .. 4 ( from 2)

add (3) and (4) to get e^x + e^-x = 2 sec (y)

or sec y = (e^ x + e^-x)/2 = cosh x (from 1)

prove (tan12)(tan48)(tan54)(tan72) = 1

we have
tanθ tan(60° - θ) tan (60° + θ) = tan3θ. (proof is below)

put θ =12 to get
tan 12 tan(60 - 12) tan (60 + 12) = tan 3 * 12
=> tan 12 tan 48 tan 72 = tan 36
=> tan 12 tan 48 tan 72 = cot (90-36) = cot 54 = 1/ tan 54
=> (tan12)(tan48)(tan54)(tan72) = 1
proved
--------------------------------------------------------------------------
to prove

tanθ tan(60° - θ) tan (60° + θ) = tan3θ

tan(60° - θ) tan (60° + θ)
= ((tan 60 - tan θ)/(1+ tan 60 tan θ))((tan 60 - tan θ)/(1+ tan 60 tan θ))
= (tan^2 60 - tan^2 θ)/(1- tan^2 60 tan^2 θ))
= (3 - tan θ)/(1 - 3 tan^2 θ))

so tanθ tan(60° - θ) tan (60° + θ) = (3 tan θ - tan^3 θ)/(1 - 3 tan^2 θ)) = tan 3θ

proved

If p and q are distinct primes and x^2 - px + q = 0 has distinct positive integral roots, then p + q has value equal to

it is got positive integral roots so roots are 1 and q as q is prime

equation is (x-1)(x-q)
= x^2 - (q+1) + q = 0

so p = q + 1

so q and q +1 are primes so q = 2 and q +1 =3 so p = 3, q = 2 and p+q = 5

On squared paper, draw a rectangle and one of its diagonals. How many grid squares are crossed by the diagonal?

Let a = (0,0) and x coordinate is to right and y coordinate downwards

Let the rectangle be ABCD as in the diagram and of the size m * n. The diagonal is AC. There are m+ 1 vertical lines and n+1 horizontal in the above diagram m = 4 and n = 3

Now let us see when it shall pass though a point which is a corner that is intersection of horizontal line and vertical line say at (p , q) coordinate

Slope of the line = m/n = p/ q where p < m , q < n .

This may pass through multiple points and let p/q is with the lowest p and q

m/p = n/q is the GCD (m,n)

So if gcd(m,n) is 1 then we do not have p/q form and the line does not pass through any point (p,q) that is a corner( we define a corner as intersection of horizontal line and vertical line)

So we take 2 cases

1) GCD(m,n) is 1 that is m,n are coprime

The diagonal from A to C shall pass through m vertical sections which shall be m squares and n-1 horizontal lines shall be cut by the diagonal ( at a point other than a corner point) so each shall give 2 squares that is addition of 1 square that making m+n-1 squares. So the diagonal pass through m+n-1 squares.

2) GCD(m,n) is not one that is m ,n are not coprimes say GCD (m,n) = p

Now m/p and n/p are coprimes and we get p parts of (m/p, n/p) rectangle through which the diagonal passes.

So number of points = p(m/p + n/p – 1) = m+ n – p

So we combining (1) and (2) get m+ n – gcd(m,n)

For example in the above figure m = 4, n= 3.

It goes through 1+ 2 + 2 + 1 ( note that in the 2^nd region and 3^rd region it is 2 as diagonal cuts the horizontal line) = 6

m+n – 1 = 4 + 3 -1 = 6

prove that tan20+4sin20=root 3

tan20+4sin20
= sin 20/ cos 20 + 4 sin 20
= ( sin 20 + 4 sin 20 cos 20) / cos 20
= (sin 20 + 2 sin 40)/ cos 20
= (sin 20 + 2 sin (60-20))/ cos 20
= ( sin 20 + 2 sin 60 cos 20 - 2 cos 60 sin 20)/ cos 20
= (sin 20+ 2 sin 60 cos 20 - sin 20)/ cos 20
= 2 sin 60cos 20/ cos 20
= 2 sin 60
= 2(√3/2) = √3

alternatively


we can proceed from (sin 20 + 2 sin 40)/ cos 20

as ( sin 20 + sin 40 + sin 40) / cos 20
= (2 sin 30 cos 10 + sin 40)/ cos 20
= (cos 10 + sin 40)/ cos 20
= ( sin 80 + sin 40)/ cos 20
= 2 sin 60 cos 20/ cos 20
= 2 sin 60
= √3

sum log(base4)2-log(base8)2+log(base 16) 2 - ..

the nth term = log( base (2^(n+1) 2 = 1/(n+1)

sign is alternating

so we get

S = 1/2 - 1/3 + 1/4 - 1/5 + ....

we have from http://www.mathkb.com/Uwe/Forum.aspx/mat…

ln 2= 1 - 1/2 + 1/3 - 1/4 ....

so ln 2 = 1- s

so S = 1 - ln 2

solve for value of X and Y if X2 +Y2=25 and X3+Y3=91?

