2022/009) Find the gcd(19! + 19, 20! + 19)

 We have $GCD(19! + 19, 20!+19)$ 

$= GCD( 19!+ 19 , 20(19! + 19) - (20!+ 19))$ using gcd(a,b) = gcd(a, na -b))

$= GCD(19! + 19, 20! + 20 * 19 - 20! -19)$

$= GCD(19! + 19, 20 * 19 - 19)$

$= GCD(!9! + 19, 361)$

Now $361 = 19^2$

$19! + 19 $is 1$9(18! + 1)$

As 19 is a prime as per wilson's theorem 18! + 1 is divisible by 19

So $19! + 19$ is divisible by $19^2 = 361$

so GCD = 361 

2022/008) Given $\sin^3 x + \cos ^3 x= \frac{1}{\sqrt{2}}$ find $\sin \, 2x$

 we have  $\sin^3 x + \cos ^3 x= \frac{1}{\sqrt{2}}$

letting $\sqrt{2} \sin\,x = s$ and $\sqrt{2} \cos\, x = c$

we have $s^3 + c^3 = 2 \sqrt{2} \frac{1}{\sqrt{2}}= 2$

or $s^3 + c^3 = 2\cdots(1)$

and $s^2 + c^2 = 2\cdots(2)$

we are interested to find $\sin\,2x$ or sc

squaring (1) we get $s^6 + c^6 + 2s^3c^3 = 4$

and cubing (2) we get $s^6 + c^6 + 3s^2c^2(s^2 + c^2) = 8$

or $s^6 + c^6 + 6s^2c^2 = 8$

from (3) and (4) we have letting sc = x

$2x^3 - 6x^2 + 4 = 0$

or $x^3 - 3x^2 + 2$   this gives x = 1

and deviding by x-1 for factoring we get $(x-1)(x^2 - 2x -2) = 0$

so x = 1 or $1\pm \sqrt{3}$ but $1 + \sqrt{3}$ being above 1 is not admissible

so x = 1 or $1-\sqrt{3}$

2022/007) If, $n^2−33,n^2−31$ and $n^2−29$ are prime numbers, then what is the number of possible values of n where n is an integer

 $n^2−33,n^2−31$ and $n^2−29$ are 3 consecutive even numbers or 3 consecutive odd numbers

they cannot be 3 consecutive even numbers as all cannot be prime

so they are 3 cosecutive odd numbers.

so one of them is divisible by 3.

so the number has to be 3 otherwise it is not prime.

so 3 numbers are 3 , 5, 7 and $n^2-33 = 3$ giving $n = \pm 6$ or number of values of n = 2

2022/006) Prove that there are infinitely many positive integers n such that n(n+1) can be represented as a sum of 2 positive integers in atleast 2 ways

 If we choose n as a square say $m^2$ then 

$n(n+1) = m^2(m^2+1) = m^4 + m^2$

now n(n+1) is reperesented as sum of 2 squares that is $(m^2)^2 + m^2$

if $m^2$ can be represented as sum of 2 squares that is $p^2 + q^2$ this is possible as as (m,p,q) form a pythagorean triple then we have

$n(n+1) = (p^2+q^2)(m^2+1) = (pm+ q)^2 + (p-qm)^2$ this is another way 

if we chose $p=x^2-y^2, q = 2xy, m= x^2+y^2, n = (x^2+y^2)^2 $ then it satisfies the condition  (p,q,m) form a pythagrean triple)

2022/005) When $x^{1000}$ is divided by $x^2-4x+3$ what is the remainder?

 When $x^{1000}$ is divided by $x^2-4x + 3$ the raminder shall be a polynomial of degree 1 that is Ax + B

So $x^{1000} = (x^2 - 4x +3) P(x) + Ax + B$ where P(x) is quotient 

We need to find A and B

Now $x^2-4x + 3 = (x-1)(x-3)$

So $x^{1000} = (x-1)(x-3) P(x) + Ax + B$

Putting x = 1 we get $1= A + B\cdots(1)$

Putting x = 3 we get $3^{1000} = 3A + B\cdots(2)$

Subtracting (1) from (2) we get 2A = $3^{1000} -1$

Or $A = \frac{3^{1000}-1}{2}$

Putting in (1) we get $B= 1 - A = \frac{1-3^{1000}}{2}$

So remainder = $\frac{3^{1000}-1}{2}x + \frac{1-3^{1000}}{2}$

2022/004) A triangle with integral sides has a perimeter 8. What will be the area of a triangle?

First let us find the lengths of sides of the trinagle. Let a,b,c be sides of triangle

By triangle inequality we have as $a+b > c$ so $a+b+c > 2c$  or $8 > 2c$ or $ c< 4$

$c = 3 => a + b = 5$ giving a = 3 and b =2 or a =2 and b = 3

$c=2=> a+b=6$ giving a = b= 3

$c=1$ is not possible as it gives $a+b= 7$ give a a =3 , b= 4 invalid triangle

sides 3,3,2 say a =  b= 3 and c = 2

and s (semiperimeter) = 8/2 = 4

so if A is area $A^2 = s(s-a)(s-b)(s-c) = 4 * 1 * 1 * 2 = 8$ or Area = $2\sqrt{2}$

 

 

2022/003) Prove $x^2+y^2+5>xy+x+3y$

 We have

$(x-y)^2 >=0$ or $x^2 + y^2 >=2xy$

$(x-1)^2 >=0$ or $x^2 + 1 >= 2x$

$(y-3)^2 >= 0$ or $y^2 + 9 >= 6y$

Adding we get $2(x^2+y^2 + 5) >= 2(xy+x + 3y)$

Or $x^2+y^2 + 5 >= xy + x + 3y$

This is equal when x = y, y = 3, x= 1 which canot be true so  $x^2+y^2 + 5 > xy + x + 3y$