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