Because we are having x^2+y^2 and x^3 + y^3 we can choose

(x+y) = a and xy = b

We get x^2+ y^2 =a^2 -2b = 25 ..1

And (x^3+y^3) = (x+y)^3 – 3xy(x+y)

or a^3 – 3ab = 91 ..2

so 2 (a^3 -3ab) – 3a(a^2- 2b) = 182-75a

or a^3 -75a + 182 = 0

this is cubic in a and by taking factors of 182(1,2,7,13,14,26,91,182) and –ve of them

we can see that a= 7 is a root and from (1) b = xy = 12

now x+ y = 7 and xy = 12 => (x-y)^2 = (x+y)^2 – 4xy = 1

so x- y = +/- 1

x-y = 1 => x= 4 and y = 3

x-y - = - 1 => x = 3 ,y = 4

so solution = (4,3) and (3,4)

find the solutions of tan x=x have a solution

x= 0 is a solution.

for each n there is exactly one solution in the range npi - pi/2 and npi+pi/2 as in this range tan x increases from - infinite to + infinite and at one point it cuts the line y = x

Find all positive integers n such that the decimal representation of n^2 consists of odd digits only

if n is even it is not possible

so n is odd

now (10k+m)^2 = 100k^2 + 20kb + m^2

m = 1 => 100k^2 + 20k + 1: k > 1 means tens digit even so k = 1 => n = 1
m= 3 => 100k^2 + 60k + 9 : k > 1 means tens digit even so k = 0 => n = 3
m = 5 => 100k^2 + 100k + 25 tens digit even
m = 7 => 100k^2 + 140k + 49 => tens digit even
m = 9 => 100k ^2 + 180k + 81 => tens digit even

so only choices n =1 and 3 whose square 1 digit numbers 1 and 9

Note: I could have used (10k+ 5), (10k+/-1) , (10k+/- 3) and could have got result with less computation

If a,b,c are the roots of x^3+4x+1=0 ,then the equation whose roots are a^2/(b+c) ,b^2/(a+c),c^2/(a+b) ?

a,b,c are roots

so we get comparing coefficients of x^2 ( that is sum of roots)

a+ b + c = 0 ..1
now
a^2/(b+c) ,b^2/(a+c),c^2/(b+a) are root of new equation

a^2/(b+c) = a^2/(-a) from (1)

so new equation has roots -a , -b , and -c

f(x) = x^3+4x + 1 = 0 has roots a,b,c

so f(-x) = - x^3 - 4x + 1 = 0 or x^3 + 4x -1 =0 has roots -a , -b , and -c ora^2/(b+c) ,b^2/(a+c),c^2/(b+a)

Show that there are infinitely many basic Pythagorean triples (x,y,z) in which z-x=2

z = (x+2)
if z (and x) even then y is also even so not basic

so z and x both are odd
let z = (2m+1) and x = (2m-1)

(2m+1)^2 - (2m-1)^2 = y^2

or 8m = y^2

so m= y^2/8

y ^2 = 16n^2 shall be a solution as y^2 is divisible by 8 and perfect square so divisible by 16

so m = 2n^2 and there is no other solution

so x = (4n^2-1), y = 4n, z = (4n^2+1) are solutions

there are infinite solutions and lowest 5 are below

n =1 gives (3,4,5)
n= 2 gives (15,8, 17)
n= 3 gives ( 35, 12, 37)
n = 4 gives (63,16,65)
n =5 gives (99,20,101)

What is value of f(x+3) , if f(x+1)=2x^2 -11x+3

we have 
x+ 3 = (x+1) + 2

as f(x+1) = 2x^2 -11x+3

so f(x+3) = 2(x+2)^2 - 11(x+ 2) + 3 = 2(x^2 + 4x + 4) - 11(x+2) + 3 = 2x^2 -3x - 11
----------------------------------------------------------------------------------------------
(rationale

f(x+1) = 2x^2 - 11x + 3

let x + 1 = y

so f(y) = 2(y-1) ^2 - 11y + 3

so f(y+2) = 2(y+1)^2 - 11(y+1) + 3

hence f(x+3) = 2(x+2)^2 - 11(x+2) + 3)

