2022/014) How do you prove that $\frac{2}{1!} + \frac{2+4}{2!} + \frac{2+4+6}{3!}+ \cdots = 3e$

 We have $n^{th}$ term 

$= \frac{2 + 4 + \cdots 2n }{n!} = \frac{n(n+1)}{n!}$

So then given exprsssion

$\frac{2}{1!} + \frac{2+4}{2!}  + \frac{2+4+6}{3!}+ \cdots$

$=\sum_{n=1}^{\infty} \frac{n(n+1)}{n!}$ putting the valut from above 

$=\sum_{n=1}^{\infty} \frac{n(n-1) + 2n }{n!}$ writing n(n+1) s n(n-1) + 2n

$=\sum_{n=1}^{\infty} (\frac{n(n-1)}{n!} + \frac{2n }{n!})$

$=\sum_{n=1}^{\infty} \frac{n(n-1)}{n!}  + \sum_{n=1}^{\infty} \frac{2n }{n!}$

$=\sum_{n=2}^{\infty} \frac{n(n-1)}{n!}  + \sum_{n=1}^{\infty} \frac{2n }{n!}$ we can convert the lowerr limit from n = 1 to n =2 because for n- 1 we have n(n-1) = 0 which can be dropped

$=\sum_{n=2}^{\infty} \frac{1}{(n-2)!}  + 2 \sum_{n=1}^{\infty} \frac{1 }{(n-1)!}$

$=\sum_{n=0}^{\infty} \frac{1}{n!}  + 2\sum_{n=0}^{\infty} \frac{1 }{n!}$ by changing the limit

 $=3 \sum_{n=0}^{\infty} \frac{1}{n!}$

= 3e by using expansion form for e   

2022/013) How do you prove that $\tan^{-1}\frac{1}{5}+ \tan^{-1} \frac{1}{7} + \tan^{-1} \frac{1}{3} + \tan^{-1} \frac{1}{8} = \frac{\pi}{4}$

Let us have $\tan^{-1}\frac{1}{5} = \alpha$

$\tan^{-1}\frac{1}{7} = \beta$

$\tan^{-1}\frac{1}{3} = \gamma$

$\tan^{-1}\frac{1}{8} = \delta$

we need to find $\alpha + \beta + \gamma + \delta$

Let us add 2 at a time $\alpha + \beta$ , $\gamma +\delta$ and then $\alpha + \beta + \gamma + \delta$

We have 

$\tan \alpha = \frac{1}{5}\cdots(1)$

and  $\tan \beta = \frac{1}{7}\cdots(2)$

So $\tan (\alpha + \beta) = \frac{\tan (\alpha) + \tan (\beta)}{1 - \tan (\alpha) \tan (\beta)}$

$= \frac{\frac{1}{5} + \frac{1}{7}}{1 - \frac{1}{5} \frac{1}{7}}$

$= \frac{\frac{12}{35}}{ \frac{34}{35}} = \frac{12}{34} = \frac{6}{17}$ 

We have 

$\tan \gamma = \frac{1}{3}\cdots(3)$

and  $\tan \delta  = \frac{1}{8}\cdots(4)$

So $\tan (\gamma + \delta) = \frac{\tan (\gamma) + \tan (\delta)}{1 - \tan (\gamma) \tan (\delta)}$

$= \frac{\frac{1}{3} + \frac{1}{8}}{1 - \frac{1}{3} \frac{1}{8}}$

$= \frac{\frac{11}{24}}{ \frac{23}{24}} = \frac{11}{23}$ 

now $\tan (\alpha + \beta + \gamma + \delta)$

$=\frac{\tan (\alpha + \beta)  + \tan (\gamma + \delta)}{1 - \tan (\alpha + beta) \tan (\gamma + \delta)}$

$= \frac{\frac{6}{17} + \frac{11}{23}}{1 - \frac{6}{17} \frac{17}{23}}$

$= \frac{\frac{6 * 23 + 11 * 17}{17 * 23}}{1-  \frac{66}{17 * 23}}$

$= \frac{6 * 23 + 11 * 17}{17 * 23 - 66}$

$=\frac{ 325}{325} = 1$

so $\tan (\alpha + \beta + \gamma + \delta) = \tan \frac{\pi}{4}$

but we need to show that $\alpha + \beta + \gamma + \delta = \frac{\pi}{4}$

as it could be $\frac{5\pi}{4}$

but as  $\alpha,\beta,\gamma, \delta$ each is less than $\frac{\pi}{4}$ so sum is less than $\pi$ so it is $\frac{\pi}{4}$ 

