Sunday, March 24, 2024

2024/020) Show that there are infinitely many composite numbers of the form $10^n +3$ (n = 1, 2, 3, ... ).

If n = 2 we have $10^2+3 = 103$ this is divisible by 103 . 

Hence  $10^2 + 3 = 0 \pmod {103}\cdots(1)$

We know that 103 is a prime but we shall use this as a starting point to show that there are infinite numbers of the form $10^n+3$ is divisible by 103.

As 103 is a prime so using Fermats little theorem

$10^{102} = - 1 \pmod {103}$

Squaring both sides we get

$10^{204} =  1 \pmod {103}$

So from (1) and above we have

$10^{204n + 2} + 3=  0 \pmod {103}$

except for n = 0 when the value is 103 all other values are above 103 and have a factor 103 so not a prime

So there are infinitely many composite numbers as they are divisible by 103. We have not taken all cases and have found infinitely many composite numbers. Hence proved

 


2024/019) Given $a_n = 6^n + 8^n$ find the remainder when $a_{83}$ is divided by 49

 We shall use Euler's theorem that is a generalization of Fermat's little theorem: For any n and any integer a coprime to n, one has 

$a^{\phi(n)} \equiv -1 \pmod n$

Where $\phi(n)$ is number of numbers that is co-prime to n,

Let us compute $\phi(49)$ 

If $n = p_1^{k_1}p_2^{k_2}p_3^{k-3}  ... $ when each $p_i$ is  prime then

 $\phi(n)=n(1-\frac{1}{p_1})(1-\frac{1}{p_2})(1-\frac{1}{p_3})$

As $49= 7^2$ so $\phi(49) = 49 * (1-\frac{1}{7}) = 49 * \frac{6}{7} = 42$

Hence  

$6^{\phi(49)} \equiv -1 \pmod {49}n$

or $6^{(42)} \equiv -1 \pmod {49}$

Squaring both sides

 $6^{(84)} \equiv 1 \pmod {49}$

Dividing by 6 we get

  $6^{(83)} \equiv \frac{1}{6} \pmod {49}\cdots(1)$

Where $\frac{1}{6}} is not fraction but inverse of 6 mod 49

Similarly

  $8^{(83)} \equiv \frac{1}{8} \pmod {49}\cdots(2)$

 Adding (1) and(2) we get 

  $6^{(83)}+ 8^{(83)} \equiv \frac{1}{6}  + \frac{1}{8} \pmod {49}\cdots(2)$

Now $\frac{1}{6}  + \frac{1}{8} \pmod {49}$

$= \frac{8 + 6}{48} \pmod{49}$

$= \frac{14}{48} \pmod{49}$

$= \frac{14}{-1} \pmod{49}$ as $48 \equiv -1 \pmod {49}$

$= -14 \pmod{49}$

$=35$ taking positive number 

Hence remainder = 35 

Sunday, March 17, 2024

2024/018) There are 1000 coins -- 999 are fair, and 1 has heads on both sides. You randomly choose a coin and flip it 10 times. Miraculously, all 10 flips turn up heads. What is the probability that you chose the unfair coin

Probability that you choose one unfair coin = $\frac{1}{1000}$

iI you choose one unfair coin the probability that all 10 heads come = 1

So   Probability that you choose one unfair coin and all heads come = $\frac{1}{1000}$

Probability that you choose one fair coin = $\frac{999}{1000}$

If you choose one fair  coin the probability that all 10 heads come = $\frac{1}{2^{20}} = \frac{1}{1024}$

So   Probability that you choose one fair coin and all heads come = $\frac{999}{1024*1000}$

 So probability that all heads come = $\frac{999}{1024*1000} + \frac{1}{1000} = \frac{2023}{1024000}$

 So probability that coin is unfiar = $\frac{\frac{1}{1000}}{\frac{2023}{1024000}}=\frac{1024}{2023}$

 

Saturday, March 2, 2024

2024/017) integrate sin ax cos bx

We have $\sin \, ax + \cos \, bx = \frac{1}{2} (\sin (a+b) x + \sin (a-b) x$

Knowing that 

$\frac{d}{dx} \ cos \, nx = - n \sin \, nx$

Hence $\int  (\sin \, ax  \cos \, bx) dx  = \int (\frac{1}{2} (\sin (a+b) x + \sin (a-b) x)dx $ 

= $\frac{\cos(a+b)x}{a+b} - \frac{\cos(a-b)x}{a-b} + C $ 


Friday, February 23, 2024

2024/016) Prove that product of 4 consecutive positive integers cannot be a perfect cube.

Proof;

We  shall use the fact that if x and y are co-prime then xy is a perfect cube if both  x and y are perfect cubes.

Let the 1st number be x so the product is $x(x+1)(x+2)(x+3)$

Either x is odd or even

If x is odd then we have $GCD(x,x+2) = GCD(x+1, x+2) = GCD(x+2,x+3) = 1$

For x + 2 to be a perfect cube minimum x is 6 

So $GCD(x+2,x(+1)(x+2)) = 1$

So we need to prove that $x(x+1)(x+3)$ is not a perfect cube.

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

We have $x(x+3) - (x-1)^2 = x^3+3x - x^2 -2x -1=  x-1 > 0$ as x is minimum 6 

So $x(x+1)(x+3) >= (x+1)^3$

  $x^3 + 4x^2 + 3$ is below  $(x+2)^3 = x^3 + 6x^2 + 12x + 8$

As it is between 2 cubes it cannot be a cube

Similarly we can prove taking $x +1$ when $x$ is even


 

Friday, February 16, 2024

2024/015) Find natual number n such that $2^n + n | 8^n + n $

 1,2,4,6 

We know that $ x+ y | x^3+y^3 $

Hence $2^n + n | (2^3)^n + n^3$

as $2^n + n | 8^n + n $

so $2^n + n | n^3-n  $

so we must have $ n^3-n =0 $ or $2^n + n <  n^3-n  $

$n^3-n= 0$ gives n = -1,0, 1 and out of theses only 1 is solution

we need to solve  $2^n + n <  n^3-n  $ or  $2^n  <  n^3-2n  $

let us find an upper bound for n putting a condition 

$2^n < n^3$ for  $n \lt 10$

putting n from 1 to 9 we see that $n \in \{1, 2,4,6\} $ satisfy the case and there is no other solution                                                                                                                                               

 

 

2024/014) Solve $x^2-y=111$ and $y^2-x=111$ for $ x\ne y$

 We are given 

 $x^2-y=111\cdots(1)$

 $y^2-x=111\cdots(2)$

From (1) and (2)

$x^2-y = y^2 -x$

Or $x^2-y^2 + x -y = 0$

Or $(x-y)(x+y) + (x-y) = 0$

Or  $(x-y)(x+y+1) = 0$

As  $ x\ne y$  dividing by x-y we get $x+y+1=0\cdots(3)$

Or $y = -(x+1)$

Putting the value in (1) we get $x^2 + (x+ 1) = 111$ or $x^2 + x - 110 = 0$

Or $(x-10)(x+11) = 0 $

Or $ x= 10$ or $x = -11$

Putting in (3) if $x= 10$ then $y = -11$

If $x= -11 $ then $y = 10$

So Solution set $x=10,y= -11$ or $x=-11,y=10$