If abc is a three digit number and a, b, c are distinct digits, what is abc if acb+bca+bac+cab+cba=3194

we have (In digit form)
abc = 100a + 10b + c
acb = 100 a + 10c + b
bac = 100b + 10a + c
bca = 100b + 10c + a
cab = 100c + 10 a + b
cba = 100c + 10 b + a

so sum = 222 ( a+ b+ c) > 3194 and 3194 + abc = 222(a+b+c)


find the lower and upper limit of a+b+ c

 a + b + c > 3194/222 = 14.38
so lower limit = 15 and let us start and lowe limit and go one by one finding abc and checking

if a + b+ c = 15 then abc = 222*15-3194 = 136 does not satisfy
a + b + c = 16 , abc = 106 + 222 = 358 yes
a+ b+ c = 17, abc = 358+ 222 = 780 no
a+ b+ c = 18, abc= 780+ 222 > 1000 = 772 no and a+b+ c > 18 is not possible

so abc = 358

Show that 2+i is a root of z^3 - 5z^2 + 9z - 5=0. Find the other 2 roots

one can put z = 2 + i and expand it but it shall be long and prone to error

you can see that it has rational coefficients so if z = 2 + i is a root then z = 2 -i must be a root

and so (z-(2+i))(z-(2 -i) ) or (z-2)^2 +1 or z^2-4z + 5 must devide z^3 - 5z^2 + 9z - 5

by division you see z^3 - 5z^2 + 9z - 5 = (z^2-4z + 5)(z-1)

so z^2-4z + 5 devides z^3 - 5z^2 + 9z - 5

so z = 2 + i is a root
other roots are 2 -i  and 1

Let x & y be positive real numbers such that x^3 + y^3 + 1/27 = xy. Find the value of 1/x.

x^3+y^3+1/27 = xy
=> x^3 + y^3 + (1/3)^3 = 3 x y (1/3)
if a^3 + b^3 + c^3 = 3abc then a + b + c = 0 or a=b= c

as (a^3+b^3+c^3) -3abc = 1/2(a+b+c) ((a-b)^2 + (b-c)^2 + (c-a)^2) 


so x + y + 1/3 = 0 or x = y= 1/3


as x+y+1/3 cannot be zero for x and y positive so x= y = 1/3 

or 1/x = 3

Find all solutions to z^2 + 4conjugate[z] + 4 = 0 where z is a complex number.

let z = a + ib

you get (a+ib)^2 + 4(a-ib) + 4 = 0

expand

a^2 - b^2 + 2aib + 4a - 4bi + 4 = 0

(a^2 - b^2 + 4a + 4) + 2abi- 4bi = 0

equate imaginary and real parts on both sides to get

(a^2 - b^2 + 4a + 4) = 0 ...1
and 2abi - 4bi = 0 => b = 0 or a = 2

solve 1 using b = 0 to get a= - 2
sollve 1 using a= 2 to get b = +/-4

so (-2,0) is a solution so z = - 2
(2, 4i) and (2,-4i) are 2 other solutions so z = 2 + 4i or 2- 4i

solve for a 3x^2 +ax -(a^2 -1) = 0 so that all solutions to the equation are real and positive

comparing with the euation

Ax^2 + Bx + C ( as a is there in given equation)

we have

roots = (- B +/- sqrt(B^2-4AC))/ (2A)

and we take the - sign for the second one as if - sign gives positive then + sign being greater adding -B gives positive.

so -B - sqrt(B^2 - 4AC) > 0

or B < - sqrt(B^2-4AC)

put B = a , A = 3 and C = a^2-1


to get a < - sqrt(a^2 + 12(a^1-1)
< - sqrt(13a^2- 12)

but a is -ve and so negate both sides to get

- a > sqrt(13a^2-12)

or a^2 > 13a^2-12 or a^2 < 1 or -1 < a < 1 but as a is -ve -1 < a ..1

as B^2 >= 4AC so a^2 > - 12(a^2- 12) or 13a^2 > 12 or a^2 > (12/13) so a < - sqrt(12/13) ..2

so a = - sqrt(12/13) is the value