2022/012) For integers x and y 34x = 43 y show that x + y is composite

 We have 34x + 34 y = 43 y + 34 y = 77 y

so 34(x+y) = 77 y

or $x+ y = \frac{77 * y}{34}$

34 is coprime to 77 , x+ y is integer so $\frac{y}{34}$ is integer  

$x+y = 77 * \frac{y}{34}$ has a factor 77 and 77 is composite so x + y is composite  

2022/011) Find the rightmost digit of $\lfloor\frac{10^{20000}}{10^{100}+3}\rfloor$

To keep the calculation simple let n = 10^{100}$

So $10^{100} + 3 = n + 3$

So $10^{20000} = n^{200}$

We need to find remainder when $n^{200}$ is divided by n+ 3

Now $n^{200} = n^{200} - 3^{200} + 3^{200}$ 

$n^{200} - 3^{200}$ is divisible by n + 3 and $3^{200} = 9^{100}$ whcih is less than $n+3$

So we need to find the unit digit of quotient of $n^{200} - 3^{200}$ divided by n + 3

we have $n^{200} - 3^{200} = (n+3)\sum_{k=0}^{199}(-1)^k n^k 3^{199-k}$

Hence

$\frac{n^{200}-3^{200}}{n+3}= \sum{k=0}^{199}(-1)^k n^k 3^{199-k}$

putting back $n= 10^{100}$

we get $\frac{10^{20000}-3^{200}}{10^{100}+3}= \sum{k=0}^{199}(-1)^k (10^{200})^k 3^{199-k}$

All the trerms except when k = 0 have a factor 100 so we get unit digit is unit digit of $-3{199}$

now $3^4$ ends with 1 so $3^{199} = (3^4)^{49} * 3^3$ ends with unit digit of $3^3$ or 7 so if we negate it it ends with 10 - 7 = 3

so ans 7 

2022/010) Solve the equation $n! = n^3 - n$ for integer n

Note: I picked the question from  https://www.youtube.com/watch?v=uqaRfADGOtg and I provide my solution for the same in a diffrerent way.

 We see that $n!$ grows at a rate much faster than $n^3$ 

For solving this we can  fix the bound for n and check on the same.

as we see $6^3 = 216$ and $6! = 720$ which is greater

by chceking the values of n from 0 to 5 we get n = 5 satsfies the equation

so Ans $n= 5$

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  $x^2+y^2 + 5 > xy + x + 3y$

2022/002) Find the smallest number whose product of digits is 10000

 Because the product of digits is 10000 so this can be represented as product os single numbers.

Let us findone such number and then see if we canmake it smaller

$10000 = 2 ^ 4 * 5 ^ 4$

So if the number has 4 2's and 4 5's then the  product of the digits is 10000. This is an 8 digit number.  this can be mad smaller if we can multiply 2 or more one digit number out of  these and getting one digit number, multiplying 3 2's we get 8 and digits are 2, 8, 5,5,5 ,5 giviing the product 1000. the smallest number that can be formed is 255558 and this is the answer   

2022/001) Solve in integers $x+y+z=3$ and $x^3+y^3 + z^3 = 3$

We are given

$x+y+ z = 3\cdots(1)$

$x^3 + y^3 + z^ 3 = 3\cdots(2)$

cubing (1) we get $(x+y+z)^3 = 3^3$

or $x^3 + y^3 + z^3 + 3(x+y)(y+z)(z+x) = 27$

or $3 + 3(x+y)(y+z) (z+x) = 27$ putting thw vaue of $x^3 + y^3 + z^3 = 3$ from (2)

or  $(x+y)(y+z)(z+ x) = 8$

now $((x+y) + (y+z) + (z+ x)) = 2 ( x + y + z) = 6$ usng (1)

we need to find 3 numbers whose sum is 6 and product is 8. They are (2,2,2) or (4,1,1) and other combinations fail

taking x + y = 2 , y + z = 2, z+ x = 2 and using (2) that is x + y + z = 3 we get x=1,y=1,z = 1

 taking x + y = 4 , y + z = 1, z+ x = 1 and using (2) that is x + y + z = 3 we get x=5,y=z = - 4

takinn n rotaton other combinations we get x=y = -4, z = 5 as one solution and z = z = -4 and y =5 as another solution

so $\{x,y,z\} = \{2,2,2\}$ or $\{5,-4,-4\}$ (that is any permuation of the same)